\displaystyle \text{MULTIPLE CHOICE QUESTIONS (MCQs)}

\displaystyle \textbf{Question 1: }\text{If }\frac{1+4p}{4},\ \frac{1-p}{2}\text{ and }\frac{1-2p}{2}\text{ are the probabilities of three}
\displaystyle \text{mutually exclusive events, then the value of }p\text{ is:}
\displaystyle \text{(a) }\frac{1}{2}\qquad\text{(b) }\frac{1}{3}\qquad\text{(c) }\frac{1}{4}\qquad\text{(d) }\frac{2}{3}
\displaystyle \text{Answer:}
\displaystyle \text{Since probabilities are non-negative, }\frac{1-2p}{2}\geq0.
\displaystyle \therefore p\leq\frac{1}{2}.
\displaystyle \text{Also, for mutually exclusive events, the sum of their probabilities cannot exceed }1.
\displaystyle \frac{1+4p}{4}+\frac{1-p}{2}+\frac{1-2p}{2}\leq1.
\displaystyle \frac{5-2p}{4}\leq1.
\displaystyle 5-2p\leq4.
\displaystyle \therefore p\geq\frac{1}{2}.
\displaystyle \therefore p=\frac{1}{2}.
\displaystyle \therefore \text{The correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Which of the following cannot be the probability of an event?}
\displaystyle \text{(a) }\frac{2}{3}\qquad\text{(b) }-\frac{1}{5}\qquad\text{(c) }15\%\qquad\text{(d) }0.7
\displaystyle \text{Answer:}
\displaystyle \text{For any event }E,\quad 0\leq P(E)\leq1.
\displaystyle -\frac{1}{5}<0,\text{ so it cannot be the probability of an event.}
\displaystyle \therefore \text{The correct option is (b) }-\frac{1}{5}.
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{Probability of an event can be:}
\displaystyle \text{(a) }-0.7\qquad\text{(b) }\frac{11}{9}\qquad\text{(c) }1.001\qquad\text{(d) }0.6
\displaystyle \text{Answer:}
\displaystyle \text{For any event }E,\quad 0\leq P(E)\leq1.
\displaystyle -0.7<0,\qquad \frac{11}{9}>1,\qquad 1.001>1.
\displaystyle \text{Only }0.6\text{ lies between }0\text{ and }1.
\displaystyle \therefore \text{The correct option is (d) }0.6.
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{In an experiment, the sum of probabilities of different events is:}
\displaystyle \text{(a) }1\qquad\text{(b) }0.5\qquad\text{(c) }-2\qquad\text{(d) }0
\displaystyle \text{Answer:}
\displaystyle \text{The sum of probabilities of all mutually exclusive and exhaustive events is }1.
\displaystyle \therefore \text{The correct option is (a) }1.
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{In rolling a die, the probability of getting number }8\text{ is:}
\displaystyle \text{(a) }0\qquad\text{(b) }1\qquad\text{(c) }-1\qquad\text{(d) }\frac{1}{2}
\displaystyle \text{Answer:}
\displaystyle \text{The possible outcomes on rolling a die are }1,2,3,4,5,6.
\displaystyle \text{Getting }8\text{ is an impossible event.}
\displaystyle \therefore P(\text{getting }8)=0.
\displaystyle \therefore \text{The correct option is (a) }0.
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{In a simultaneous throw of 2 coins, the probability of having 2 heads is:}
\displaystyle \text{(a) }\frac{1}{4}\qquad\text{(b) }\frac{1}{2}\qquad\text{(c) }\frac{1}{8}\qquad\text{(d) }\frac{1}{6}
\displaystyle \text{Answer:}
\displaystyle S=\{HH,HT,TH,TT\}.
\displaystyle n(S)=4.
\displaystyle \text{The favourable outcome for getting 2 heads is }HH.
\displaystyle \therefore n(E)=1.
\displaystyle P(E)=\frac{n(E)}{n(S)}=\frac{1}{4}.
\displaystyle \therefore \text{The correct option is (a) }\frac{1}{4}.
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{The probability of getting sum more than }7\text{ when a pair of dice are thrown is:}
\displaystyle \text{(a) }\frac{7}{36}\qquad\text{(b) }\frac{5}{12}\qquad\text{(c) }\frac{7}{12}\qquad\text{(d) None of these}
\displaystyle \text{Answer:}
\displaystyle \text{Total number of possible outcomes}=6\times6=36.
\displaystyle \text{For a sum greater than }7,\text{ the possible sums are }8,9,10,11,12.
\displaystyle \text{Number of favourable outcomes}=5+4+3+2+1=15.
\displaystyle P(\text{sum}>7)=\frac{15}{36}=\frac{5}{12}.
\displaystyle \therefore \text{The correct option is (b) }\frac{5}{12}.
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{The probability of raining on day 1 is }0.2\text{ and on day 2 is }0.3.
\displaystyle \text{Assuming the events are independent, the probability of raining on both the days is:}
\displaystyle \text{(a) }0.2\qquad\text{(b) }0.1\qquad\text{(c) }0.06\qquad\text{(d) }0.25
\displaystyle \text{Answer:}
\displaystyle P(\text{rain on both days})=P(\text{rain on day 1})P(\text{rain on day 2}).
\displaystyle =0.2\times0.3=0.06.
\displaystyle \therefore \text{The correct option is (c) }0.06.
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{If }A\text{ and }B\text{ are two events, such that }P(A\cup B)=\frac{3}{4},
\displaystyle P(A\cap B)=\frac{1}{4},\ P(A')=\frac{2}{3},\text{ where }A'\text{ stands for the complementary event of }A,
\displaystyle \text{then }P(B)\text{ is given by:}
\displaystyle \text{(a) }\frac{1}{3}\qquad\text{(b) }\frac{2}{3}\qquad\text{(c) }\frac{1}{9}\qquad\text{(d) }\frac{2}{9}
\displaystyle \text{Answer:}
\displaystyle P(A')=\frac{2}{3}.
\displaystyle \therefore P(A)=1-P(A')=1-\frac{2}{3}=\frac{1}{3}.
\displaystyle P(A\cup B)=P(A)+P(B)-P(A\cap B).
\displaystyle \frac{3}{4}=\frac{1}{3}+P(B)-\frac{1}{4}.
\displaystyle P(B)=\frac{3}{4}-\frac{1}{3}+\frac{1}{4}.
\displaystyle =1-\frac{1}{3}=\frac{2}{3}.
\displaystyle \therefore \text{The correct option is (b) }\frac{2}{3}.
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{In the following Venn diagram, circles }A\text{ and }B\text{ represent two events.}
\displaystyle \text{The probability of the union of the shaded regions will be:}
\displaystyle \text{(a) }P(A)+P(B)-2P(A\cap B)
\displaystyle \text{(b) }P(A)+P(B)-P(A\cap B)
\displaystyle \text{(c) }P(A)+P(B)
\displaystyle \text{(d) }2P(A)+2P(B)-P(A\cap B)
\displaystyle \text{Answer:}
\displaystyle \text{The shaded regions are }A-B\text{ and }B-A.
\displaystyle P(A-B)=P(A)-P(A\cap B).
\displaystyle P(B-A)=P(B)-P(A\cap B).
\displaystyle \therefore P(\text{shaded region})=P(A-B)+P(B-A).
\displaystyle =P(A)+P(B)-2P(A\cap B).
\displaystyle \therefore \text{The correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{A single letter is selected at random from the word ``PROBABILITY''. The}
\displaystyle \text{probability that the selected letter is a vowel is:}
\displaystyle \text{(a) }\frac{1}{11}\qquad\text{(b) }\frac{3}{11}\qquad\text{(c) }\frac{4}{11}\qquad\text{(d) }0
\displaystyle \text{Answer:}
\displaystyle \text{The word PROBABILITY has }11\text{ letters.}
\displaystyle \text{The vowels are }O,A,I,I,\text{ so the number of vowels is }4.
\displaystyle P(\text{vowel})=\frac{\text{Number of vowels}}{\text{Total number of letters}}=\frac{4}{11}.
\displaystyle \therefore \text{The correct option is (c) }\frac{4}{11}.
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{A bag contains 10 balls, out of which 4 balls are white and the others are}
\displaystyle \text{non-white. The probability of getting a non-white ball is:}
\displaystyle \text{(a) }\frac{2}{5}\qquad\text{(b) }\frac{3}{5}\qquad\text{(c) }\frac{1}{2}\qquad\text{(d) }\frac{2}{3}
\displaystyle \text{Answer:}
\displaystyle \text{Number of non-white balls}=10-4=6.
\displaystyle P(\text{non-white ball})=\frac{6}{10}=\frac{3}{5}.
\displaystyle \therefore \text{The correct option is (b) }\frac{3}{5}.
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{Two dice are thrown together. The probability of getting the sum of digits as}
\displaystyle \text{a multiple of }4\text{ is:}
\displaystyle \text{(a) }\frac{1}{9}\qquad\text{(b) }\frac{1}{3}\qquad\text{(c) }\frac{1}{4}\qquad\text{(d) }\frac{5}{9}
\displaystyle \text{Answer:}
\displaystyle \text{Total number of possible outcomes}=6\times6=36.
\displaystyle \text{The possible sums which are multiples of }4\text{ are }4,8\text{ and }12.
\displaystyle \text{Number of outcomes with sum }4=3.
\displaystyle \text{Number of outcomes with sum }8=5.
\displaystyle \text{Number of outcomes with sum }12=1.
\displaystyle \therefore \text{Number of favourable outcomes}=3+5+1=9.
\displaystyle P(\text{sum is a multiple of }4)=\frac{9}{36}=\frac{1}{4}.
\displaystyle \therefore \text{The correct option is (c) }\frac{1}{4}.
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{If the probabilities for }A\text{ to fail in an examination is }0.2\text{ and that for }B\text{ is }0.3,
\displaystyle \text{then the probability that either }A\text{ or }B\text{ fails is:}
\displaystyle \text{(a) }>0.5\qquad\text{(b) }0.5\qquad\text{(c) }\leq0.5\qquad\text{(d) }0
\displaystyle \text{Answer:}
\displaystyle P(A\cup B)=P(A)+P(B)-P(A\cap B).
\displaystyle =0.2+0.3-P(A\cap B).
\displaystyle =0.5-P(A\cap B).
\displaystyle \text{Since }P(A\cap B)\geq0,\quad P(A\cup B)\leq0.5.
\displaystyle \therefore \text{The correct option is (c) }\leq0.5.
\displaystyle \\

