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

\displaystyle \textbf{Question 1: }\text{One card is drawn from a pack of 52 cards. The}
\displaystyle \text{probability that it is the card of a king or spade is}
\displaystyle \text{(a) }\frac{1}{26}\qquad\text{(b) }\frac{3}{26}\qquad\text{(c) }\frac{4}{13}\qquad\text{(d) }\frac{3}{13}
\displaystyle \text{Answer:}
\displaystyle \text{There are }4\text{ kings and }13\text{ spades, with }1\text{ king of spades.}
\displaystyle \therefore P(\text{king or spade})=\frac{4+13-1}{52} =\frac{16}{52}=\frac{4}{13}.
\displaystyle \text{Hence, the correct option is (c).}
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Two dice are thrown together. The probability that}
\displaystyle \text{at least one will show its digit greater than }3\text{ is}
\displaystyle \text{(a) }\frac14\qquad\text{(b) }\frac34\qquad\text{(c) }\frac12\qquad\text{(d) }\frac18
\displaystyle \text{Answer:}
\displaystyle P(\text{both digits}\leq3)=\frac{3}{6}\times\frac{3}{6}=\frac14.
\displaystyle \therefore P(\text{at least one digit}>3)=1-\frac14=\frac34.
\displaystyle \text{Hence, the correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{Two dice are thrown simultaneously. The probability}
\displaystyle \text{of obtaining a total score of }5\text{ is}
\displaystyle \text{(a) }\frac{1}{18}\qquad\text{(b) }\frac{1}{12}\qquad\text{(c) }\frac{1}{9}\qquad\text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle \text{Favourable outcomes are }(1,4),(2,3),(3,2),(4,1).
\displaystyle \therefore P(\text{total score }5)=\frac{4}{36}=\frac{1}{9}.
\displaystyle \text{Hence, the correct option is (c).}
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{Two dice are thrown simultaneously. The probability}
\displaystyle \text{of obtaining total score of seven is}
\displaystyle \text{(a) }\frac{5}{36}\qquad\text{(b) }\frac{6}{36}\qquad\text{(c) }\frac{7}{36}\qquad\text{(d) }\frac{8}{36}
\displaystyle \text{Answer:}
\displaystyle \text{Favourable outcomes are }(1,6),(2,5),(3,4),(4,3),(5,2),(6,1).
\displaystyle \therefore P(\text{total score }7)=\frac{6}{36}=\frac{1}{6}.
\displaystyle \text{Hence, the correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{The probability of getting a total of }10\text{ in a}
\displaystyle \text{single throw of two dice is}
\displaystyle \text{(a) }\frac19\qquad\text{(b) }\frac{1}{12}\qquad\text{(c) }\frac16\qquad\text{(d) }\frac{5}{36}
\displaystyle \text{Answer:}
\displaystyle \text{Favourable outcomes are }(4,6),(5,5),(6,4).
\displaystyle \therefore P(\text{total }10)=\frac{3}{36}=\frac{1}{12}.
\displaystyle \text{Hence, the correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{A card is drawn at random from a pack of }100\text{ cards}
\displaystyle \text{numbered }1\text{ to }100.\text{ The probability of drawing a number which is a square is}
\displaystyle \text{(a) }\frac15\qquad\text{(b) }\frac25\qquad\text{(c) }\frac{1}{10}\qquad\text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle \text{Square numbers are }1^2,2^2,\ldots,10^2,\text{ i.e., }10\text{ numbers.}
\displaystyle \therefore P(\text{square number})=\frac{10}{100}=\frac{1}{10}.
\displaystyle \text{Hence, the correct option is (c).}
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{A bag contains 3 red, 4 white and 5 blue balls.}
\displaystyle \text{All balls are different. Two balls are drawn at random. The probability}
\displaystyle \text{that they are of different colour is}
\displaystyle \text{(a) }\frac{47}{66}\qquad\text{(b) }\frac{10}{33}\qquad\text{(c) }\frac13\qquad\text{(d) }1
\displaystyle \text{Answer:}
\displaystyle \text{Total ways}={}^{12}C_2=66.
\displaystyle \text{Same-colour ways}={}^{3}C_2+{}^{4}C_2+{}^{5}C_2=3+6+10=19.
\displaystyle \therefore P(\text{different colours})=1-\frac{19}{66}=\frac{47}{66}.
\displaystyle \text{Hence, the correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{Two dice are thrown together. The probability that}
\displaystyle \text{neither they show equal digits nor the sum of their digits is }9\text{ will be}
\displaystyle \text{(a) }\frac{13}{15}\qquad\text{(b) }\frac{13}{18}\qquad\text{(c) }\frac19\qquad\text{(d) }\frac89
\displaystyle \text{Answer:}
\displaystyle \text{Equal digits occur in }6\text{ outcomes.}
\displaystyle \text{Sum }9\text{ occurs in }(3,6),(4,5),(5,4),(6,3),\text{ i.e., }4\text{ outcomes.}
\displaystyle \text{These two events have no common outcome.}
\displaystyle \therefore\text{Excluded outcomes}=6+4=10.
\displaystyle \therefore P=\frac{36-10}{36}=\frac{26}{36}=\frac{13}{18}.
