\displaystyle \textbf{Question 1. }\text{One card is drawn from a pack of }52\text{ cards. The probability that it is the} \\ \text{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{Number of kings}=4
\displaystyle \text{Number of spades}=13
\displaystyle \text{One card is both king and spade, i.e., king of spades}
\displaystyle \therefore \text{Favourable outcomes}=4+13-1=16
\displaystyle \therefore P(\text{king or spade})=\frac{16}{52}=\frac{4}{13}
\displaystyle \therefore\text{Correct option is (c).}
\\

\displaystyle \textbf{Question 2. }\text{Two dice are thrown together. The probability that at least one will show} \\ \text{its digit greater than }3\text{ is}
\displaystyle \text{(a) }\frac{1}{4}\qquad \text{(b) }\frac{3}{4}\qquad \text{(c) }\frac{1}{2}\qquad \text{(d) }\frac{1}{8}
\displaystyle \text{Answer:}
\displaystyle \text{Total outcomes}=36
\displaystyle \text{At least one digit greater than }3=1-\text{neither digit greater than }3
\displaystyle \text{Neither digit greater than }3\Rightarrow \text{both digits are from }1,2,3
\displaystyle \therefore \text{Number of such outcomes}=3\times3=9
\displaystyle \therefore P=1-\frac{9}{36}=\frac{27}{36}=\frac{3}{4}
\displaystyle \therefore\text{Correct option is (b).}
\\

\displaystyle \textbf{Question 3. }\text{Two dice are thrown simultaneously. The probability of obtaining a total} \\ \text{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{Total outcomes}=36
\displaystyle \text{Outcomes with sum }5\text{ are }(1,4),(2,3),(3,2),(4,1)
\displaystyle \therefore \text{Favourable outcomes}=4
\displaystyle \therefore P(\text{sum }5)=\frac{4}{36}=\frac{1}{9}
\displaystyle \therefore\text{Correct option is (c).}
\\

\displaystyle \textbf{Question 4. }\text{Two dice are thrown simultaneously. The probability of obtaining total} \\ \text{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{Total outcomes}=36
\displaystyle \text{Outcomes with sum }7\text{ are }(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)
\displaystyle \therefore \text{Favourable outcomes}=6
\displaystyle \therefore P(\text{sum }7)=\frac{6}{36}
\displaystyle \therefore\text{Correct option is (b).}
\\

\displaystyle \textbf{Question 5. }\text{The probability of getting a total of }10\text{ in a single throw of two dice is}
\displaystyle \text{(a) }\frac{1}{9}\qquad \text{(b) }\frac{1}{12}\qquad \text{(c) }\frac{1}{6}\qquad \text{(d) }\frac{5}{36}
\displaystyle \text{Answer:}
\displaystyle \text{Total outcomes}=36
\displaystyle \text{Outcomes with sum }10\text{ are }(4,6),(5,5),(6,4)
\displaystyle \therefore \text{Favourable outcomes}=3
\displaystyle \therefore P(\text{sum }10)=\frac{3}{36}=\frac{1}{12}
\displaystyle \therefore\text{Correct option is (b).}
\\

\displaystyle \textbf{Question 6. }\text{A card is drawn at random from a pack of }100\text{ cards numbered } \\ 1\text{ to }100.\text{ The probability of drawing a number which is a square is}
\displaystyle \text{(a) }\frac{1}{5}\qquad \text{(b) }\frac{2}{5}\qquad \text{(c) }\frac{1}{10}\qquad \text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle \text{Square numbers from }1\text{ to }100\text{ are }1^2,2^2,3^2,\ldots,10^2
\displaystyle \therefore \text{Number of favourable outcomes}=10
\displaystyle \text{Total outcomes}=100
\displaystyle \therefore P(\text{square number})=\frac{10}{100}=\frac{1}{10}
\displaystyle \therefore\text{Correct option is (c).}
\\