\displaystyle \textbf{Question 15: }\text{If }\frac{2}{11}\text{ is the probability of an event, then the probability of the event}
\displaystyle \text{`not A' is:}
\displaystyle \text{(a) }\frac{9}{11}\qquad\text{(b) }\frac{11}{2}\qquad\text{(c) }\frac{11}{9}\qquad\text{(d) }\frac{2}{11}
\displaystyle \text{Answer:}
\displaystyle P(A)=\frac{2}{11}.
\displaystyle P(A')=1-P(A).
\displaystyle =1-\frac{2}{11}=\frac{9}{11}.
\displaystyle \therefore \text{The correct option is (a) }\frac{9}{11}.
\displaystyle \\

\displaystyle \textbf{Question 16: }\text{An experiment is called a random experiment, if it:}
\displaystyle \text{(a) has more than one possible outcome}
\displaystyle \text{(b) is not possible to predict the outcome in advance}
\displaystyle \text{(c) Both (a) and (b)}
\displaystyle \text{(d) None of the above}
\displaystyle \text{Answer:}
\displaystyle \text{A random experiment has more than one possible outcome.}
\displaystyle \text{Also, the particular outcome cannot be predicted with certainty in advance.}
\displaystyle \therefore \text{Both (a) and (b) are true.}
\displaystyle \therefore \text{The correct option is (c).}
\displaystyle \\

\displaystyle \textbf{Question 17: }\text{An event can be classified into various types on the basis of the:}
\displaystyle \text{(a) experiment}\qquad\text{(b) sample space}\qquad\text{(c) elements}\qquad\text{(d) None of the above}
\displaystyle \text{Answer:}
\displaystyle \text{Events can be classified on the basis of the number and nature of elements they contain.}
\displaystyle \therefore \text{The correct option is (c) elements.}
\displaystyle \\

\displaystyle \textbf{Question 18: }\text{An event which has only one sample point of a sample space is called a}
\displaystyle \text{simple event.}
\displaystyle \text{(a) two}\qquad\text{(b) three}\qquad\text{(c) one}\qquad\text{(d) zero}
\displaystyle \text{Answer:}
\displaystyle \text{A simple or elementary event contains exactly one sample point.}
\displaystyle \therefore \text{The correct option is (c) one.}
\displaystyle \\

\displaystyle \textbf{Question 19: }\text{If an event has more than one sample point, then it is called a/an:}
\displaystyle \text{(a) simple event}\qquad\text{(b) elementary event}
\displaystyle \text{(c) compound event}\qquad\text{(d) None of these}
\displaystyle \text{Answer:}
\displaystyle \text{An event containing more than one sample point is called a compound event.}
\displaystyle \therefore \text{The correct option is (c) compound event.}
\displaystyle \\

\displaystyle \textbf{Question 20: }\text{A die is rolled. Let }E\text{ be the event ``die shows 4'' and }F\text{ be the event}
\displaystyle \text{``die shows even number''. Then, }E\text{ and }F\text{ are:}
\displaystyle \text{(a) mutually exclusive}\qquad\text{(b) exhaustive}
\displaystyle \text{(c) mutually exclusive and exhaustive}\qquad\text{(d) None of the above}
\displaystyle \text{Answer:}
\displaystyle S=\{1,2,3,4,5,6\},\qquad E=\{4\},\qquad F=\{2,4,6\}.
\displaystyle E\cap F=\{4\}\neq\varnothing.
\displaystyle \therefore E\text{ and }F\text{ are not mutually exclusive.}
\displaystyle E\cup F=\{2,4,6\}\neq S.
\displaystyle \therefore E\text{ and }F\text{ are not exhaustive.}
\displaystyle \therefore \text{The correct option is (d) None of the above.}
\displaystyle \\

\displaystyle \text{CASE BASED/SOURCE BASED / PASSAGE BASED}

\displaystyle \textbf{Question 21: }\text{If three coins are tossed, all possible outcomes are}
\displaystyle \{HHH,\ HHT,\ HTH,\ THH,\ TTH,\ HTT,\ THT,\ TTT\}.
\displaystyle \text{If }E_1,E_2,\ldots,E_n\text{ are events of a sample space }S\text{ and if }
\displaystyle E_1\cup E_2\cup E_3\cup\cdots\cup E_n=S,\text{ then }E_1,E_2,\ldots,E_n\text{ are called exhaustive events.}
\displaystyle \text{Further, if }E_i\cap E_j=\phi\text{ for }i\neq j,\text{ i.e. events }E_i\text{ and }E_j\text{ are pairwise disjoint}
\displaystyle \text{and }\bigcup E_i=S,\text{ then events }E_1,E_2,\ldots,E_n\text{ are called mutually exclusive and exhaustive events.}
\displaystyle \text{Now answer the following:}
\displaystyle \text{(i) Two events which are mutually exclusive.}
\displaystyle \text{(ii) Three events which are mutually exclusive and exhaustive.}
\displaystyle \text{(iii) Two events, which are not mutually exclusive.}
\displaystyle \text{(iv) Two events which are mutually exclusive but not exhaustive.}
\displaystyle \text{(v) Three events which are mutually exclusive but not exhaustive.}
\displaystyle \text{Answer:}
\displaystyle S=\{HHH,HHT,HTH,THH,TTH,HTT,THT,TTT\}.
\displaystyle \text{(i) Let }E_1=\{HHH\}\text{ and }E_2=\{TTT\}.
\displaystyle E_1\cap E_2=\phi.
\displaystyle \therefore E_1\text{ and }E_2\text{ are mutually exclusive events.}
\displaystyle \text{(ii) Let }E_1=\{HHH,HHT,HTH,THH\},
\displaystyle E_2=\{HTT,THT,TTH\},\qquad E_3=\{TTT\}.
\displaystyle E_1\cap E_2=E_2\cap E_3=E_1\cap E_3=\phi.
\displaystyle E_1\cup E_2\cup E_3=S.
\displaystyle \therefore E_1,E_2,E_3\text{ are mutually exclusive and exhaustive events.}
\displaystyle \text{(iii) Let }E_1=\{HHH,HHT,HTH,HTT\}\text{ and }
\displaystyle E_2=\{HHH,HHT,THH,THT\}.
\displaystyle E_1\cap E_2=\{HHH,HHT\}\neq\phi.
\displaystyle \therefore E_1\text{ and }E_2\text{ are not mutually exclusive.}
\displaystyle \text{(iv) Let }E_1=\{HHH\}\text{ and }E_2=\{TTT\}.
\displaystyle E_1\cap E_2=\phi,
\displaystyle \text{but }E_1\cup E_2=\{HHH,TTT\}\neq S.
\displaystyle \therefore E_1\text{ and }E_2\text{ are mutually exclusive but not exhaustive.}
\displaystyle \text{(v) Let }E_1=\{HHH\},\quad E_2=\{HHT\},\quad E_3=\{HTH\}.
\displaystyle E_1,E_2,E_3\text{ are pairwise disjoint,}
\displaystyle \text{but }E_1\cup E_2\cup E_3=\{HHH,HHT,HTH\}\neq S.
\displaystyle \therefore E_1,E_2,E_3\text{ are mutually exclusive but not exhaustive.}
\displaystyle \\

