\displaystyle \textbf{Question 1: }\text{(a) If }A\text{ and }B\text{ are mutually exclusive events associated with a random}
\displaystyle \text{experiment such that }P(A)=0.4\text{ and }P(B)=0.5,\text{ find:}
\displaystyle \text{(i) }P(A\cup B)\qquad\text{(ii) }P(\overline{A}\cap\overline{B})\qquad\text{(iii) }P(\overline{A}\cap B)
\displaystyle \text{(iv) }P(A\cap\overline{B}).
\displaystyle \text{(b) If }P(A)=0.54,\ P(B)=0.69\text{ and }P(A\cap B)=0.35,\text{ find:}
\displaystyle \text{(i) }P(A\cup B)\qquad\text{(ii) }P(\overline{A}\cup\overline{B})\qquad\text{(iii) }P(A\cap\overline{B})
\displaystyle \text{(iv) }P(B\cap A).
\displaystyle \text{(c) Fill in the blanks in the following table:}
\displaystyle \begin{array}{c|c|c|c|c} & P(A)&P(B)&P(A\cap B)&P(A\cup B)\\ \hline \text{(i)}&\frac{1}{3}&\frac{1}{5}&\frac{1}{15}&\cdots\\ \text{(ii)}&0.35&\cdots&0.25&0.6\\ \text{(iii)}&0.5&0.35&\cdots&0.7 \end{array}
\displaystyle \text{Answer:}

\displaystyle \text{(a) Given }P(A)=0.4\text{ and }P(B)=0.5.
\displaystyle \text{Since }A\text{ and }B\text{ are mutually exclusive, }P(A\cap B)=0.
\displaystyle \text{(i) }P(A\cup B)=P(A)+P(B)-P(A\cap B).
\displaystyle =0.4+0.5-0=0.9.
\displaystyle \text{(ii) }P(\overline{A}\cap\overline{B})=P(\overline{A\cup B}).
\displaystyle =1-P(A\cup B)=1-0.9=0.1.
\displaystyle \text{(iii) }P(\overline{A}\cap B)=P(B)-P(A\cap B).
\displaystyle =0.5-0=0.5.
\displaystyle \text{(iv) }P(A\cap\overline{B})=P(A)-P(A\cap B).
\displaystyle =0.4-0=0.4.

\displaystyle \text{(b) Given }P(A)=0.54,\ P(B)=0.69\text{ and }P(A\cap B)=0.35.
\displaystyle \text{(i) }P(A\cup B)=P(A)+P(B)-P(A\cap B).
\displaystyle =0.54+0.69-0.35=0.88.
\displaystyle \text{(ii) By De Morgan's law, }\overline{A}\cup\overline{B}=\overline{A\cap B}.
\displaystyle P(\overline{A}\cup\overline{B})=1-P(A\cap B).
\displaystyle =1-0.35=0.65.
\displaystyle \text{(iii) }P(A\cap\overline{B})=P(A)-P(A\cap B).
\displaystyle =0.54-0.35=0.19.
\displaystyle \text{(iv) }P(B\cap A)=P(A\cap B)=0.35.