\displaystyle \text{Hence, the correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{Four persons are selected at random out of 3 men,}
\displaystyle 2\text{ women and }4\text{ children. The probability that there are exactly}
\displaystyle 2\text{ children in the selection is}
\displaystyle \text{(a) }\frac{11}{21}\qquad\text{(b) }\frac{9}{21}\qquad\text{(c) }\frac{10}{21}\qquad\text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle \text{Total ways}={}^{9}C_4=126.
\displaystyle \text{Favourable ways}={}^{4}C_2\times{}^{5}C_2=6\times10=60.
\displaystyle \therefore P(\text{exactly 2 children})=\frac{60}{126}=\frac{10}{21}.
\displaystyle \text{Hence, the correct option is (c).}
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{The probabilities of happening of two events }A
\displaystyle \text{and }B\text{ are }0.25\text{ and }0.50\text{ respectively. If the probability of}
\displaystyle \text{happening of }A\text{ and }B\text{ together is }0.14,\text{ then probability that}
\displaystyle \text{neither }A\text{ nor }B\text{ happens is}
\displaystyle \text{(a) }0.39\qquad\text{(b) }0.25\qquad\text{(c) }0.11\qquad\text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle P(A\cup B)=P(A)+P(B)-P(A\cap B).
\displaystyle =0.25+0.50-0.14=0.61.
\displaystyle \therefore P(\text{neither }A\text{ nor }B)=1-0.61=0.39.
\displaystyle \text{Hence, the correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{A die is rolled, then the probability that a number}
\displaystyle 1\text{ or }6\text{ may appear is}
\displaystyle \text{(a) }\frac23\qquad\text{(b) }\frac56\qquad\text{(c) }\frac13\qquad\text{(d) }\frac12
\displaystyle \text{Answer:}
\displaystyle \text{Favourable outcomes}=\{1,6\},\quad\text{total outcomes}=6.
\displaystyle \therefore P(1\text{ or }6)=\frac{2}{6}=\frac13.
\displaystyle \text{Hence, the correct option is (c).}
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{Six boys and six girls sit in a row randomly. The}
\displaystyle \text{probability that all girls sit together is}
\displaystyle \text{(a) }\frac{1}{122}\qquad\text{(b) }\frac{1}{112}\qquad\text{(c) }\frac{1}{102}\qquad\text{(d) }\frac{1}{132}
\displaystyle \text{Answer:}
\displaystyle \text{Total arrangements}=12!.
\displaystyle \text{Treat the }6\text{ girls as one unit. Thus, there are }7\text{ units.}
\displaystyle \text{Favourable arrangements}=7!\times6!.
\displaystyle \therefore P(\text{all girls together})=\frac{7!\,6!}{12!}=\frac{1}{132}.
\displaystyle \text{Hence, the correct option is (d).}
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{The probabilities of three mutually exclusive events }A,B
\displaystyle \text{and }C\text{ are given by }\frac23,\frac14\text{ and }\frac16\text{ respectively. The statement}
\displaystyle \text{(a) is true}\qquad\text{(b) is false}\qquad\text{(c) nothing can be said}\qquad\text{(d) could be either}
\displaystyle \text{Answer:}
\displaystyle P(A)+P(B)+P(C)=\frac23+\frac14+\frac16=\frac{13}{12}>1.
\displaystyle \text{For mutually exclusive events, their probability sum cannot exceed }1.
\displaystyle \therefore\text{The statement is false. Hence, the correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{If }\frac{1-3p}{2},\frac{1+4p}{3},\frac{1+p}{6}\text{ are the probabilities}
\displaystyle \text{of three mutually exclusive and exhaustive events, then the set of all}
\displaystyle \text{values of }p\text{ is}
\displaystyle \text{(a) }(0,1)\qquad\text{(b) }\left(-\frac14,\frac13\right)\qquad\text{(c) }\left(0,\frac13\right)\qquad\text{(d) }(0,\infty)
\displaystyle \text{Answer:}
\displaystyle \frac{1-3p}{2}+\frac{1+4p}{3}+\frac{1+p}{6}=1.
\displaystyle \text{This identity holds for every }p,\text{ so each probability must be positive.}
\displaystyle 1-3p>0\Rightarrow p<\frac13,\qquad1+4p>0\Rightarrow p>-\frac14.
\displaystyle 1+p>0\Rightarrow p>-1.
\displaystyle \therefore-\frac14<p<\frac13.
\displaystyle \text{Hence, the correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 15: }\text{A pack of cards contains 4 aces, 4 kings, 4 queens}
\displaystyle \text{and 4 jacks. Two cards are drawn at random. The probability that at least}
\displaystyle \text{one of them is an ace is}
\displaystyle \text{(a) }\frac15\qquad\text{(b) }\frac{3}{16}\qquad\text{(c) }\frac{9}{20}\qquad\text{(d) }\frac19
\displaystyle \text{Answer:}
\displaystyle \text{Total cards}=16,\qquad\text{non-aces}=12.
\displaystyle P(\text{at least one ace})=1-\frac{{}^{12}C_2}{{}^{16}C_2}.
\displaystyle =1-\frac{66}{120}=\frac{54}{120}=\frac{9}{20}.
\displaystyle \text{Hence, the correct option is (c).}
\displaystyle \\