\displaystyle \textbf{Question 22: }\text{Grandmother of Reena, Simi and Aryan has a bag, which contains 9 discs of}
\displaystyle \text{different beautiful colors, of which 4 are red, 3 are blue and 2 are yellow.}
\displaystyle \text{The discs are similar in shape and size. Children asked grandmother to give them discs and}
\displaystyle \text{grandmother draws a disc at random from the bag. Calculate the probability that it will be}
\displaystyle \text{(i) red, (ii) yellow, (iii) blue, (iv) not blue, (v) either red or yellow.}
\displaystyle \text{Answer:}
\displaystyle \text{Total number of discs}=9.
\displaystyle \text{Number of red discs}=4,\quad\text{blue discs}=3,\quad\text{yellow discs}=2.
\displaystyle \text{(i) }P(\text{red})=\frac{4}{9}.
\displaystyle \text{(ii) }P(\text{yellow})=\frac{2}{9}.
\displaystyle \text{(iii) }P(\text{blue})=\frac{3}{9}=\frac{1}{3}.
\displaystyle \text{(iv) }P(\text{not blue})=1-P(\text{blue}).
\displaystyle =1-\frac{1}{3}=\frac{2}{3}.
\displaystyle \text{(v) Since a disc cannot be both red and yellow, the two events are mutually exclusive.}
\displaystyle P(\text{red or yellow})=P(\text{red})+P(\text{yellow}).
\displaystyle =\frac{4}{9}+\frac{2}{9}=\frac{6}{9}=\frac{2}{3}.
\displaystyle \\

\displaystyle \textbf{Question 23: }\text{An urn contains twenty white slips of paper numbered from 1 through 20,}
\displaystyle \text{ten red slips of paper numbered from 1 through 10, forty yellow slips of paper numbered}
\displaystyle \text{from 1 through 40 and ten blue slips of paper numbered from 1 through 10. These 80 slips}
\displaystyle \text{of paper are thoroughly shuffled so that each slip has an equal chance of being drawn.}
\displaystyle \text{A slip is drawn at random from the urn.}
\displaystyle \text{Based on the above information, answer the following questions:}
\displaystyle \text{(i) What is the probability that slip drawn is blue or white?}
\displaystyle \text{(ii) What is the probability that slip drawn is numbered 1, 2, 3, 4 or 5?}
\displaystyle \text{(iii) What is the probability that slip drawn is red or yellow and numbered 1, 2, 3 or 4?}
\displaystyle \text{(iv) What is the probability that slip drawn is numbered 5, 15, 25 or 35?}
\displaystyle \text{(v) What is the probability that slip drawn is white and number higher than 12 or yellow}
\displaystyle \text{and numbered higher than 26?}
\displaystyle \text{Answer:}
\displaystyle \text{Total number of slips}=20+10+40+10=80.
\displaystyle \text{(i) Number of blue or white slips}=10+20=30.
\displaystyle P(\text{blue or white})=\frac{30}{80}=\frac{3}{8}.
\displaystyle \text{(ii) For each colour, slips numbered }1,2,3,4,5\text{ are available.}
\displaystyle \text{Number of favourable slips}=5+5+5+5=20.
\displaystyle P(\text{numbered }1,2,3,4\text{ or }5)=\frac{20}{80}=\frac{1}{4}.
\displaystyle \text{(iii) Red slips numbered }1,2,3,4=4.
\displaystyle \text{Yellow slips numbered }1,2,3,4=4.
\displaystyle \text{Number of favourable slips}=4+4=8.
\displaystyle P(\text{red or yellow and numbered }1,2,3\text{ or }4)=\frac{8}{80}=\frac{1}{10}.
\displaystyle \text{(iv) White slips numbered }5,15=2.
\displaystyle \text{Red slips numbered }5=1.
\displaystyle \text{Yellow slips numbered }5,15,25,35=4.
\displaystyle \text{Blue slips numbered }5=1.
\displaystyle \text{Number of favourable slips}=2+1+4+1=8.
\displaystyle P(\text{numbered }5,15,25\text{ or }35)=\frac{8}{80}=\frac{1}{10}.
\displaystyle \text{(v) White slips numbered higher than }12\text{ are }13,14,\ldots,20.
\displaystyle \text{Their number}=20-12=8.
\displaystyle \text{Yellow slips numbered higher than }26\text{ are }27,28,\ldots,40.
\displaystyle \text{Their number}=40-26=14.
\displaystyle \text{Number of favourable slips}=8+14=22.
\displaystyle P(\text{required event})=\frac{22}{80}=\frac{11}{40}.
\displaystyle \\

\displaystyle \textbf{Question 24: }\text{The given Venn diagram shows three events }A,B\text{ and }C\text{ and also the}
\displaystyle \text{probabilities of the various intersections (for }P(A\cap B)=0.07\text{).}
\displaystyle \text{Based on the above information, answer the following questions:}
\displaystyle \text{(i) The value of }P(B\cap C')\text{ is}
\displaystyle \text{(ii) The value of }P(A\cap B)\text{ is}
\displaystyle \text{(iii) The value of }P(A\cap B')\text{ is}
\displaystyle \text{(iv) The value of }P(B\cap C)\text{ is}
\displaystyle \text{(v) The probability of occurrence of exactly one of the events }A,B\text{ or }C\text{ is}
\displaystyle \text{Answer:}
\displaystyle \text{From the Venn diagram,}
\displaystyle P(A\text{ only})=0.13,\quad P(A\cap B)=0.07,\quad P(B\text{ only})=0.10,
\displaystyle P(B\cap C)=0.15,\quad P(C\text{ only})=0.28.
\displaystyle \text{(i) }B\cap C'\text{ represents the part of }B\text{ which does not belong to }C.
\displaystyle P(B\cap C')=0.10+0.07=0.17.
\displaystyle \text{(ii) From the diagram, }P(A\cap B)=0.07.
\displaystyle \text{(iii) }A\cap B'\text{ represents the part of }A\text{ which does not belong to }B.
\displaystyle P(A\cap B')=0.13.
\displaystyle \text{(iv) From the diagram, }P(B\cap C)=0.15.
\displaystyle \text{(v) Exactly one of }A,B\text{ or }C\text{ means the regions belonging only to }A,B\text{ or }C.
\displaystyle P(\text{exactly one})=0.13+0.10+0.28.
\displaystyle =0.51.
\displaystyle \\