\displaystyle \text{(c) We use }P(A\cup B)=P(A)+P(B)-P(A\cap B).
\displaystyle \text{(i) }P(A\cup B)=\frac{1}{3}+\frac{1}{5}-\frac{1}{15}.
\displaystyle =\frac{5+3-1}{15}=\frac{7}{15}.
\displaystyle \text{(ii) }P(B)=P(A\cup B)-P(A)+P(A\cap B).
\displaystyle =0.6-0.35+0.25=0.5.
\displaystyle \text{(iii) }P(A\cap B)=P(A)+P(B)-P(A\cup B).
\displaystyle =0.5+0.35-0.7=0.15.
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{If }A\text{ and }B\text{ are two events associated with a random experiment such}
\displaystyle \text{that }P(A)=0.3,\ P(B)=0.4\text{ and }P(A\cup B)=0.5,\text{ find }P(A\cap B).
\displaystyle \text{Answer:}
\displaystyle \text{Given }P(A)=0.3,\ P(B)=0.4\text{ and }P(A\cup B)=0.5.
\displaystyle P(A\cup B)=P(A)+P(B)-P(A\cap B).
\displaystyle \therefore P(A\cap B)=P(A)+P(B)-P(A\cup B).
\displaystyle =0.3+0.4-0.5=0.2.
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{If }A\text{ and }B\text{ are two events associated with a random experiment such}
\displaystyle \text{that }P(A)=0.5,\ P(B)=0.3\text{ and }P(A\cap B)=0.2,\text{ find }P(A\cup B).
\displaystyle \text{Answer:}
\displaystyle \text{Given }P(A)=0.5,\ P(B)=0.3\text{ and }P(A\cap B)=0.2.
\displaystyle P(A\cup B)=P(A)+P(B)-P(A\cap B).
\displaystyle =0.5+0.3-0.2=0.6.
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{If }A\text{ and }B\text{ are two events associated with a random experiment such}
\displaystyle \text{that }P(A\cup B)=0.8,\ P(A\cap B)=0.3\text{ and }P(\overline{A})=0.5,\text{ find }P(B).
\displaystyle \text{Answer:}
\displaystyle \text{Given }P(A\cup B)=0.8,\ P(A\cap B)=0.3\text{ and }P(\overline{A})=0.5.
\displaystyle P(A)=1-P(\overline{A})=1-0.5=0.5.
\displaystyle P(A\cup B)=P(A)+P(B)-P(A\cap B).
\displaystyle \therefore P(B)=P(A\cup B)-P(A)+P(A\cap B).
\displaystyle =0.8-0.5+0.3=0.6.
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{Given two mutually exclusive events }A\text{ and }B\text{ such that}
\displaystyle P(A)=\frac{1}{2}\text{ and }P(B)=\frac{1}{3},\text{ find }P(A\cup B).
\displaystyle \text{Answer:}
\displaystyle \text{Given }P(A)=\frac{1}{2}\text{ and }P(B)=\frac{1}{3}.
\displaystyle \text{Since }A\text{ and }B\text{ are mutually exclusive, }P(A\cap B)=0.
\displaystyle P(A\cup B)=P(A)+P(B)-P(A\cap B).
\displaystyle =\frac{1}{2}+\frac{1}{3}-0=\frac{5}{6}.
\displaystyle \therefore P(A\cup B)=\frac{5}{6}.
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{There are three events }A,\ B\text{ and }C,\text{ of which exactly one must occur.}
\displaystyle \text{The odds against }A\text{ are }8:3\text{ and the odds against }B\text{ are }5:2.\text{ Find the odds}
\displaystyle \text{against }C.
\displaystyle \text{Answer:}
\displaystyle \text{Since exactly one of the events }A,\ B\text{ and }C\text{ occurs, they are mutually exclusive and exhaustive.}
\displaystyle \therefore P(A\cup B\cup C)=P(A)+P(B)+P(C)=1.
\displaystyle \text{Odds against }A=8:3.
\displaystyle \therefore \frac{P(\overline{A})}{P(A)}=\frac{8}{3}.
\displaystyle \frac{1-P(A)}{P(A)}=\frac{8}{3}.
\displaystyle 8P(A)=3-3P(A).
\displaystyle 11P(A)=3.
\displaystyle \therefore P(A)=\frac{3}{11}.
\displaystyle \text{Odds against }B=5:2.
\displaystyle \therefore \frac{P(\overline{B})}{P(B)}=\frac{5}{2}.
\displaystyle \frac{1-P(B)}{P(B)}=\frac{5}{2}.
\displaystyle 5P(B)=2-2P(B).
\displaystyle 7P(B)=2.
\displaystyle \therefore P(B)=\frac{2}{7}.
\displaystyle P(C)=1-P(A)-P(B).
\displaystyle =1-\frac{3}{11}-\frac{2}{7}=\frac{34}{77}.
\displaystyle \therefore P(\overline{C})=1-\frac{34}{77}=\frac{43}{77}.
\displaystyle \text{Hence, the odds against }C=\frac{P(\overline{C})}{P(C)}=\frac{\frac{43}{77}}{\frac{34}{77}}=\frac{43}{34}.
\displaystyle \therefore \text{The required odds against }C\text{ are }43:34.
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{One of the two events must happen. Given that the chance of one is }
\displaystyle \text{two-third of the other, find the odds in favour of the other.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the two events be }A\text{ and }B.
\displaystyle \text{Since exactly one of the two events occurs, they are mutually exclusive and exhaustive.}
\displaystyle \therefore P(A\cap B)=0\text{ and }P(A\cup B)=P(A)+P(B)=1.
\displaystyle \text{Given }P(A)=\frac{2}{3}P(B).
\displaystyle \therefore \frac{2}{3}P(B)+P(B)=1.
\displaystyle \frac{5}{3}P(B)=1.
\displaystyle \therefore P(B)=\frac{3}{5}.
\displaystyle P(\overline{B})=1-\frac{3}{5}=\frac{2}{5}.
\displaystyle \therefore \text{Odds in favour of }B=\frac{P(B)}{P(\overline{B})}=\frac{\frac{3}{5}}{\frac{2}{5}}=\frac{3}{2}.
\displaystyle \therefore \text{The required odds in favour of the other event are }3:2.
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{A card is drawn at random from a well-shuffled deck of }52\text{ cards. Find the}
\displaystyle \text{probability that it is a spade or a king.}
\displaystyle \text{Answer:}
\displaystyle P(\text{Spade})=\frac{13}{52}=\frac{1}{4}.
\displaystyle P(\text{King})=\frac{4}{52}=\frac{1}{13}.
\displaystyle P(\text{Spade}\cap\text{King})=\frac{1}{52}.
\displaystyle P(\text{Spade}\cup\text{King})=P(\text{Spade})+P(\text{King})-P(\text{Spade}\cap\text{King}).
\displaystyle =\frac{1}{4}+\frac{1}{13}-\frac{1}{52}=\frac{13+4-1}{52}=\frac{16}{52}=\frac{4}{13}.
\displaystyle \therefore \text{The required probability is }\frac{4}{13}.
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{In a single throw of two dice, find the probability that neither a doublet}