\displaystyle \textbf{Question 16: }\text{If three dice are thrown simultaneously, then the}
\displaystyle \text{probability of getting a score of }5\text{ is}
\displaystyle \text{(a) }\frac{5}{216}\qquad\text{(b) }\frac16\qquad\text{(c) }\frac{1}{36}\qquad\text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle \text{Total outcomes}=6^3=216.
\displaystyle \text{For sum }5,\text{ triples are permutations of }(1,1,3),(1,2,2).
\displaystyle \text{Favourable outcomes}=3+3=6.
\displaystyle \therefore P(\text{score }5)=\frac{6}{216}=\frac{1}{36}.
\displaystyle \text{Hence, the correct option is (c).}
\displaystyle \\

\displaystyle \textbf{Question 17: }\text{One of the two events must occur. If the chance of}
\displaystyle \text{one is }\frac23\text{ of the other, then odds in favour of the other are}
\displaystyle \text{(a) }1:3\qquad\text{(b) }3:1\qquad\text{(c) }2:3\qquad\text{(d) }3:2
\displaystyle \text{Answer:}
\displaystyle \text{Let the chances of the two events be }2x\text{ and }3x.
\displaystyle 2x+3x=1\Rightarrow x=\frac15.
\displaystyle \therefore\text{Probability of the other event}=3x=\frac35.
\displaystyle \text{Odds in favour}=\frac35:\frac25=3:2.
\displaystyle \text{Hence, the correct option is (d).}
\displaystyle \\

\displaystyle \textbf{Question 18: }\text{The probability that a leap year will have}
\displaystyle 53\text{ Fridays or }53\text{ Saturdays is}
\displaystyle \text{(a) }\frac27\qquad\text{(b) }\frac37\qquad\text{(c) }\frac47\qquad\text{(d) }\frac17
\displaystyle \text{Answer:}
\displaystyle 366\text{ days}=52\text{ weeks}+2\text{ days.}
\displaystyle \text{The two extra days must be Thu-Fri, Fri-Sat or Sat-Sun.}
\displaystyle \therefore P(53\text{ Fridays or }53\text{ Saturdays})=\frac37.
\displaystyle \text{Hence, the correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 19: }\text{A person write 4 letters and addresses 4 envelopes.}
\displaystyle \text{If the letters are placed in the envelopes at random, then the}
\displaystyle \text{probability that all letters are not placed in the right envelopes, is}
\displaystyle \text{(a) }\frac14\qquad\text{(b) }\frac{11}{24}\qquad\text{(c) }\frac{15}{24}\qquad\text{(d) }\frac{23}{24}
\displaystyle \text{Answer:}
\displaystyle \text{Total arrangements}=4!=24.
\displaystyle \text{Only one arrangement places all letters in the right envelopes.}
\displaystyle \therefore P(\text{all not correctly placed})=1-\frac1{24}=\frac{23}{24}.
\displaystyle \text{Hence, the correct option is (d).}
\displaystyle \\