\displaystyle \textbf{Question 25: }\text{The letters of the word `ASSASSINATION' are arranged at random.}
\displaystyle \text{Based on the above information, answer the following questions:}
\displaystyle \text{(i) The probability that four S's come consecutively is}
\displaystyle \text{(ii) The probability that two I's and two N's come together is}
\displaystyle \text{(iii) The probability that all A's are not coming together is}
\displaystyle \text{(iv) The probability that no two A's are coming together is}
\displaystyle \text{(v) The probability that all vowels are coming together is}
\displaystyle \text{Answer:}
\displaystyle \text{In ASSASSINATION, }A=3,\ S=4,\ I=2,\ N=2,\ T=1,\ O=1.
\displaystyle \text{Total number of distinct arrangements}=\frac{13!}{3!4!2!2!}.
\displaystyle \text{(i) Treat the four S's as one block.}
\displaystyle \text{Number of favourable arrangements}=\frac{10!}{3!2!2!}.
\displaystyle P(\text{four S's together})=\frac{\frac{10!}{3!2!2!}}{\frac{13!}{3!4!2!2!}}.
\displaystyle =\frac{4!\,10!}{13!}=\frac{2}{143}.
\displaystyle \text{(ii) Treat }II\text{ and }NN\text{ as two separate blocks.}
\displaystyle \text{Number of favourable arrangements}=\frac{11!}{3!4!}.
\displaystyle P(II\text{ and }NN\text{ together})=\frac{\frac{11!}{3!4!}}{\frac{13!}{3!4!2!2!}}.
\displaystyle =\frac{2!2!\,11!}{13!}=\frac{1}{39}.
\displaystyle \text{(iii) First find the probability that all three A's are together.}
\displaystyle \text{Treat }AAA\text{ as one block.}
\displaystyle \text{Number of favourable arrangements}=\frac{11!}{4!2!2!}.
\displaystyle P(\text{all A's together})=\frac{\frac{11!}{4!2!2!}}{\frac{13!}{3!4!2!2!}}.
\displaystyle =\frac{3!\,11!}{13!}=\frac{1}{26}.
\displaystyle \therefore P(\text{all A's are not together})=1-\frac{1}{26}=\frac{25}{26}.
\displaystyle \text{(iv) Arrange the 10 non-A letters first.}
\displaystyle \text{Number of arrangements of non-A letters}=\frac{10!}{4!2!2!}.
\displaystyle \text{These 10 letters create }11\text{ gaps in which the three A's can be placed.}
\displaystyle \text{Choose 3 of these 11 gaps, so the number of choices is }\binom{11}{3}.
\displaystyle \text{Number of favourable arrangements}=\frac{10!}{4!2!2!}\binom{11}{3}.
\displaystyle P(\text{no two A's together})=\frac{\frac{10!}{4!2!2!}\binom{11}{3}}{\frac{13!}{3!4!2!2!}}.
\displaystyle =\frac{3!\,10!}{13!}\binom{11}{3}=\frac{15}{26}.
\displaystyle \text{(v) The vowels are }A,A,A,I,I,O,\text{ i.e. }6\text{ vowels.}
\displaystyle \text{Number of arrangements within the vowel block}=\frac{6!}{3!2!}.
\displaystyle \text{Treating all vowels as one block, there are }8\text{ objects to arrange.}
\displaystyle \text{Number of arrangements of these objects}=\frac{8!}{4!2!}.
\displaystyle \text{Number of favourable arrangements}=\frac{8!}{4!2!}\times\frac{6!}{3!2!}.
\displaystyle P(\text{all vowels together})=\frac{\frac{8!}{4!2!}\times\frac{6!}{3!2!}}{\frac{13!}{3!4!2!2!}}.
\displaystyle =\frac{8!\,6!}{13!}=\frac{2}{429}.
\displaystyle \\

\displaystyle \text{ASSERTION REASONING TYPE QUESTIONS}

\displaystyle \text{ASSERTION-REASON TYPE QUESTIONS}
\displaystyle \text{Assertion and Reason. Each of these questions also has four alternative choices, only one of which}
\displaystyle \text{is the correct answer. You have to select one of the codes (a), (b), (c) and (d) given below.}
\displaystyle \text{(a) Assertion is correct, reason is correct; reason is a correct explanation for assertion.}
\displaystyle \text{(b) Assertion is correct, reason is correct; reason is not a correct explanation for assertion.}
\displaystyle \text{(c) Assertion is correct, reason is incorrect.}
\displaystyle \text{(d) Assertion is incorrect, reason is correct.}
\displaystyle \\

\displaystyle \textbf{Question 26: }\text{Assertion: Probability of getting a head in a toss of an unbiased coin is }\frac{1}{2}.
\displaystyle \text{Reason: In a simultaneous toss of two coins, the probability of getting `no tails' is }\frac{1}{4}.
\displaystyle \text{Answer:}
\displaystyle \text{For one unbiased coin, }S=\{H,T\}.
\displaystyle P(\text{head})=\frac{1}{2}.
\displaystyle \therefore \text{The Assertion is correct.}
\displaystyle \text{For two coins, }S=\{HH,HT,TH,TT\}.
\displaystyle \text{`No tails' means the outcome }HH.
\displaystyle P(\text{no tails})=\frac{1}{4}.
\displaystyle \therefore \text{The Reason is also correct.}
\displaystyle \text{However, the Reason does not explain the Assertion.}
\displaystyle \therefore \text{The correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 26: }\text{Assertion: Probability of getting a head in a toss of an unbiased coin is }\frac{1}{2}.
\displaystyle \text{Reason: In a simultaneous toss of two coins, the probability of getting `no tails' is }\frac{1}{4}.
\displaystyle \text{Answer:}
\displaystyle \text{For an unbiased coin, }S=\{H,T\}.
\displaystyle P(\text{head})=\frac{1}{2}.
\displaystyle \therefore \text{The Assertion is correct.}
\displaystyle \text{For two coins, }S=\{HH,HT,TH,TT\}.
\displaystyle \text{The event `no tails' is }\{HH\}.
\displaystyle \therefore P(\text{no tails})=\frac{1}{4}.
\displaystyle \therefore \text{The Reason is correct.}
\displaystyle \text{However, the Reason is not the correct explanation of the Assertion.}
\displaystyle \therefore \text{The correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 26: }\text{Assertion: Probability of getting a head in a toss of an unbiased coin is }\frac{1}{2}.
\displaystyle \text{Reason: In a simultaneous toss of two coins, the probability of getting `no tails' is }\frac{1}{4}.
\displaystyle \text{Answer:}
\displaystyle \text{For an unbiased coin, }S=\{H,T\}.
\displaystyle P(\text{head})=\frac{1}{2}.
\displaystyle \therefore \text{The Assertion is correct.}
\displaystyle \text{For two coins, }S=\{HH,HT,TH,TT\}.
\displaystyle \text{The event `no tails' is }\{HH\}.
\displaystyle \therefore P(\text{no tails})=\frac{1}{4}.
\displaystyle \therefore \text{The Reason is correct.}
\displaystyle \text{However, the Reason is not the correct explanation of the Assertion.}
\displaystyle \therefore \text{The correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 28: }\text{Assertion: A letter is chosen at random from the word NAGATATION.}
\displaystyle \text{Then, the total number of outcomes is }10.
\displaystyle \text{Reason: A letter is chosen at random from the word `ASSASSINATION'.}
\displaystyle \text{Then, the total number of outcomes is }13.
\displaystyle \text{Answer:}
\displaystyle \text{The word NAGATATION contains }10\text{ letters.}
\displaystyle \therefore \text{The total number of outcomes for choosing one letter is }10.
\displaystyle \therefore \text{The Assertion is correct.}
\displaystyle \text{The word ASSASSINATION contains }13\text{ letters.}
\displaystyle \therefore \text{The total number of outcomes for choosing one letter is }13.
\displaystyle \therefore \text{The Reason is also correct.}
\displaystyle \text{However, the Reason is not the correct explanation of the Assertion.}
\displaystyle \therefore \text{The correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 29: }\text{Consider a single throw of die and two events.}
\displaystyle A=\text{the number is even}=\{2,4,6\}.
\displaystyle B=\text{the number is a multiple of }3=\{3,6\}.
\displaystyle \text{Assertion: }P(A\cup B)=\frac{4}{6}=\frac{2}{3}\text{ and }P(A\cap B)=\frac{1}{6}.
\displaystyle \text{Reason: }P(A'\cap B')=P((A\cup B)')=1-\frac{2}{3}=\frac{1}{3}.
\displaystyle \text{Answer:}
\displaystyle A\cup B=\{2,3,4,6\}.
\displaystyle \therefore P(A\cup B)=\frac{4}{6}=\frac{2}{3}.
\displaystyle A\cap B=\{6\}.
\displaystyle \therefore P(A\cap B)=\frac{1}{6}.
\displaystyle \therefore \text{The Assertion is correct.}
\displaystyle \text{By De Morgan's law, }A'\cap B'=(A\cup B)'.
\displaystyle P(A'\cap B')=1-P(A\cup B).
\displaystyle =1-\frac{2}{3}=\frac{1}{3}.
\displaystyle \therefore \text{The Reason is also correct.}
\displaystyle \text{However, the Reason is not the correct explanation of the Assertion.}
\displaystyle \therefore \text{The correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 30: }\text{Assertion: Two dice are thrown simultaneously. There are 11 possible}
\displaystyle \text{outcomes and each of them has a probability }\frac{1}{11}.
\displaystyle \text{Reason: Probability of an event }E\text{ is defined as}
\displaystyle P(E)=\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}.
\displaystyle \text{Answer:}
\displaystyle \text{When two dice are thrown, the total number of equally likely outcomes}=6\times6=36.
\displaystyle \text{There are 11 possible sums, namely }2,3,4,\ldots,12,
\displaystyle \text{but these sums are not equally likely.}
\displaystyle P(\text{sum }2)=\frac{1}{36},\qquad P(\text{sum }7)=\frac{6}{36}=\frac{1}{6}.
\displaystyle \therefore \text{The Assertion is incorrect.}
\displaystyle \text{For equally likely outcomes, }P(E)=\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}.
\displaystyle \therefore \text{The Reason is correct.}
\displaystyle \therefore \text{The correct option is (d).}
\displaystyle \\