\displaystyle \textbf{Question 7. }\text{A bag contains }3\text{ red, }4\text{ white and }5\text{ blue balls. All balls are } \\ \text{different. Two balls are drawn at random. The probability that they are of} \\ \text{different colour is}
\displaystyle \text{(a) }\frac{47}{66}\qquad \text{(b) }\frac{10}{33}\qquad \text{(c) }\frac{1}{3}\qquad \text{(d) }1
\displaystyle \text{Answer:}
\displaystyle \text{Total balls}=3+4+5=12
\displaystyle \text{Total ways of drawing }2\text{ balls}={}^{12}C_2=66
\displaystyle \text{Favourable ways}=3\times4+4\times5+5\times3
\displaystyle =12+20+15=47
\displaystyle \therefore P(\text{different colours})=\frac{47}{66}
\displaystyle \therefore\text{Correct option is (a).}
\\

\displaystyle \textbf{Question 8. }\text{Two dice are thrown together. The probability that neither they show equal} \\ \text{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) }\frac{1}{9}\qquad \text{(d) }\frac{8}{9}
\displaystyle \text{Answer:}
\displaystyle \text{Total outcomes}=36
\displaystyle \text{Equal digit outcomes are }(1,1),(2,2),(3,3),(4,4),(5,5),(6,6)
\displaystyle \therefore \text{Number of equal digit outcomes}=6
\displaystyle \text{Outcomes with sum }9\text{ are }(3,6),(4,5),(5,4),(6,3)
\displaystyle \therefore \text{Number of outcomes with sum }9=4
\displaystyle \text{There is no common outcome in the above two cases}
\displaystyle \therefore \text{Number of outcomes to be excluded}=6+4=10
\displaystyle \therefore \text{Favourable outcomes}=36-10=26
\displaystyle \therefore P=\frac{26}{36}=\frac{13}{18}
\displaystyle \therefore\text{Correct option is (b).}
\\

\displaystyle \textbf{Question 9. }\text{Four persons are selected at random out of }3\text{ men, }2\text{ women and} \\ 4\text{ children. The probability that there are exactly }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 persons}=3+2+4=9
\displaystyle \text{Total ways of selecting }4\text{ persons}={}^{9}C_4=126
\displaystyle \text{For exactly }2\text{ children, select }2\text{ children from }4\text{ and }2\text{ others from }5
\displaystyle \text{Favourable ways}={}^{4}C_2\times{}^{5}C_2=6\times10=60
\displaystyle \therefore P=\frac{60}{126}=\frac{10}{21}
\displaystyle \therefore\text{Correct option is (c).}
\\

\displaystyle \textbf{Question 10. }\text{The probabilities of happening of two events }A\text{ and }B\text{ are }0.25\text{ and } \\ 0.50\text{ respectively. If the probability of happening of }A\text{ and }B\text{ together is }0.14,\text{ then} \\ \text{probability that 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)=0.25,\quad P(B)=0.50,\quad P(A\cap B)=0.14
\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-P(A\cup B)
\displaystyle =1-0.61=0.39
\displaystyle \therefore\text{Correct option is (a).}
\\

\displaystyle \textbf{Question 11. }\text{A die is rolled, then the probability that a number }1\text{ or }6\text{ may appear is}
\displaystyle \text{(a) }\frac{2}{3}\qquad \text{(b) }\frac{5}{6}\qquad \text{(c) }\frac{1}{3}\qquad \text{(d) }\frac{1}{2}
\displaystyle \text{Answer:}
\displaystyle \text{Total outcomes}=6
\displaystyle \text{Favourable outcomes are }1\text{ and }6
\displaystyle \therefore \text{Number of favourable outcomes}=2
\displaystyle \therefore P=\frac{2}{6}=\frac{1}{3}
\displaystyle \therefore\text{Correct option is (c).}
\\