\displaystyle \textbf{Question 20: }A\text{ and }B\text{ are two events such that }P(A)=0.25
\displaystyle \text{and }P(B)=0.50.\text{ The probability of both happening together is }0.14.
\displaystyle \text{The probability of both }A\text{ and }B\text{ not happening is}
\displaystyle \text{(a) }0.39\qquad\text{(b) }0.25\qquad\text{(c) }0.11\qquad\text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle P(A\cup B)=0.25+0.50-0.14=0.61.
\displaystyle \therefore P(\overline A\cap\overline B)=1-P(A\cup B)=1-0.61=0.39.
\displaystyle \text{Hence, the correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 21: }\text{If the probability of }A\text{ to fail in an examination}
\displaystyle \text{is }\frac15\text{ and that of }B\text{ is }\frac{3}{10}.\text{ Then, the probability that either}
\displaystyle A\text{ or }B\text{ fails is}
\displaystyle \text{(a) }\frac12\qquad\text{(b) }\frac{11}{25}\qquad\text{(c) }\frac{19}{50}\qquad\text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle P(A\text{ fails, }B\text{ passes})=\frac15\times\frac7{10}=\frac7{50}.
\displaystyle P(A\text{ passes, }B\text{ fails})=\frac45\times\frac3{10}=\frac{12}{50}.
\displaystyle \therefore P(\text{either }A\text{ or }B\text{ fails})=\frac7{50}+\frac{12}{50}=\frac{19}{50}.
\displaystyle \text{Hence, the correct option is (c).}
\displaystyle \\

\displaystyle \textbf{Question 22: }\text{A box contains 10 good articles and 6 defective}
\displaystyle \text{articles. One item is drawn at random. The probability that it is either}
\displaystyle \text{good or has a defect, is}
\displaystyle \text{(a) }\frac{64}{64}\qquad\text{(b) }\frac{49}{64}\qquad\text{(c) }\frac{40}{64}\qquad\text{(d) }\frac{24}{64}
\displaystyle \text{Answer:}
\displaystyle \text{Every article is either good or defective.}
\displaystyle \therefore P(\text{good or defective})=\frac{10+6}{16}=1=\frac{64}{64}.
\displaystyle \text{Hence, the correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 23: }\text{Three integers are chosen at random from the first}
\displaystyle 20\text{ integers. The probability that their product is even is}
\displaystyle \text{(a) }\frac{2}{19}\qquad\text{(b) }\frac{3}{29}\qquad\text{(c) }\frac{17}{19}\qquad\text{(d) }\frac{4}{19}
\displaystyle \text{Answer:}
\displaystyle \text{There are }10\text{ odd and }10\text{ even integers from }1\text{ to }20.
\displaystyle P(\text{product even})=1-P(\text{all three are odd}).
\displaystyle =1-\frac{{}^{10}C_3}{{}^{20}C_3} =1-\frac{120}{1140}=1-\frac{2}{19}=\frac{17}{19}.
\displaystyle \text{Hence, the correct option is (c).}
\displaystyle \\

\displaystyle \textbf{Question 24: }\text{Out of 30 consecutive integers, 2 are chosen at}
\displaystyle \text{random. The probability that their sum is odd, is}
\displaystyle \text{(a) }\frac{14}{29}\qquad\text{(b) }\frac{16}{29}\qquad\text{(c) }\frac{15}{29}\qquad\text{(d) }\frac{10}{29}
\displaystyle \text{Answer:}
\displaystyle \text{There are }15\text{ odd and }15\text{ even integers.}
\displaystyle \text{For an odd sum, one integer must be odd and the other even.}
\displaystyle P(\text{odd sum})=\frac{15\times15}{{}^{30}C_2} =\frac{225}{435}=\frac{15}{29}.
\displaystyle \text{Hence, the correct option is (c).}
\displaystyle \\

\displaystyle \textbf{Question 25: }\text{A bag contains 5 black balls, 4 white balls and 3}
\displaystyle \text{red balls. If a ball is selected randomwise, the probability that it is}
\displaystyle \text{black or red ball is}
\displaystyle \text{(a) }\frac13\qquad\text{(b) }\frac14\qquad\text{(c) }\frac{5}{12}\qquad\text{(d) }\frac23
\displaystyle \text{Answer:}
\displaystyle \text{Total balls}=5+4+3=12.
\displaystyle \text{Black or red balls}=5+3=8.
\displaystyle \therefore P(\text{black or red})=\frac8{12}=\frac23.
\displaystyle \text{Hence, the correct option is (d).}
\displaystyle \\