\displaystyle \text{VERY SHORT ANSWER QUESTIONS}

\displaystyle \textbf{Question 31: }\text{The numbers }1,\,2,\,3\text{ and }4\text{ are written separately on four slips of paper.}
\displaystyle \text{The slips are put in a box and mixed thoroughly. A person draws two slips from the box,}
\displaystyle \text{one after the other, without replacement. Describe the sample space for the experiment.}
\displaystyle \text{Answer:}
\displaystyle \text{Since the slips are drawn one after the other, the order of drawing matters.}
\displaystyle \text{The possible ordered outcomes are:}
\displaystyle S=\{(1,2),(1,3),(1,4),(2,1),(2,3),(2,4),
\displaystyle \qquad (3,1),(3,2),(3,4),(4,1),(4,2),(4,3)\}.
\displaystyle \therefore n(S)=4\times3=12.
\displaystyle \\

\displaystyle \textbf{Question 32: }\text{An experiment consists of rolling a die and then tossing a coin once if the}
\displaystyle \text{number on the die is even. If the number on the die is odd, the coin is tossed twice.}
\displaystyle \text{Write the sample space for this experiment.}
\displaystyle \text{Answer:}
\displaystyle \text{For an even number }2,4\text{ or }6,\text{ the coin is tossed once.}
\displaystyle \text{The possible outcomes are }(2,H),(2,T),(4,H),(4,T),(6,H),(6,T).
\displaystyle \text{For an odd number }1,3\text{ or }5,\text{ the coin is tossed twice.}
\displaystyle \text{The possible outcomes are }(1,HH),(1,HT),(1,TH),(1,TT),
\displaystyle (3,HH),(3,HT),(3,TH),(3,TT),
\displaystyle (5,HH),(5,HT),(5,TH),(5,TT).
\displaystyle \therefore S=\{(1,HH),(1,HT),(1,TH),(1,TT),(2,H),(2,T),
\displaystyle \qquad (3,HH),(3,HT),(3,TH),(3,TT),(4,H),(4,T),
\displaystyle \qquad (5,HH),(5,HT),(5,TH),(5,TT),(6,H),(6,T)\}.
\displaystyle \therefore n(S)=18.
\displaystyle \\

\displaystyle \textbf{Question 33: }\text{A box contains }1\text{ red and }3\text{ identical white balls. Two balls are drawn at}
\displaystyle \text{random in succession without replacement. Write the sample space for this experiment.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }R\text{ denote a red ball and }W\text{ denote a white ball.}
\displaystyle \text{Since there is only one red ball, }RR\text{ is not possible.}
\displaystyle \text{The possible outcomes are }RW,\ WR\text{ and }WW.
\displaystyle \therefore S=\{RW,WR,WW\}.
\displaystyle \\

\displaystyle \textbf{Question 34: }\text{If }\frac{3}{4}\text{ is the probability of an event, what is the probability of the event}
\displaystyle \text{``not }A\text{''?}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }P(A)=\frac{3}{4}.
\displaystyle \text{We know that }P(A')=1-P(A).
\displaystyle \therefore P(A')=1-\frac{3}{4}=\frac{1}{4}.
\displaystyle \therefore \text{The probability of the event ``not }A\text{'' is }\frac{1}{4}.
\displaystyle \\

\displaystyle \textbf{Question 35: }\text{There are four men and six women on the city council. If one council}
\displaystyle \text{member is selected for a committee at random, how likely is it that it is a woman?}
\displaystyle \text{Answer:}
\displaystyle \text{Total number of council members}=4+6=10.
\displaystyle \text{Number of women}=6.
\displaystyle P(\text{selecting a woman})=\frac{\text{Number of women}}{\text{Total number of council members}}
\displaystyle =\frac{6}{10}=\frac{3}{5}.
\displaystyle \therefore \text{The probability of selecting a woman is }\frac{3}{5}.
\displaystyle \\

\displaystyle \textbf{Question 36: }\text{If two cards are drawn from a well shuffled pack, what is the probability}
\displaystyle \text{that at least one of the two is heart?}
\displaystyle \text{Answer:}
\displaystyle \text{A standard pack has }52\text{ cards, of which }13\text{ are hearts and }39\text{ are non-hearts.}
\displaystyle \text{Let }A=\text{the event that at least one of the two cards is a heart.}
\displaystyle \text{Then }A'=\text{the event that neither of the two cards is a heart.}
\displaystyle P(A')=\frac{{}^{39}C_2}{{}^{52}C_2}
\displaystyle =\frac{39\times38}{52\times51}=\frac{19}{34}.
\displaystyle P(A)=1-P(A')
\displaystyle =1-\frac{19}{34}=\frac{15}{34}.
\displaystyle \therefore \text{The probability that at least one of the two cards is a heart is }\frac{15}{34}.
\displaystyle \\

\displaystyle \textbf{Question 37: }\text{What is the probability that a leap year will have }53\text{ Sundays?}
\displaystyle \text{Answer:}
\displaystyle \text{A leap year has }366\text{ days}=52\text{ weeks}+2\text{ days.}
\displaystyle \text{Thus, every day occurs }52\text{ times and two consecutive days occur }53\text{ times.}
\displaystyle \text{The possible pairs of extra days are:}
\displaystyle (Mon,Tue),(Tue,Wed),(Wed,Thu),(Thu,Fri),(Fri,Sat),(Sat,Sun),(Sun,Mon).
\displaystyle \text{For }53\text{ Sundays, the favourable pairs are }(Sat,Sun)\text{ and }(Sun,Mon).
\displaystyle \therefore P(53\text{ Sundays})=\frac{2}{7}.
\displaystyle \therefore \text{The probability that a leap year has }53\text{ Sundays is }\frac{2}{7}.
\displaystyle \\

\displaystyle \textbf{Question 38: }\text{What is the probability of getting a total of }10\text{ in a single throw of two}
\displaystyle \text{dice?}
\displaystyle \text{Answer:}
\displaystyle \text{When two dice are thrown, the total number of equally likely outcomes}=6\times6=36.
\displaystyle \text{The outcomes giving a total of }10\text{ are }(4,6),(5,5)\text{ and }(6,4).
\displaystyle \therefore \text{Number of favourable outcomes}=3.
\displaystyle P(\text{total of }10)=\frac{3}{36}=\frac{1}{12}.
\displaystyle \therefore \text{The probability of getting a total of }10\text{ is }\frac{1}{12}.
\displaystyle \\

\displaystyle \textbf{Question 39: }\text{Two dice are rolled simultaneously. What is the probability that the}
\displaystyle \text{numbers on them are different?}
\displaystyle \text{Answer:}
\displaystyle \text{When two dice are rolled, the total number of equally likely outcomes}=6\times6=36.
\displaystyle \text{The outcomes having the same number are }(1,1),(2,2),(3,3),(4,4),(5,5),(6,6).
\displaystyle \therefore \text{Number of outcomes having the same number}=6.
\displaystyle \therefore \text{Number of outcomes having different numbers}=36-6=30.
\displaystyle P(\text{different numbers})=\frac{30}{36}=\frac{5}{6}.
\displaystyle \therefore \text{The probability that the numbers on the dice are different is }\frac{5}{6}.
\displaystyle \\