\displaystyle \textbf{Question 12. }\text{Six boys and six girls sit in a row randomly. The probability that all girls sit} \\ \text{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 of }12\text{ persons}=12!
\displaystyle \text{Treat all }6\text{ girls as one block}
\displaystyle \text{Then }6\text{ boys and }1\text{ girls' block give }7\text{ objects}
\displaystyle \text{These can be arranged in }7!\text{ ways}
\displaystyle \text{The }6\text{ girls can be arranged among themselves in }6!\text{ ways}
\displaystyle \therefore \text{Favourable arrangements}=7!\times6!
\displaystyle \therefore P=\frac{7!\times6!}{12!}
\displaystyle =\frac{1}{132}
\displaystyle \therefore\text{Correct option is (d).}
\\

\displaystyle \textbf{Question 13. }\text{The probabilities of three mutually exclusive events }A,B\text{ and }C \\ \text{ are given by }\frac{2}{3},\frac{1}{4}\text{ and }\frac{1}{6}\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 \text{For mutually exclusive events,}
\displaystyle P(A\cup B\cup C)=P(A)+P(B)+P(C)
\displaystyle =\frac{2}{3}+\frac{1}{4}+\frac{1}{6}
\displaystyle =\frac{8+3+2}{12}=\frac{13}{12}
\displaystyle \text{But probability cannot be greater than }1
\displaystyle \therefore \text{The given statement is false}
\displaystyle \therefore\text{Correct option is (b).}
\\

\displaystyle \textbf{Question 14. }\text{If }\frac{1-3p}{2},\frac{1+4p}{3},\frac{1+p}{6}\text{ are the probabilities of} \\ \text{three mutually exclusive and exhaustive events, then the set of all values of }p\text{ is}
\displaystyle \text{(a) }(0,1)\qquad \text{(b) }\left(-\frac{1}{4},\frac{1}{3}\right)\qquad \text{(c) }\left(0,\frac{1}{3}\right)\qquad \text{(d) }(0,\infty)
\displaystyle \text{Answer:}
\displaystyle \text{Each probability must be positive.}
\displaystyle \frac{1-3p}{2}>0\Rightarrow p<\frac{1}{3}
\displaystyle \frac{1+4p}{3}>0\Rightarrow p>-\frac{1}{4}
\displaystyle \frac{1+p}{6}>0\Rightarrow p>-1
\displaystyle \therefore -\frac{1}{4}<p<\frac{1}{3}
\displaystyle \text{Also, }\frac{1-3p}{2}+\frac{1+4p}{3}+\frac{1+p}{6}=1
\displaystyle \text{Hence, the required set is }\left(-\frac{1}{4},\frac{1}{3}\right)
\displaystyle \therefore\text{Correct option is (b).}
\\

\displaystyle \textbf{Question 15. }\text{A pack of cards contains }4\text{ aces, }4\text{ kings, }4\text{ queens and }4\text{ jacks. Two} \\ \text{cards are drawn at random. The probability that at least one of them is an ace is}
\displaystyle \text{(a) }\frac{1}{5}\qquad \text{(b) }\frac{3}{16}\qquad \text{(c) }\frac{9}{20}\qquad \text{(d) }\frac{1}{9}
\displaystyle \text{Answer:}
\displaystyle \text{Total cards}=16
\displaystyle \text{Total ways of drawing }2\text{ cards}={}^{16}C_2=120
\displaystyle \text{At least one ace}=1-\text{no ace}
\displaystyle \text{Number of non-ace cards}=12
\displaystyle \therefore P(\text{no ace})=\frac{{}^{12}C_2}{{}^{16}C_2}=\frac{66}{120}
\displaystyle \therefore P(\text{at least one ace})=1-\frac{66}{120}=\frac{54}{120}=\frac{9}{20}
\displaystyle \therefore\text{Correct option is (c).}
\\