\displaystyle \textbf{Question 26: }\text{Two dice are thrown simultaneously. The probability}
\displaystyle \text{of getting a pair of aces is}
\displaystyle \text{(a) }\frac{1}{36}\qquad\text{(b) }\frac13\qquad\text{(c) }\frac16\qquad\text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle \text{A pair of aces means both dice show }1.
\displaystyle P(\text{pair of aces})=\frac16\times\frac16=\frac{1}{36}.
\displaystyle \text{Hence, the correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 27: }\text{An urn contains 9 balls two of which are red, three}
\displaystyle \text{blue and four black. Three balls are drawn at random. The probability}
\displaystyle \text{that they are of the same colour is}
\displaystyle \text{(a) }\frac{5}{84}\qquad\text{(b) }\frac39\qquad\text{(c) }\frac37\qquad\text{(d) }\frac7{17}
\displaystyle \text{Answer:}
\displaystyle \text{Total ways}={}^{9}C_3=84.
\displaystyle \text{Same-colour ways}={}^{3}C_3+{}^{4}C_3=1+4=5.
\displaystyle \therefore P(\text{same colour})=\frac{5}{84}.
\displaystyle \text{Hence, the correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 28: }\text{Five persons entered the lift cabin on the ground}
\displaystyle \text{floor of an 8 floor house. Suppose that each of them independently and}
\displaystyle \text{with equal probability can leave the cabin at any floor beginning with}
\displaystyle \text{the first, then the probability of all 5 persons leaving at different floor is}
\displaystyle \text{(a) }\frac{{}^7P_5}{7^5}\qquad\text{(b) }\frac{7^5}{{}^7P_5}\qquad\text{(c) }\frac6{{}^6P_5}\qquad\text{(d) }\frac{{}^5P_5}{5^5}
\displaystyle \text{Answer:}
\displaystyle \text{There are }7\text{ possible floors for each person.}
\displaystyle \text{Total possibilities}=7^5.
\displaystyle \text{Ways of choosing different floors for }5\text{ persons}={}^{7}P_5.
\displaystyle \therefore P=\frac{{}^{7}P_5}{7^5}.
\displaystyle \text{Hence, the correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 29: }\text{A box contains 10 good articles and 6 with defects.}
\displaystyle \text{One item is drawn at random. The probability that it is either good or}
\displaystyle \text{has a defect is}
\displaystyle \text{(a) }\frac{64}{64}\qquad\text{(b) }\frac{49}{64}\qquad\text{(c) }\frac{40}{64}\qquad\text{(d) }\frac{24}{64}
\displaystyle \text{Answer:}
\displaystyle \text{Every article is either good or defective.}
\displaystyle \therefore P(\text{good or defective})=\frac{10+6}{16}=1=\frac{64}{64}.
\displaystyle \text{Hence, the correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 30: }\text{A box contains 6 nails and 10 nuts. Half of the nails}
\displaystyle \text{and half of the nuts are rusted. If one item is chosen at random, the}
\displaystyle \text{probability that it is rusted or is a nail is}
\displaystyle \text{(a) }\frac{3}{16}\qquad\text{(b) }\frac{5}{16}\qquad\text{(c) }\frac{11}{16}\qquad\text{(d) }\frac{14}{16}
\displaystyle \text{Answer:}
\displaystyle \text{Rusted items}=3+5=8,\quad\text{nails}=6,\quad\text{rusted nails}=3.
\displaystyle P(\text{rusted or nail})=\frac{8+6-3}{16}=\frac{11}{16}.
\displaystyle \text{Hence, the correct option is (c).}
\displaystyle \\

\displaystyle \textbf{Question 31: }\text{If }S\text{ is the sample space and }P(A)=\frac13P(B)
\displaystyle \text{and }S=A\cup B,\text{ where }A\text{ and }B\text{ are two mutually exclusive events,}
\displaystyle \text{then }P(A)=
\displaystyle \text{(a) }\frac14\qquad\text{(b) }\frac12\qquad\text{(c) }\frac34\qquad\text{(d) }\frac38
\displaystyle \text{Answer:}
\displaystyle P(A)+P(B)=1.
\displaystyle P(A)=\frac13P(B)\Rightarrow P(B)=3P(A).
\displaystyle \therefore P(A)+3P(A)=1\Rightarrow P(A)=\frac14.
\displaystyle \text{Hence, the correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 32: }\text{One mapping is selected at random from all the mappings}
\displaystyle \text{of the set }A=\{1,2,3,\ldots,n\}\text{ into itself. The probability that the}
\displaystyle \text{mapping selected is one to one is}
\displaystyle \text{(a) }\frac{1}{n^n}\qquad\text{(b) }\frac{n!}{n^n}\qquad\text{(c) }\frac{(n-1)!}{n^{n-1}}\qquad\text{(d) None of these}
\displaystyle \text{Answer:}
\displaystyle \text{Total mappings}=n^n,\qquad\text{one-one mappings}=n!.
\displaystyle \therefore P=\frac{n!}{n^n} =\frac{n(n-1)!}{n\cdot n^{n-1}} =\frac{(n-1)!}{n^{n-1}}.
\displaystyle \text{Hence, the book gives the correct option as (c).}
\displaystyle \\