\displaystyle \textbf{Question 40: }\text{At random all the letters of the word ``ARTICLE'' are arranged in all}
\displaystyle \text{possible ways. What is the probability that the arrangement begins with a}
\displaystyle \text{vowel and ends with a consonant?}
\displaystyle \text{Answer:}
\displaystyle \text{The word ``ARTICLE'' has }7\text{ distinct letters.}
\displaystyle \text{Number of vowels}=3\quad(A,I,E),\qquad \text{Number of consonants}=4.
\displaystyle \text{Total number of arrangements}=7!.
\displaystyle \text{For favourable arrangements, the first letter can be chosen in }3\text{ ways.}
\displaystyle \text{The last letter can be chosen in }4\text{ ways.}
\displaystyle \text{The remaining }5\text{ letters can be arranged in }5!\text{ ways.}
\displaystyle \therefore \text{Number of favourable arrangements}=3\times4\times5!.
\displaystyle P(\text{begins with a vowel and ends with a consonant})
\displaystyle =\frac{3\times4\times5!}{7!}=\frac{12}{42}=\frac{2}{7}.
\displaystyle \therefore \text{The required probability is }\frac{2}{7}.
\displaystyle \\

\displaystyle \text{SHORT ANSWER QUESTIONS}

\displaystyle \textbf{Question 41: }\text{A letter is chosen at random from the word ``ASSASSINATION''. Find the}
\displaystyle \text{probability that the letter is (i) a vowel (ii) a consonant (iii) letter S.}
\displaystyle \text{Answer:}
\displaystyle \text{The word ``ASSASSINATION'' contains }13\text{ letters.}
\displaystyle \text{The vowels are }A,A,A,I,I,O.
\displaystyle \therefore \text{Number of vowels}=6.
\displaystyle \text{(i) }P(\text{a vowel})=\frac{6}{13}.

\displaystyle \text{Number of consonants}=13-6=7.
\displaystyle \text{(ii) }P(\text{a consonant})=\frac{7}{13}.

\displaystyle \text{The letter }S\text{ occurs }4\text{ times.}
\displaystyle \text{(iii) }P(\text{letter }S)=\frac{4}{13}.
\displaystyle \\

\displaystyle \textbf{Question 42: }\text{A and B are events such that }P(A)=0.42,\ P(B)=0.48\text{ and}
\displaystyle P(A\text{ and }B)=0.16.\text{ Determine (i) }P(\text{not }A),\text{ (ii) }P(\text{not }B)
\displaystyle \text{and (iii) }P(A\text{ or }B).
\displaystyle \text{Answer:}
\displaystyle \text{Given, }P(A)=0.42,\quad P(B)=0.48,\quad P(A\cap B)=0.16.
\displaystyle \text{(i) }P(\text{not }A)=1-P(A)
\displaystyle =1-0.42=0.58.

\displaystyle \text{(ii) }P(\text{not }B)=1-P(B)
\displaystyle =1-0.48=0.52.

\displaystyle \text{(iii) }P(A\text{ or }B)=P(A\cup B)
\displaystyle =P(A)+P(B)-P(A\cap B)
\displaystyle =0.42+0.48-0.16=0.74.
\displaystyle \\

\displaystyle \textbf{Question 43: }\text{In Class XI of a school }40\%\text{ of the students study Mathematics and }30\%
\displaystyle \text{study Biology. }10\%\text{ of the class study both Mathematics and Biology. If a student is}
\displaystyle \text{selected at random from the class, find the probability that he will be studying}
\displaystyle \text{Mathematics or Biology.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }M=\text{event that the student studies Mathematics and }B=\text{event that the student studies Biology.}
\displaystyle \text{Given, }P(M)=0.40,\quad P(B)=0.30,\quad P(M\cap B)=0.10.
\displaystyle \text{Using }P(M\cup B)=P(M)+P(B)-P(M\cap B),
\displaystyle P(M\cup B)=0.40+0.30-0.10=0.60=\frac{3}{5}.
\displaystyle \therefore \text{The probability that the student studies Mathematics or Biology is }\frac{3}{5}.
\displaystyle \\

\displaystyle \textbf{Question 44: }\text{A die has two faces each with number }1\text{, three faces each with number }2
\displaystyle \text{and one face with number }3.\text{ If the die is rolled once, determine (i) }P(2)
\displaystyle \text{(ii) }P(1\text{ or }3)\text{ (iii) }P(\text{not }3).
\displaystyle \text{Answer:}
\displaystyle \text{The die has }6\text{ equally likely faces.}

\displaystyle \text{(i) There are }3\text{ faces numbered }2.
\displaystyle \therefore P(2)=\frac{3}{6}=\frac{1}{2}.

\displaystyle \text{(ii) There are }2\text{ faces numbered }1\text{ and }1\text{ face numbered }3.
\displaystyle \therefore P(1\text{ or }3)=\frac{2+1}{6}=\frac{3}{6}=\frac{1}{2}.

\displaystyle \text{(iii) There are }5\text{ faces which are not numbered }3.
\displaystyle \therefore P(\text{not }3)=\frac{5}{6}.
\displaystyle \\

\displaystyle \textbf{Question 45: }\text{In a certain lottery }10,000\text{ tickets are sold and ten equal prizes are}
\displaystyle \text{awarded. What is the probability of not getting a prize if you buy (a) one ticket}
\displaystyle \text{(b) two tickets (c) }10\text{ tickets?}
\displaystyle \text{Answer:}
\displaystyle \text{Number of winning tickets}=10.
\displaystyle \text{Number of non-winning tickets}=10,000-10=9,990.

\displaystyle \text{(a) }P(\text{no prize with one ticket})=\frac{9990}{10000}
\displaystyle =\frac{999}{1000}=0.999.

\displaystyle \text{(b) }P(\text{no prize with two tickets})=\frac{{}^{9990}C_2}{{}^{10000}C_2}
\displaystyle =\frac{9990}{10000}\times\frac{9989}{9999}
\displaystyle \approx0.998001.

\displaystyle \text{(c) }P(\text{no prize with }10\text{ tickets})=\frac{{}^{9990}C_{10}}{{}^{10000}C_{10}}
\displaystyle =\frac{9990}{10000}\times\frac{9989}{9999}\times\frac{9988}{9998}\times\cdots\times\frac{9981}{9991}
\displaystyle \approx0.990040.
\displaystyle \\

\displaystyle \textbf{Question 46: }\text{A five digit number without repetition is formed by the digits }1,2,3,4,
\displaystyle 5,6,7,8.\text{ What is the probability that the number has even digits at both ends?}
\displaystyle \text{Answer:}
\displaystyle \text{Total number of five digit numbers}={}^{8}P_5.
\displaystyle \text{The even digits are }2,4,6,8.
\displaystyle \text{For the first digit, there are }4\text{ choices.}
\displaystyle \text{For the last digit, there are }3\text{ choices from the remaining even digits.}
\displaystyle \text{The middle three positions can be filled from the remaining }6\text{ digits in }{}^{6}P_3\text{ ways.}
\displaystyle \therefore \text{Number of favourable numbers}=4\times3\times{}^{6}P_3.
\displaystyle P(\text{even digits at both ends})=\frac{4\times3\times{}^{6}P_3}{{}^{8}P_5}
\displaystyle =\frac{4\times3\times120}{6720}=\frac{1440}{6720}=\frac{3}{14}.
\displaystyle \therefore \text{The required probability is }\frac{3}{14}.
\displaystyle \\

\displaystyle \textbf{Question 47: }\text{Four cards are drawn from a well-shuffled deck of }52\text{ cards. What is the}
\displaystyle \text{probability of obtaining }3\text{ diamonds and one spade?}
\displaystyle \text{Answer:}
\displaystyle \text{Total number of ways of drawing }4\text{ cards from }52\text{ cards}={}^{52}C_4.
\displaystyle \text{There are }13\text{ diamonds and }13\text{ spades in a deck.}
\displaystyle \text{Number of ways of choosing }3\text{ diamonds}={}^{13}C_3.
\displaystyle \text{Number of ways of choosing one spade}={}^{13}C_1.
\displaystyle \therefore \text{Number of favourable outcomes}={}^{13}C_3\times{}^{13}C_1.
\displaystyle P(3\text{ diamonds and one spade})=\frac{{}^{13}C_3\times{}^{13}C_1}{{}^{52}C_4}
\displaystyle =\frac{286\times13}{270725}=\frac{3718}{270725}=\frac{286}{20825}.
\displaystyle \therefore \text{The required probability is }\frac{286}{20825}.
\displaystyle \\

\displaystyle \textbf{Question 48: }\text{A and B are events such that }P(A)=0.42,\ P(B)=0.48\text{ and}
\displaystyle P(A\text{ and }B)=0.16.\text{ Determine (i) }P(\text{not }A),\text{ (ii) }P(\text{not }B)
\displaystyle \text{and (iii) }P(A\text{ or }B).
\displaystyle \text{Answer:}
\displaystyle \text{Given, }P(A)=0.42,\quad P(B)=0.48,\quad P(A\cap B)=0.16.
\displaystyle \text{(i) }P(\text{not }A)=1-P(A)
\displaystyle =1-0.42=0.58.