\displaystyle \textbf{Question 16. }\text{If three dice are thrown simultaneously, then the probability of getting a score} \\ \text{of }5\text{ is}
\displaystyle \text{(a) }\frac{5}{216}\qquad \text{(b) }\frac{1}{6}\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{ possible triples are }(1,1,3),(1,3,1),(3,1,1)
\displaystyle \text{and }(1,2,2),(2,1,2),(2,2,1)
\displaystyle \therefore \text{Favourable outcomes}=6
\displaystyle \therefore P(\text{sum }5)=\frac{6}{216}=\frac{1}{36}
\displaystyle \therefore\text{Correct option is (c).}
\\

\displaystyle \textbf{Question 17. }\text{One of the two events must occur. If the chance of one is }\frac{2}{3}\text{ of the} \\ \text{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 probabilities of the two events be }P(A)\text{ and }P(B)
\displaystyle \text{Since one of the two events must occur,}
\displaystyle P(A)+P(B)=1
\displaystyle \text{Given }P(A)=\frac{2}{3}P(B)
\displaystyle \frac{2}{3}P(B)+P(B)=1
\displaystyle \frac{5}{3}P(B)=1
\displaystyle P(B)=\frac{3}{5}
\displaystyle \therefore P(A)=\frac{2}{5}
\displaystyle \text{Odds in favour of the other event }B=P(B):P(A)
\displaystyle =\frac{3}{5}:\frac{2}{5}=3:2
\displaystyle \therefore\text{Correct option is (d).}
\\

\displaystyle \textbf{Question 18. }\text{The probability that a leap year will have }53\text{ Fridays or }53\text{ Saturdays is}
\displaystyle \text{(a) }\frac{2}{7}\qquad \text{(b) }\frac{3}{7}\qquad \text{(c) }\frac{4}{7}\qquad \text{(d) }\frac{1}{7}
\displaystyle \text{Answer:}
\displaystyle \text{A leap year has }366\text{ days }=52\text{ weeks }+2\text{ days}
\displaystyle \text{The two extra days can be }(Mon,Tue),(Tue,Wed),\ldots,(Sun,Mon)
\displaystyle \text{For }53\text{ Fridays or }53\text{ Saturdays, the favourable pairs are}
\displaystyle (Thu,Fri),(Fri,Sat),(Sat,Sun)
\displaystyle \therefore \text{Favourable outcomes}=3
\displaystyle \text{Total outcomes}=7
\displaystyle \therefore P=\frac{3}{7}
\displaystyle \therefore\text{Correct option is (b).}
\\

\displaystyle \textbf{Corrected Question 19. }\text{A person writes }4\text{ letters and addresses }4\text{ envelopes. If the} \\ \text{letters are placed in the envelopes at random, then the probability that at least one letter} \\ \text{is not placed in the right envelope, is}
\displaystyle \text{(a) }\frac{1}{4}\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 of }4\text{ letters in }4\text{ envelopes}=4!=24
\displaystyle \text{Only one arrangement has all letters in the correct envelopes}
\displaystyle \therefore P(\text{all letters in correct envelopes})=\frac{1}{24}
\displaystyle \therefore P(\text{at least one letter not in correct envelope})=1-\frac{1}{24}
\displaystyle =\frac{23}{24}
\displaystyle \therefore\text{Correct option is (d).}
\\

\displaystyle \textbf{Question 20. }\text{A and B are two events such that }P(A)=0.25\text{ and } \\ P(B)=0.50. \text{ The probability of both happening together is }0.14.\text{ The probability of} \\ \text{both A and B 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)=P(A)+P(B)-P(A\cap B)
\displaystyle =0.25+0.50-0.14
\displaystyle =0.61
\displaystyle \therefore P(\text{neither }A\text{ nor }B)=1-0.61=0.39
\displaystyle \therefore\text{Correct option is (a).}
\\