\displaystyle \text{nor a total of }9\text{ appears.}
\displaystyle \text{Answer:}
\displaystyle \text{Total number of possible outcomes}=6\times6=36.
\displaystyle \text{Let }A\text{ be the event of getting a doublet.}
\displaystyle A=\{(1,1),(2,2),(3,3),(4,4),(5,5),(6,6)\}.
\displaystyle \therefore P(A)=\frac{6}{36}=\frac{1}{6}.
\displaystyle \text{Let }B\text{ be the event of getting a total of }9.
\displaystyle B=\{(3,6),(4,5),(5,4),(6,3)\}.
\displaystyle \therefore P(B)=\frac{4}{36}=\frac{1}{9}.
\displaystyle \text{Since }A\text{ and }B\text{ are mutually exclusive, }P(A\cap B)=0.
\displaystyle P(A\cup B)=P(A)+P(B)-P(A\cap B).
\displaystyle =\frac{1}{6}+\frac{1}{9}=\frac{5}{18}.
\displaystyle \therefore P(\text{neither a doublet nor a total of }9)=1-\frac{5}{18}=\frac{13}{18}.
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{A natural number is chosen at random from the first }500\text{ natural numbers.}
\displaystyle \text{Find the probability that it is divisible by }3\text{ or }5.
\displaystyle \text{Answer:}
\displaystyle \text{Total number of possible outcomes}=500.
\displaystyle \therefore n(S)=500.
\displaystyle \text{Multiples of }3\text{ are }3,6,9,\ldots,498.
\displaystyle \text{Number of multiples of }3=166.
\displaystyle P(3)=\frac{166}{500}.
\displaystyle \text{Multiples of }5\text{ are }5,10,15,\ldots,500.
\displaystyle \text{Number of multiples of }5=100.
\displaystyle P(5)=\frac{100}{500}.
\displaystyle \text{Multiples of both }3\text{ and }5\text{ are }15,30,45,\ldots,495.
\displaystyle \text{Number of multiples of }15=33.
\displaystyle P(3\cap5)=\frac{33}{500}.
\displaystyle P(3\cup5)=P(3)+P(5)-P(3\cap5).
\displaystyle =\frac{166}{500}+\frac{100}{500}-\frac{33}{500}=\frac{233}{500}.
\displaystyle \therefore \text{The required probability is }\frac{233}{500}.
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{A die is thrown twice. Find the probability that at least one of the two throws}
\displaystyle \text{shows the number }3.
\displaystyle \text{Answer:}
\displaystyle \text{When a die is thrown twice, the total number of possible outcomes}=6\times6=36.
\displaystyle \therefore n(S)=36.
\displaystyle \text{Let }A\text{ be the event that the first throw is }3.
\displaystyle A=\{(3,1),(3,2),(3,3),(3,4),(3,5),(3,6)\}.
\displaystyle \therefore P(A)=\frac{6}{36}=\frac{1}{6}.
\displaystyle \text{Let }B\text{ be the event that the second throw is }3.
\displaystyle B=\{(1,3),(2,3),(3,3),(4,3),(5,3),(6,3)\}.
\displaystyle \therefore P(B)=\frac{6}{36}=\frac{1}{6}.
\displaystyle \text{The outcome }(3,3)\text{ is common to both events.}
\displaystyle \therefore P(A\cap B)=\frac{1}{36}.
\displaystyle P(A\cup B)=P(A)+P(B)-P(A\cap B).
\displaystyle =\frac{1}{6}+\frac{1}{6}-\frac{1}{36}=\frac{11}{36}.
\displaystyle \therefore \text{The required probability is }\frac{11}{36}.
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{A card is drawn from a well-shuffled deck of }52\text{ cards. Find the}
\displaystyle \text{probability of getting an ace or a spade.}
\displaystyle \text{Answer:}
\displaystyle P(\text{Ace})=\frac{4}{52}=\frac{1}{13}.
\displaystyle P(\text{Spade})=\frac{13}{52}=\frac{1}{4}.
\displaystyle P(\text{Ace}\cap\text{Spade})=\frac{1}{52}.
\displaystyle P(\text{Ace}\cup\text{Spade})=P(\text{Ace})+P(\text{Spade})-P(\text{Ace}\cap\text{Spade}).
\displaystyle =\frac{1}{13}+\frac{1}{4}-\frac{1}{52}=\frac{4+13-1}{52}=\frac{16}{52}=\frac{4}{13}.
\displaystyle \therefore \text{The required probability is }\frac{4}{13}.
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{The probability that a student passes both English and Hindi is }0.5,
\displaystyle \text{the probability of passing neither is }0.1,\text{ and the probability of passing English}
\displaystyle \text{is }0.75.\text{ Find the probability of passing Hindi.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }E\text{ and }H\text{ denote the events of passing English and Hindi respectively.}
\displaystyle \text{Given }P(E\cap H)=0.5,\ P(\overline{E}\cap\overline{H})=0.1\text{ and }P(E)=0.75.
\displaystyle P(E\cup H)=1-P(\overline{E}\cap\overline{H})=1-0.1=0.9.
\displaystyle P(E\cup H)=P(E)+P(H)-P(E\cap H).
\displaystyle \therefore P(H)=P(E\cup H)-P(E)+P(E\cap H).
\displaystyle =0.9-0.75+0.5=0.65.
\displaystyle \therefore \text{The probability of passing Hindi is }0.65.
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{One number is chosen at random from the numbers }1\text{ to }100.\text{ Find the}
\displaystyle \text{probability that it is divisible by }4\text{ or }6.
\displaystyle \text{Answer:}
\displaystyle \text{Total number of possible outcomes}=100.
\displaystyle \therefore n(S)=100.
\displaystyle \text{Multiples of }4\text{ are }4,8,12,\ldots,100.
\displaystyle \text{Number of multiples of }4=25.
\displaystyle P(4)=\frac{25}{100}.
\displaystyle \text{Multiples of }6\text{ are }6,12,18,\ldots,96.
\displaystyle \text{Number of multiples of }6=16.
\displaystyle P(6)=\frac{16}{100}.
\displaystyle \text{Multiples of both }4\text{ and }6\text{ are multiples of }12.
\displaystyle \text{They are }12,24,36,\ldots,96.
\displaystyle \text{Number of multiples of }12=8.
\displaystyle P(4\cap6)=\frac{8}{100}.
\displaystyle P(4\cup6)=P(4)+P(6)-P(4\cap6).
\displaystyle =\frac{25}{100}+\frac{16}{100}-\frac{8}{100}=\frac{33}{100}.
\displaystyle \therefore \text{The required probability is }\frac{33}{100}.
\displaystyle \\