\displaystyle \text{(ii) }P(\text{not }B)=1-P(B)
\displaystyle =1-0.48=0.52.

\displaystyle \text{(iii) }P(A\text{ or }B)=P(A\cup B)
\displaystyle =P(A)+P(B)-P(A\cap B)
\displaystyle =0.42+0.48-0.16=0.74.
\displaystyle \\

\displaystyle \textbf{Question 49: }\text{In Class XI of a school }40\%\text{ of the students study Mathematics and }30\%
\displaystyle \text{study Biology. }10\%\text{ of the class study both Mathematics and Biology. If a student is}
\displaystyle \text{selected at random from the class, find the probability that he will be studying}
\displaystyle \text{Mathematics or Biology.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }M=\text{event that the student studies Mathematics and }B=\text{event that the student studies Biology.}
\displaystyle \text{Given, }P(M)=0.40,\quad P(B)=0.30,\quad P(M\cap B)=0.10.
\displaystyle P(M\cup B)=P(M)+P(B)-P(M\cap B)
\displaystyle =0.40+0.30-0.10=0.60=\frac{3}{5}.
\displaystyle \therefore \text{The probability that the student studies Mathematics or Biology is }\frac{3}{5}.
\displaystyle \\

\displaystyle \textbf{Question 50: }\text{A fair coin is tossed four times, and a person wins Re. }1\text{ for each head and}
\displaystyle \text{loses Rs. }1.50\text{ for each tail that turns up. From the sample space calculate how many}
\displaystyle \text{different amounts of money you can have after four tosses and the probability of}
\displaystyle \text{having each of these amounts.}
\displaystyle \text{Answer:}
\displaystyle \text{For four tosses of a fair coin, the total number of equally likely outcomes}=2^4=16.
\displaystyle \text{If there are }h\text{ heads, the number of tails}=4-h.
\displaystyle \text{Net amount}=h-1.50(4-h).

\displaystyle \text{For }h=0,\quad \text{net amount}=-6,\quad P=\frac{{}^{4}C_0}{16}=\frac{1}{16}.

\displaystyle \text{For }h=1,\quad \text{net amount}=-3.50,\quad P=\frac{{}^{4}C_1}{16}=\frac{4}{16}=\frac{1}{4}.

\displaystyle \text{For }h=2,\quad \text{net amount}=-1,\quad P=\frac{{}^{4}C_2}{16}=\frac{6}{16}=\frac{3}{8}.

\displaystyle \text{For }h=3,\quad \text{net amount}=1.50,\quad P=\frac{{}^{4}C_3}{16}=\frac{4}{16}=\frac{1}{4}.

\displaystyle \text{For }h=4,\quad \text{net amount}=4,\quad P=\frac{{}^{4}C_4}{16}=\frac{1}{16}.
\displaystyle \therefore \text{There are }5\text{ different possible amounts: Rs. }-6,\text{ Rs. }-3.50,\text{ Rs. }-1,
\displaystyle \text{Rs. }1.50\text{ and Rs. }4.
\displaystyle \\

\displaystyle \text{LONG ANSWER QUESTIONS}

\displaystyle \textbf{Question 51: }\text{In a class of }60\text{ students, }30\text{ opted for NCC, }32\text{ opted for NSS and }24
\displaystyle \text{opted for both NCC and NSS. If one of these students is selected at random, find the}
\displaystyle \text{probability that:}
\displaystyle \text{(i) The student opted for NCC or NSS.}
\displaystyle \text{(ii) The student has opted neither NCC nor NSS.}
\displaystyle \text{(iii) The student has opted NSS but not NCC.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }A=\text{event that the student opted for NCC and }B=\text{event that the student opted for NSS.}
\displaystyle n(A)=30,\quad n(B)=32,\quad n(A\cap B)=24,\quad n(S)=60.

\displaystyle \text{(i) }n(A\cup B)=n(A)+n(B)-n(A\cap B)
\displaystyle =30+32-24=38.
\displaystyle \therefore P(A\cup B)=\frac{38}{60}=\frac{19}{30}.

\displaystyle \text{(ii) Number of students who opted for neither NCC nor NSS}=60-38=22.
\displaystyle \therefore P(\text{neither NCC nor NSS})=\frac{22}{60}=\frac{11}{30}.

\displaystyle \text{(iii) Number of students who opted for NSS but not NCC}=32-24=8.
\displaystyle \therefore P(\text{NSS but not NCC})=\frac{8}{60}=\frac{2}{15}.
\displaystyle \\

\displaystyle \textbf{Question 52: }\text{Out of }100\text{ students, two sections of }40\text{ and }60\text{ are formed. If you and your}
\displaystyle \text{friend are among the }100\text{ students, what is the probability that (a) you both enter the}
\displaystyle \text{same section? (b) you both enter the different sections?}
\displaystyle \text{Answer:}

\displaystyle \text{(a) }P(\text{both enter the same section})
\displaystyle =P(\text{both in the section of }40)+P(\text{both in the section of }60)
\displaystyle =\frac{40}{100}\times\frac{39}{99}+\frac{60}{100}\times\frac{59}{99}
\displaystyle =\frac{1560+3540}{9900}=\frac{5100}{9900}=\frac{17}{33}.
\displaystyle \therefore P(\text{both enter the same section})=\frac{17}{33}.

\displaystyle \text{(b) }P(\text{both enter the different sections})
\displaystyle =1-P(\text{both enter the same section})
\displaystyle =1-\frac{17}{33}=\frac{16}{33}.
\displaystyle \therefore P(\text{both enter the different sections})=\frac{16}{33}.
\displaystyle \\

\displaystyle \textbf{Question 53: }\text{Three letters are dictated to three persons and an envelope is addressed to}
\displaystyle \text{each of them. The letters are inserted into the envelopes at random so that each}
\displaystyle \text{envelope contains exactly one letter. Find the probability that at least one letter is}
\displaystyle \text{in its proper envelope.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the letters be }L_1,L_2,L_3\text{ and the corresponding envelopes be }E_1,E_2,E_3.
\displaystyle \text{Total number of ways of placing the }3\text{ letters in the }3\text{ envelopes}=3!=6.
\displaystyle \text{There are }2\text{ arrangements in which no letter is in its proper envelope.}
\displaystyle \therefore \text{Number of arrangements with at least one letter in its proper envelope}=6-2=4.
\displaystyle P(\text{at least one letter in its proper envelope})=\frac{4}{6}=\frac{2}{3}.
\displaystyle \therefore \text{The required probability is }\frac{2}{3}.
\displaystyle \\

\displaystyle \textbf{Question 54: }\text{One card is drawn from a well shuffled deck of }52\text{ cards. If each outcome is}
\displaystyle \text{equally likely, calculate the probability that the card will be (i) a diamond (ii) not an ace}
\displaystyle \text{(iii) a black card (i.e., a club or a spade) (iv) not a diamond (v) not a black card.}
\displaystyle \text{Answer:}
\displaystyle \text{Total number of cards in a well shuffled deck}=52.

\displaystyle \text{(i) There are }13\text{ diamonds in a deck.}
\displaystyle \therefore P(\text{a diamond})=\frac{13}{52}=\frac{1}{4}.

\displaystyle \text{(ii) There are }4\text{ aces, so the number of cards which are not aces}=52-4=48.
\displaystyle \therefore P(\text{not an ace})=\frac{48}{52}=\frac{12}{13}.

\displaystyle \text{(iii) There are }13\text{ clubs and }13\text{ spades, so there are }26\text{ black cards.}
\displaystyle \therefore P(\text{a black card})=\frac{26}{52}=\frac{1}{2}.

\displaystyle \text{(iv) There are }52-13=39\text{ cards which are not diamonds.}
\displaystyle \therefore P(\text{not a diamond})=\frac{39}{52}=\frac{3}{4}.