\displaystyle \textbf{Corrected Question 21. }\text{If the probability of }A\text{ to fail in an examination is }\frac{1}{5}\text{ and that of } \\ B\text{ is }\frac{3}{10}.\text{ If both fail independently, then the probability that either }A\text{ or }B\text{ fails is}
\displaystyle \text{(a) }\frac{1}{2}\qquad \text{(b) }\frac{11}{25}\qquad \text{(c) }\frac{19}{50}\qquad \text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle P(A)=\frac{1}{5},\qquad P(B)=\frac{3}{10}
\displaystyle \text{Probability that exactly one fails is}
\displaystyle P(A\cap B')+P(A'\cap B)
\displaystyle =P(A)\,P(B')+P(A')\,P(B)
\displaystyle =\frac{1}{5}\times\frac{7}{10}+\frac{4}{5}\times\frac{3}{10}
\displaystyle =\frac{7}{50}+\frac{12}{50}
\displaystyle =\frac{19}{50}
\displaystyle \therefore\text{Correct option is (c).}
\\

\displaystyle \textbf{Question 22. }\text{A box contains }10\text{ good articles and }6\text{ defective articles. One item is} \\ \text{drawn at random. The probability that it is either 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})=1
\displaystyle =\frac{64}{64}
\displaystyle \therefore\text{Correct option is (a).}
\\

\displaystyle \textbf{Question 23. }\text{Three integers are chosen at random from the first }20\text{ integers. The} \\ \text{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{Total ways of choosing }3\text{ integers from }20={}^{20}C_3
\displaystyle \text{Product is odd only when all three integers are odd}
\displaystyle \text{There are }10\text{ odd integers from }1\text{ to }20
\displaystyle P(\text{product is odd})=\frac{{}^{10}C_3}{{}^{20}C_3}
\displaystyle \therefore P(\text{product is even})=1-\frac{{}^{10}C_3}{{}^{20}C_3}
\displaystyle =1-\frac{120}{1140}
\displaystyle =1-\frac{2}{19}=\frac{17}{19}
\displaystyle \therefore\text{Correct option is (c).}
\\

\displaystyle \textbf{Question 24. }\text{Out of }30\text{ consecutive integers, }2\text{ are chosen at random. The} \\ \text{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{In }30\text{ consecutive integers, there are }15\text{ odd and }15\text{ even integers}
\displaystyle \text{Sum is odd when one number is odd and the other is even}
\displaystyle \text{Favourable ways}=15\times15=225
\displaystyle \text{Total ways}={}^{30}C_2=435
\displaystyle \therefore P=\frac{225}{435}=\frac{15}{29}
\displaystyle \therefore\text{Correct option is (c).}
\\

\displaystyle \textbf{Question 25. }\text{A bag contains }5\text{ black balls, }4\text{ white balls and }3\text{ red balls.} \\ \text{If a ball is selected randomwise, the probability that it is black or red ball is}
\displaystyle \text{(a) }\frac{1}{3}\qquad \text{(b) }\frac{1}{4}\qquad \text{(c) }\frac{5}{12}\qquad \text{(d) }\frac{2}{3}
\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})=\frac{8}{12}=\frac{2}{3}
\displaystyle \therefore\text{Correct option is (d).}
\\

\displaystyle \textbf{Question 26. }\text{Two dice are thrown simultaneously. The probability of getting a pair of} \\ \text{aces is}
\displaystyle \text{(a) }\frac{1}{36}\qquad \text{(b) }\frac{1}{3}\qquad \text{(c) }\frac{1}{6}\qquad \text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle \text{In dice, ace means the number }1
\displaystyle \text{A pair of aces means }(1,1)
\displaystyle \text{Total outcomes}=36
\displaystyle \text{Favourable outcome}=1
\displaystyle \therefore P=\frac{1}{36}
\displaystyle \therefore\text{Correct option is (a).}
\\