\displaystyle \textbf{Question 33: }\text{If }A,B,C\text{ are three mutually exclusive and exhaustive}
\displaystyle \text{events of an experiment such that }3P(A)=2P(B)=P(C),\text{ then }P(A)
\displaystyle \text{is equal to}
\displaystyle \text{(a) }\frac{1}{11}\qquad\text{(b) }\frac{2}{11}\qquad\text{(c) }\frac{5}{11}\qquad\text{(d) }\frac{6}{11}
\displaystyle \text{Answer:}
\displaystyle \text{Let }3P(A)=2P(B)=P(C)=k.
\displaystyle P(A)=\frac{k}{3},\quad P(B)=\frac{k}{2},\quad P(C)=k.
\displaystyle \frac{k}{3}+\frac{k}{2}+k=1\Rightarrow\frac{11k}{6}=1\Rightarrow k=\frac6{11}.
\displaystyle \therefore P(A)=\frac{k}{3}=\frac{2}{11}.
\displaystyle \text{Hence, the correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 34: }\text{If }A\text{ and }B\text{ are mutually exclusive events then}
\displaystyle \text{(a) }P(A)\leq P(\overline B)\qquad\text{(b) }P(A)\geq P(\overline B)
\displaystyle \text{(c) }P(A)<P(\overline B)\qquad\text{(d) None of these}
\displaystyle \text{Answer:}
\displaystyle A\cap B=\phi\Rightarrow A\subseteq\overline B.
\displaystyle \therefore P(A)\leq P(\overline B).
\displaystyle \text{Hence, the correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 35: }\text{If }P(A\cup B)=P(A\cap B)\text{ for any two events}
\displaystyle A\text{ and }B,\text{ then}
\displaystyle \text{(a) }P(A)=P(B)\qquad\text{(b) }P(A)>P(B)
\displaystyle \text{(c) }P(A)<P(B)\qquad\text{(d) None of these}
\displaystyle \text{Answer:}
\displaystyle P(A\cup B)=P(A)+P(B)-P(A\cap B).
\displaystyle \therefore P(A)+P(B)=2P(A\cap B).
\displaystyle P(A\cap B)\leq P(A),P(B)\Rightarrow P(A)=P(B)=P(A\cap B).
\displaystyle \therefore P(A)=P(B).
\displaystyle \text{Hence, the correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 36: }\text{Three numbers are chosen from }1\text{ to }20.\text{ The}
\displaystyle \text{probability that they are not consecutive is}
\displaystyle \text{(a) }\frac{186}{190}\qquad\text{(b) }\frac{187}{190}\qquad\text{(c) }\frac{188}{190}\qquad\text{(d) }\frac{18}{{}^{20}C_3}
\displaystyle \text{Answer:}
\displaystyle \text{Total selections}={}^{20}C_3=1140.
\displaystyle \text{Consecutive triples}=(1,2,3),(2,3,4),\ldots,(18,19,20).
\displaystyle \therefore\text{Number of consecutive triples}=18.
\displaystyle P(\text{not consecutive})=1-\frac{18}{1140} =1-\frac{3}{190}=\frac{187}{190}.
\displaystyle \text{Hence, the correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 37: }6\text{ boys and }6\text{ girls sit in a row at random. The}
\displaystyle \text{probability that all the girls sit together is}
\displaystyle \text{(a) }\frac{1}{432}\qquad\text{(b) }\frac{12}{431}\qquad\text{(c) }\frac{1}{132}\qquad\text{(d) None of these}
\displaystyle \text{Answer:}
\displaystyle \text{Total arrangements}=12!.
\displaystyle \text{Treating }6\text{ girls as one unit, favourable arrangements}=7!\times6!.
\displaystyle \therefore P(\text{all girls together})=\frac{7!\,6!}{12!}=\frac{1}{132}.
\displaystyle \text{Hence, the correct option is (c).}
\displaystyle \\