\displaystyle \textbf{Question 15: }\text{From a well-shuffled deck of }52\text{ cards, }4\text{ cards are drawn at random.}
\displaystyle \text{Find the probability that all the drawn cards are of the same colour.}
\displaystyle \text{Answer:}
\displaystyle \text{A deck of }52\text{ cards has }26\text{ black cards and }26\text{ red cards.}
\displaystyle \text{Total number of ways of drawing }4\text{ cards}={}^{52}C_4.
\displaystyle \therefore n(S)={}^{52}C_4.
\displaystyle \text{Let }A\text{ be the event that all }4\text{ cards are black.}
\displaystyle n(A)={}^{26}C_4.
\displaystyle \text{Let }B\text{ be the event that all }4\text{ cards are red.}
\displaystyle n(B)={}^{26}C_4.
\displaystyle \text{Since }A\text{ and }B\text{ are mutually exclusive,}
\displaystyle P(A\cup B)=\frac{{}^{26}C_4+{}^{26}C_4}{{}^{52}C_4}.
\displaystyle =\frac{2\times{}^{26}C_4}{{}^{52}C_4}=\frac{92}{833}.
\displaystyle \therefore \text{The required probability is }\frac{92}{833}.
\displaystyle \\

\displaystyle \textbf{Question 16: }\text{Out of }100\text{ students who appeared for two examinations, }60\text{ passed the}
\displaystyle \text{first, }50\text{ passed the second and }30\text{ passed both. Find the probability that a}
\displaystyle \text{student selected at random has passed at least one examination.}
\displaystyle \text{Answer:}
\displaystyle \text{Total number of students}=100.
\displaystyle \text{Let }A\text{ be the event that a student passes the first examination.}
\displaystyle P(A)=\frac{60}{100}=0.6.
\displaystyle \text{Let }B\text{ be the event that a student passes the second examination.}
\displaystyle P(B)=\frac{50}{100}=0.5.
\displaystyle P(A\cap B)=\frac{30}{100}=0.3.
\displaystyle \text{Using the addition rule,}
\displaystyle P(A\cup B)=P(A)+P(B)-P(A\cap B).
\displaystyle =0.6+0.5-0.3=0.8.
\displaystyle \therefore \text{The probability that the student has passed at least one examination is }0.8.
\displaystyle \\