\displaystyle \textbf{Question 27. }\text{An urn contains }9\text{ balls, two of which are red, three blue and four black.} \\ \text{Three balls are drawn at random. The probability that they are of the same colour is}
\displaystyle \text{(a) }\frac{5}{84}\qquad \text{(b) }\frac{3}{9}\qquad \text{(c) }\frac{3}{7}\qquad \text{(d) }\frac{7}{17}
\displaystyle \text{Answer:}
\displaystyle \text{Total ways of drawing }3\text{ balls}={}^{9}C_3=84
\displaystyle \text{For same colour, possible selections are }3\text{ blue or }3\text{ black}
\displaystyle \text{Favourable ways}={}^{3}C_3+{}^{4}C_3=1+4=5
\displaystyle \therefore P=\frac{5}{84}
\displaystyle \therefore\text{Correct option is (a).}
\\

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

\displaystyle \textbf{Question 29. }\text{A box contains }10\text{ good articles and }6\text{ with defects. One item is} \\ \text{drawn at random. The probability that it is either 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})=1
\displaystyle =\frac{64}{64}
\displaystyle \therefore\text{Correct option is (a).}
\\

\displaystyle \textbf{Question 30. }\text{A box contains }6\text{ nails and }10\text{ nuts. Half of the nails and half of} \\ \text{the nuts are rusted. If one item is chosen at random, the probability that it} \\ \text{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{Total items}=6+10=16
\displaystyle \text{Rusted nails}=3,\qquad \text{rusted nuts}=5
\displaystyle \therefore \text{Total rusted items}=3+5=8
\displaystyle \text{Number of nails}=6
\displaystyle \text{Rusted nails are common in both groups}=3
\displaystyle \therefore \text{Favourable items}=8+6-3=11
\displaystyle \therefore P=\frac{11}{16}
\displaystyle \therefore\text{Correct option is (c).}
\\

\displaystyle \textbf{Question 31. }\text{If }S\text{ is the sample space and }P(A)=\frac{1}{3}P(B)\text{ and }S=A\cup B,\text{ where }A\text{ and }B\text{ are two mutually exclusive events, then }P(A)=
\displaystyle \text{(a) }\frac{1}{4}\qquad \text{(b) }\frac{1}{2}\qquad \text{(c) }\frac{3}{4}\qquad \text{(d) }\frac{3}{8}
\displaystyle \text{Answer:}
\displaystyle S=A\cup B\text{ and }A,B\text{ are mutually exclusive}
\displaystyle \therefore P(A)+P(B)=1
\displaystyle P(A)=\frac{1}{3}P(B)
\displaystyle \therefore \frac{1}{3}P(B)+P(B)=1
\displaystyle \frac{4}{3}P(B)=1
\displaystyle P(B)=\frac{3}{4}
\displaystyle \therefore P(A)=\frac{1}{3}\times\frac{3}{4}=\frac{1}{4}
\displaystyle \therefore\text{Correct option is (a).}
\\

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

\displaystyle \textbf{Question 33. }\text{If }A,B,C\text{ are three mutually exclusive and exhaustive events of an} \\ \text{experiment such that }3P(A)=2P(B)=P(C),\text{ then }P(A)\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 \therefore P(A)=\frac{k}{3},\qquad P(B)=\frac{k}{2},\qquad P(C)=k
\displaystyle \text{Since }A,B,C\text{ are mutually exclusive and exhaustive,}
\displaystyle P(A)+P(B)+P(C)=1
\displaystyle \frac{k}{3}+\frac{k}{2}+k=1
\displaystyle \frac{2k+3k+6k}{6}=1
\displaystyle \frac{11k}{6}=1
\displaystyle k=\frac{6}{11}
\displaystyle \therefore P(A)=\frac{k}{3}=\frac{6}{33}=\frac{2}{11}
\displaystyle \therefore\text{Correct option is (b).}
\\