\displaystyle \text{(v) There are }52-26=26\text{ cards which are not black.}
\displaystyle \therefore P(\text{not a black card})=\frac{26}{52}=\frac{1}{2}.
\displaystyle \\

\displaystyle \textbf{Question 55: }\text{A bag contains }9\text{ discs of which }4\text{ are red, }3\text{ are blue and }2\text{ are yellow. The}
\displaystyle \text{discs are similar in shape and size. A disc is drawn at random from the bag. Calculate the}
\displaystyle \text{probability that it will be (i) red, (ii) yellow, (iii) blue, (iv) not blue,}
\displaystyle \text{(v) either red or blue.}
\displaystyle \text{Answer:}
\displaystyle \text{Total number of discs}=9.

\displaystyle \text{(i) Number of red discs}=4.
\displaystyle \therefore P(\text{red})=\frac{4}{9}.

\displaystyle \text{(ii) Number of yellow discs}=2.
\displaystyle \therefore P(\text{yellow})=\frac{2}{9}.

\displaystyle \text{(iii) Number of blue discs}=3.
\displaystyle \therefore P(\text{blue})=\frac{3}{9}=\frac{1}{3}.

\displaystyle \text{(iv) Number of discs which are not blue}=4+2=6.
\displaystyle \therefore P(\text{not blue})=\frac{6}{9}=\frac{2}{3}.

\displaystyle \text{(v) Number of discs which are either red or blue}=4+3=7.
\displaystyle \therefore P(\text{either red or blue})=\frac{7}{9}.
\displaystyle \\

\displaystyle \textbf{Question 56: }\text{Two students Anil and Ashima appeared in an examination. The probability}
\displaystyle \text{that Anil will qualify the examination is }0.05\text{ and that Ashima will qualify the}
\displaystyle \text{examination is }0.10.\text{ The probability that both will qualify the examination is }0.02.
\displaystyle \text{Find the probability that:}
\displaystyle \text{(a) Both Anil and Ashima will not qualify the examination.}
\displaystyle \text{(b) At least one of them will not qualify the examination.}
\displaystyle \text{(c) Only one of them will qualify the examination.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }A=\text{event that Anil qualifies and }B=\text{event that Ashima qualifies.}
\displaystyle \text{Given, }P(A)=0.05,\quad P(B)=0.10,\quad P(A\cap B)=0.02.

\displaystyle \text{(a) }P(A\cup B)=P(A)+P(B)-P(A\cap B)
\displaystyle =0.05+0.10-0.02=0.13.
\displaystyle P(\text{both do not qualify})=1-P(A\cup B)
\displaystyle =1-0.13=0.87.
\displaystyle \therefore \text{The probability that both will not qualify is }0.87.

\displaystyle \text{(b) }P(\text{at least one does not qualify})=1-P(A\cap B)
\displaystyle =1-0.02=0.98.
\displaystyle \therefore \text{The probability that at least one will not qualify is }0.98.

\displaystyle \text{(c) }P(\text{only Anil qualifies})=P(A)-P(A\cap B)
\displaystyle =0.05-0.02=0.03.
\displaystyle P(\text{only Ashima qualifies})=P(B)-P(A\cap B)
\displaystyle =0.10-0.02=0.08.
\displaystyle P(\text{only one qualifies})=0.03+0.08=0.11.
\displaystyle \therefore \text{The probability that only one of them will qualify is }0.11.
\displaystyle \\

\displaystyle \textbf{Question 57: }\text{A committee of two persons is selected from two men and two women.}
\displaystyle \text{What is the probability that the committee will have (a) no man? (b) one man?}
\displaystyle \text{(c) two men?}
\displaystyle \text{Answer:}
\displaystyle \text{Total number of ways of selecting a committee of }2\text{ persons}={}^{4}C_2=6.

\displaystyle \text{(a) For no man, both members of the committee must be women.}
\displaystyle \text{Number of favourable committees}={}^{2}C_2=1.
\displaystyle \therefore P(\text{no man})=\frac{1}{6}.

\displaystyle \text{(b) For one man, choose }1\text{ man from }2\text{ and }1\text{ woman from }2.
\displaystyle \text{Number of favourable committees}={}^{2}C_1\times{}^{2}C_1=4.
\displaystyle \therefore P(\text{one man})=\frac{4}{6}=\frac{2}{3}.

\displaystyle \text{(c) For two men, both members of the committee must be men.}
\displaystyle \text{Number of favourable committees}={}^{2}C_2=1.
\displaystyle \therefore P(\text{two men})=\frac{1}{6}.
\displaystyle \\

\displaystyle \textbf{Question 58: }\text{The number lock of a suitcase has }4\text{ wheels, each labelled with ten digits,}
\displaystyle \text{i.e., from }0\text{ to }9.\text{ The lock opens with a sequence of four digits with no repeats. What is}
\displaystyle \text{the probability of a person getting the right sequence to open the suitcase?}
\displaystyle \text{Answer:}
\displaystyle \text{There are }10\text{ digits, from }0\text{ to }9,\text{ and no digit can be repeated.}
\displaystyle \text{Total number of possible sequences}={}^{10}P_4
\displaystyle =10\times9\times8\times7=5040.
\displaystyle \text{There is only one sequence that opens the suitcase.}
\displaystyle \therefore P(\text{getting the right sequence})=\frac{1}{5040}.
\displaystyle \therefore \text{The required probability is }\frac{1}{5040}.
\displaystyle \\

\displaystyle \textbf{Question 59: }\text{If }4\text{-digit numbers greater than }5,000\text{ are randomly formed from the digits}
\displaystyle 0,1,3,5\text{ and }7,\text{ what is the probability of forming a number divisible by }5\text{ when,}
\displaystyle \text{(i) the digits are repeated? (ii) the repetition of digits is not allowed?}
\displaystyle \text{Answer:}

\displaystyle \text{(i) When repetition of digits is allowed:}
\displaystyle \text{For a number greater than }5000,\text{ the thousands digit can be }5\text{ or }7.
\displaystyle \therefore \text{Total number of possible numbers}=2\times5\times5\times5=250.
\displaystyle \text{For the number to be divisible by }5,\text{ the units digit must be }0\text{ or }5.
\displaystyle \therefore \text{Number of favourable numbers}=2\times5\times5\times2=100.
\displaystyle P(\text{number divisible by }5)=\frac{100}{250}=\frac{2}{5}.
\displaystyle \therefore \text{The required probability is }\frac{2}{5}.

\displaystyle \text{(ii) When repetition of digits is not allowed:}
\displaystyle \text{The thousands digit can be }5\text{ or }7.
\displaystyle \text{For each choice, the remaining }3\text{ places can be filled in }{}^{4}P_3\text{ ways.}
\displaystyle \therefore \text{Total number of possible numbers}=2\times{}^{4}P_3=2\times24=48.
\displaystyle \text{If the thousands digit is }5,\text{ the units digit must be }0.
\displaystyle \text{Number of favourable numbers}={}^{3}P_2=6.
\displaystyle \text{If the thousands digit is }7,\text{ the units digit can be }0\text{ or }5.
\displaystyle \text{Number of favourable numbers}=2\times{}^{3}P_2=12.
\displaystyle \therefore \text{Total number of favourable numbers}=6+12=18.
\displaystyle P(\text{number divisible by }5)=\frac{18}{48}=\frac{3}{8}.
\displaystyle \therefore \text{The required probability is }\frac{3}{8}.
\displaystyle \\

\displaystyle \textbf{Question 60: }\text{Three letters are dictated to three persons and an envelope is addressed to}
\displaystyle \text{each of them. The letters are inserted into the envelopes at random so that each}
\displaystyle \text{envelope contains exactly one letter. Find the probability that at least one letter is}
\displaystyle \text{in its proper envelope.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the letters be }L_1,L_2,L_3\text{ and the corresponding envelopes be }E_1,E_2,E_3.
\displaystyle \text{Total number of ways of placing the }3\text{ letters in the }3\text{ envelopes}=3!=6.
\displaystyle \text{There are }2\text{ arrangements in which no letter is in its proper envelope.}
\displaystyle \therefore \text{Number of arrangements with at least one letter in its proper envelope}=6-2=4.
\displaystyle P(\text{at least one letter in its proper envelope})=\frac{4}{6}=\frac{2}{3}.
\displaystyle \therefore \text{The required probability is }\frac{2}{3}.
\displaystyle \\

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