\displaystyle \textbf{Question 17: }\text{A box contains }10\text{ white, }6\text{ red and }10\text{ black balls. A ball is}
\displaystyle \text{drawn at random. Find the probability that it is either white or red.}
\displaystyle \text{Answer:}
\displaystyle P(\text{White})=\frac{10}{26}=\frac{5}{13}.
\displaystyle P(\text{Red})=\frac{6}{26}=\frac{3}{13}.
\displaystyle P(\text{White}\cap\text{Red})=0.
\displaystyle P(\text{White}\cup\text{Red})=P(\text{White})+P(\text{Red})-P(\text{White}\cap\text{Red}).
\displaystyle =\frac{5}{13}+\frac{3}{13}-0=\frac{8}{13}.
\displaystyle \therefore \text{The required probability is }\frac{8}{13}.
\displaystyle \\

\displaystyle \textbf{Question 18: }\text{In a race, the odds in favour of the horses }A,\ B,\ C\text{ and }D\text{ are}
\displaystyle \text{respectively }1:3,\ 1:4,\ 1:5\text{ and }1:6.\text{ Find the probability that one of them wins}
\displaystyle \text{the race.}
\displaystyle \text{Answer:}
\displaystyle \text{Odds in favour of }A=1:3.
\displaystyle \therefore P(A)=\frac{1}{1+3}=\frac{1}{4}.
\displaystyle \text{Odds in favour of }B=1:4.
\displaystyle \therefore P(B)=\frac{1}{1+4}=\frac{1}{5}.
\displaystyle \text{Odds in favour of }C=1:5.
\displaystyle \therefore P(C)=\frac{1}{1+5}=\frac{1}{6}.
\displaystyle \text{Odds in favour of }D=1:6.
\displaystyle \therefore P(D)=\frac{1}{1+6}=\frac{1}{7}.
\displaystyle \text{Since only one horse can win the race, the events are mutually exclusive.}
\displaystyle P(A\cup B\cup C\cup D)=P(A)+P(B)+P(C)+P(D).
\displaystyle =\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}=\frac{105+84+70+60}{420}=\frac{319}{420}.
\displaystyle \therefore \text{The required probability is }\frac{319}{420}.
\displaystyle \\

\displaystyle \textbf{Question 19: }\text{The probability that a person travels by plane is }\frac{3}{5}\text{ and that}
\displaystyle \text{the person travels by train is }\frac{1}{4}.\text{ Find the probability that the person}
\displaystyle \text{travels by plane or train.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }P\text{ be the event of travelling by plane and }T\text{ the event of travelling by train.}
\displaystyle \text{Given }P(P)=\frac{3}{5}\text{ and }P(T)=\frac{1}{4}.
\displaystyle \text{Since the person cannot travel by both plane and train simultaneously,}
\displaystyle P(P\cap T)=0.
\displaystyle P(P\cup T)=P(P)+P(T)-P(P\cap T).
\displaystyle =\frac{3}{5}+\frac{1}{4}-0=\frac{12+5}{20}=\frac{17}{20}.
\displaystyle \therefore \text{The required probability is }\frac{17}{20}.
\displaystyle \\