\displaystyle \textbf{Question 38: }\text{Without repetition of the numbers, four digit numbers}
\displaystyle \text{are formed with the numbers }0,2,3,5.\text{ The probability of such a}
\displaystyle \text{number divisible by }5\text{ is}
\displaystyle \text{(a) }\frac15\qquad\text{(b) }\frac45\qquad\text{(c) }\frac{1}{30}\qquad\text{(d) }\frac59
\displaystyle \text{Answer:}
\displaystyle \text{Total four digit numbers}=4!-3!=18.
\displaystyle \text{Ending in }0:\ 3!=6,\qquad\text{ending in }5:\ 3!-2!=4.
\displaystyle \therefore P(\text{divisible by }5)=\frac{6+4}{18}=\frac59.
\displaystyle \text{Hence, the correct option is (d).}
\displaystyle \\

\displaystyle \textbf{Question 39: }\text{If the probability for }A\text{ to fail in an examination}
\displaystyle \text{is }0.2\text{ and that for }B\text{ is }0.3,\text{ then the probability that either }A\text{ or}
\displaystyle B\text{ fails is}
\displaystyle \text{(a) }>0.5\qquad\text{(b) }0.5\qquad\text{(c) }<0.5\qquad\text{(d) }0
\displaystyle \text{Answer:}
\displaystyle P(\text{either fails})\leq P(A)+P(B)=0.2+0.3=0.5.
\displaystyle \text{Since both can fail together, }P(\text{either fails})<0.5.
\displaystyle \text{Hence, the correct option is (c).}
\displaystyle \\

\displaystyle \textbf{Question 40: }\text{Three digit numbers are formed using the digits}
\displaystyle 0,2,4,6,8.\text{ A number is chosen at random out of these numbers. What is the}
\displaystyle \text{probability that this number has the same digits?}
\displaystyle \text{(a) }\frac{1}{16}\qquad\text{(b) }\frac{16}{25}\qquad\text{(c) }\frac{1}{645}\qquad\text{(d) }\frac{1}{25}
\displaystyle \text{Answer:}
\displaystyle \text{Total three digit numbers}=4\times5\times5=100.
\displaystyle \text{Numbers with same digits are }222,444,666,888,\text{ i.e., }4.
\displaystyle \therefore P(\text{same digits})=\frac{4}{100}=\frac{1}{25}.
\displaystyle \text{Hence, the correct option is (d).}
\displaystyle \\

\displaystyle \text{VERY SHORT ANSWER QUESTIONS}

\displaystyle \textbf{Question 1: }\text{Three numbers are chosen at random from numbers}
\displaystyle 1\text{ to }30.\text{ Write the probability that the chosen numbers are consecutive.}
\displaystyle \text{Answer:}
\displaystyle \text{Total number of ways of choosing 3 numbers from 30} ={}^{30}C_3.
\displaystyle \text{The consecutive triples are }(1,2,3),(2,3,4),\ldots,(28,29,30).
\displaystyle \therefore\text{Number of favourable outcomes}=28.
\displaystyle \therefore P(\text{chosen numbers are consecutive}) =\frac{28}{{}^{30}C_3}.
\displaystyle =\frac{28}{4060}=\frac{1}{145}.
\displaystyle \\