\displaystyle \textbf{Question 34. }\text{If }A\text{ and }B\text{ are mutually exclusive events then}
\displaystyle \text{(a) }P(A)\leq P(B)\qquad \text{(b) }P(A)\geq P(B)\qquad \text{(c) }P(A)<P(B)\qquad \text{(d) None of these}
\displaystyle \text{Answer:}
\displaystyle \text{If }A\text{ and }B\text{ are mutually exclusive, then }P(A\cap B)=0
\displaystyle \text{But there is no fixed relation between }P(A)\text{ and }P(B)
\displaystyle \therefore\text{Correct option is (d).}
\\

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

\displaystyle \textbf{Question 36. }\text{Three numbers are chosen from }1\text{ to }20.\text{ The probability that they are} \\ \text{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 ways of choosing }3\text{ numbers from }20={}^{20}C_3=1140
\displaystyle \text{Consecutive triples are }(1,2,3),(2,3,4),\ldots,(18,19,20)
\displaystyle \therefore \text{Number of consecutive triples}=18
\displaystyle \therefore P(\text{consecutive})=\frac{18}{1140}=\frac{3}{190}
\displaystyle \therefore P(\text{not consecutive})=1-\frac{3}{190}=\frac{187}{190}
\displaystyle \therefore\text{Correct option is (b).}
\\

\displaystyle \textbf{Question 37. }6\text{ boys and }6\text{ girls sit in a row at random. The probability that all} \\ \text{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 of }12\text{ persons}=12!
\displaystyle \text{Treat all }6\text{ girls as one block}
\displaystyle \text{Then }6\text{ boys and }1\text{ girls' block give }7\text{ objects}
\displaystyle \text{These can be arranged in }7!\text{ ways}
\displaystyle \text{The }6\text{ girls can be arranged among themselves in }6!\text{ ways}
\displaystyle \therefore \text{Favourable arrangements}=7!\times6!
\displaystyle \therefore P=\frac{7!\times6!}{12!}
\displaystyle =\frac{1}{132}
\displaystyle \therefore\text{Correct option is (c).}
\\

\displaystyle \textbf{Question 38. }\text{Without repetition of the numbers, four digit numbers are formed with the} \\ \text{numbers }0,2,3,5.\text{ The probability of such a number divisible by }5\text{ is}
\displaystyle \text{(a) }\frac{1}{5}\qquad \text{(b) }\frac{4}{5}\qquad \text{(c) }\frac{1}{30}\qquad \text{(d) }\frac{5}{9}
\displaystyle \text{Answer:}
\displaystyle \text{Total four digit numbers using }0,2,3,5\text{ without repetition}=4!-3!=24-6=18
\displaystyle \text{A number divisible by }5\text{ must end in }0\text{ or }5
\displaystyle \text{Case 1: Last digit is }0
\displaystyle \text{Remaining }3\text{ digits can be arranged in }3!=6\text{ ways}
\displaystyle \text{Case 2: Last digit is }5
\displaystyle \text{First digit cannot be }0,\text{ so first digit can be chosen in }2\text{ ways}
\displaystyle \text{Remaining }2\text{ digits can be arranged in }2!=2\text{ ways}
\displaystyle \therefore \text{Favourable numbers}=6+2\times2=10
\displaystyle \therefore P=\frac{10}{18}=\frac{5}{9}
\displaystyle \therefore\text{Correct option is (d).}
\\

\displaystyle \textbf{Question 39. }\text{If the probability for }A\text{ to fail in an examination is }0.2\text{ and that for } \\ B\text{ is }0.3,\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)=0.2,\qquad P(B)=0.3
\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,
\displaystyle P(A\cup B)\leq0.5
\displaystyle \therefore\text{Correct option is (c).}
\\

\displaystyle \textbf{Question 40. }\text{Three digit numbers are formed using the digits }0,2,4,6,8.\text{ A number} \\ \text{is chosen at random out of these numbers what is the probability that this number} \\ \text{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 all same digits are }222,444,666,888
\displaystyle \therefore \text{Favourable numbers}=4
\displaystyle \therefore P=\frac{4}{100}=\frac{1}{25}
\displaystyle \therefore\text{Correct option is (d).}
\\

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