\displaystyle \textbf{Question 2: }n\,(>3)\text{ persons are sitting in a row. Two of them}
\displaystyle \text{are selected. Write the probability that they are together.}
\displaystyle \text{Answer:}
\displaystyle \text{Total number of ways of selecting 2 persons from }n ={}^{n}C_2.
\displaystyle \text{There are }n-1\text{ adjacent pairs in a row of }n\text{ persons.}
\displaystyle \therefore\text{Number of favourable outcomes}=n-1.
\displaystyle \therefore P(\text{the selected persons are together}) =\frac{n-1}{{}^{n}C_2}.
\displaystyle =\frac{n-1}{\frac{n(n-1)}{2}}=\frac{2}{n}.
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{A single letter is selected at random from the word}
\displaystyle \text{'PROBABILITY'. What is the probability that it is a vowel?}
\displaystyle \text{Answer:}
\displaystyle \text{The word PROBABILITY has }11\text{ letters.}
\displaystyle \text{The vowels are O, A, I, I, i.e., }4\text{ vowels.}
\displaystyle \therefore P(\text{vowel})=\frac{4}{11}.
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{What is the probability that a leap year will have}
\displaystyle 53\text{ Fridays or }53\text{ Saturdays?}
\displaystyle \text{Answer:}
\displaystyle \text{A leap year has }366\text{ days}=52\text{ weeks}+2\text{ days.}
\displaystyle \text{The two extra days are consecutive days of the week.}
\displaystyle \text{For }53\text{ Fridays or }53\text{ Saturdays, the extra days must be}
\displaystyle \text{Thu-Fri, Fri-Sat or Sat-Sun.}
\displaystyle \therefore\text{Favourable cases}=3,\quad\text{total cases}=7.
\displaystyle \therefore P(53\text{ Fridays or }53\text{ Saturdays})=\frac{3}{7}.
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{Three dice are thrown simultaneously. What is the}
\displaystyle \text{probability of getting }15\text{ as the sum?}
\displaystyle \text{Answer:}
\displaystyle \text{Total number of outcomes}=6^3=216.
\displaystyle \text{Favourable triples are permutations of }(3,6,6),(4,5,6),(5,5,5).
\displaystyle \therefore\text{Number of favourable outcomes}=3+6+1=10.
\displaystyle \therefore P(\text{sum }15)=\frac{10}{216}=\frac{5}{108}.
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{If the letters of the word 'MISSISSIPPI' are written}
\displaystyle \text{down at random in a row, what is the probability that four S's come together?}
\displaystyle \text{Answer:}
\displaystyle \text{MISSISSIPPI has }11\text{ letters: M}^{1},\text{ I}^{4},\text{ S}^{4},\text{ P}^{2}.
\displaystyle \text{Total arrangements}=\frac{11!}{4!\,4!\,2!}.
\displaystyle \text{Treat the four S's as one unit. Then there are }8\text{ units.}
\displaystyle \text{Favourable arrangements}=\frac{8!}{4!\,2!}.
\displaystyle \therefore P=\frac{8!}{4!\,2!}\times\frac{4!\,4!\,2!}{11!}.
\displaystyle =\frac{8!\,4!}{11!}=\frac{1}{165}.
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{What is the probability that the 13th day of a}
\displaystyle \text{randomly chosen month is Friday?}
\displaystyle \text{Answer:}
\displaystyle \text{There are }12\text{ months and }7\text{ possible days for the 13th.}
\displaystyle \therefore\text{Total possible cases}=12\times7=84.
\displaystyle \text{For a chosen month, the favourable case is that its 13th is Friday.}
\displaystyle \therefore P=\frac{1}{84}.
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{Three of the six vertices of a regular hexagon are}
\displaystyle \text{chosen at random. What is the probability that the triangle with these}
\displaystyle \text{vertices is equilateral.}
\displaystyle \text{Answer:}
\displaystyle \text{Total number of triangles}={}^{6}C_3=20.
\displaystyle \text{There are }2\text{ equilateral triangles formed by alternate vertices.}
\displaystyle \therefore P(\text{equilateral triangle})=\frac{2}{20}=\frac{1}{10}.
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{If }E_1\text{ and }E_2\text{ are independent events, write the}
\displaystyle \text{value of }P\left((E_1\cup E_2)\cap(\overline{E_1}\cap\overline{E_2})\right).
\displaystyle \text{Answer:}
\displaystyle \overline{E_1}\cap\overline{E_2} =\overline{(E_1\cup E_2)}.
\displaystyle \therefore (E_1\cup E_2)\cap(\overline{E_1}\cap\overline{E_2}) =(E_1\cup E_2)\cap\overline{(E_1\cup E_2)}=\phi.
\displaystyle \therefore P\left((E_1\cup E_2)\cap (\overline{E_1}\cap\overline{E_2})\right)=0.
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{If }A\text{ and }B\text{ are two independent events such that}
\displaystyle P(A\cap B)=\frac{1}{6}\text{ and }P(\overline A\cap\overline B)=\frac{1}{3}, \text{ then write the values of }
\displaystyle P(A)\text{ and }P(B).
\displaystyle \text{Answer:}
\displaystyle \text{Let }P(A)=p\text{ and }P(B)=q.
\displaystyle \text{Since }A\text{ and }B\text{ are independent, }pq=\frac{1}{6}.
\displaystyle (1-p)(1-q)=\frac{1}{3}.
\displaystyle 1-p-q+pq=\frac{1}{3}.
\displaystyle \therefore p+q=1+\frac{1}{6}-\frac{1}{3}=\frac{5}{6}.
\displaystyle \therefore p,q\text{ are roots of }x^2-\frac{5}{6}x+\frac{1}{6}=0.
\displaystyle 6x^2-5x+1=0\Rightarrow(3x-1)(2x-1)=0.
\displaystyle \therefore \{P(A),P(B)\}=\left\{\frac12,\frac13\right\}.
\displaystyle \text{Hence, }P(A)=\frac12,\ P(B)=\frac13\text{ or vice versa.}
\displaystyle \\

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