\displaystyle \textbf{Question 20: }\text{Two cards are drawn from a well-shuffled pack of }52\text{ cards. Find the}
\displaystyle \text{probability that either both are black or both are kings.}
\displaystyle \text{Answer:}
\displaystyle \text{Total number of ways of drawing }2\text{ cards}={}^{52}C_2.
\displaystyle \therefore n(S)={}^{52}C_2.
\displaystyle \text{Let }A\text{ be the event that both cards are black.}
\displaystyle n(A)={}^{26}C_2.
\displaystyle \text{Let }B\text{ be the event that both cards are kings.}
\displaystyle n(B)={}^{4}C_2.
\displaystyle \text{The common event }A\cap B\text{ is drawing the two black kings.}
\displaystyle \therefore n(A\cap B)={}^{2}C_2=1.
\displaystyle P(A\cup B)=P(A)+P(B)-P(A\cap B).
\displaystyle =\frac{{}^{26}C_2}{{}^{52}C_2}+\frac{{}^{4}C_2}{{}^{52}C_2}-\frac{{}^{2}C_2}{{}^{52}C_2}.
\displaystyle =\frac{26\times25}{52\times51}+\frac{4\times3}{52\times51}-\frac{2}{52\times51}.
\displaystyle =\frac{600}{2652}=\frac{55}{221}.
\displaystyle \therefore \text{The required probability is }\frac{55}{221}.
\displaystyle \\

\displaystyle \textbf{Question 21: }\text{In an entrance test graded on the basis of two examinations, the probability}
\displaystyle \text{that a student passes the first examination is }0.8,\text{ the second examination is }0.7,
\displaystyle \text{and at least one examination is }0.95.\text{ Find the probability of passing both.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }A\text{ be the event that the student passes the first examination.}
\displaystyle \text{Let }B\text{ be the event that the student passes the second examination.}
\displaystyle \text{Given }P(A)=0.8,\ P(B)=0.7\text{ and }P(A\cup B)=0.95.
\displaystyle P(A\cap B)=P(A)+P(B)-P(A\cup B).
\displaystyle =0.8+0.7-0.95=0.55.
\displaystyle \therefore \text{The probability of passing both examinations is }0.55.
\displaystyle \\

\displaystyle \textbf{Question 22: }\text{A box contains }30\text{ bolts and }40\text{ nuts. Half of the bolts and half of the}
\displaystyle \text{nuts are rusted. If two items are drawn at random, find the probability that either both}
\displaystyle \text{are rusted or both are bolts.}
\displaystyle \text{Answer:}
\displaystyle \text{Number of bolts}=30,\qquad \text{Number of nuts}=40.
\displaystyle \text{Rusted bolts}=15,\qquad \text{Rusted nuts}=20.
\displaystyle \text{Hence, total number of rusted items}=15+20=35.
\displaystyle \text{Total number of items}=30+40=70.
\displaystyle \text{Total number of ways of drawing }2\text{ items}={}^{70}C_2.
\displaystyle \text{Let }R\text{ be the event that both selected items are rusted.}
\displaystyle \text{Let }B\text{ be the event that both selected items are bolts.}
\displaystyle \text{The events }R\text{ and }B\text{ are not mutually exclusive.}
\displaystyle \text{Their common event is selecting two rusted bolts.}
\displaystyle \therefore P(R\cup B)=P(R)+P(B)-P(R\cap B).
\displaystyle =\frac{{}^{35}C_2}{{}^{70}C_2}+\frac{{}^{30}C_2}{{}^{70}C_2}-\frac{{}^{15}C_2}{{}^{70}C_2}.
\displaystyle =\frac{35\times34}{70\times69}+\frac{30\times29}{70\times69}-\frac{15\times14}{70\times69}.
\displaystyle =\frac{1850}{4830}=\frac{185}{483}.
\displaystyle \therefore \text{The required probability is }\frac{185}{483}.
\displaystyle \\

\displaystyle \textbf{Question 23: }\text{An integer is chosen at random from the first }200\text{ positive integers.}
\displaystyle \text{Find the probability that it is divisible by }6\text{ or }8.
\displaystyle \text{Answer:}
\displaystyle \text{Total number of possible outcomes}=200.
\displaystyle \therefore n(S)=200.
\displaystyle \text{Let }A\text{ be the event that the chosen integer is divisible by }6.
\displaystyle \text{Multiples of }6\text{ are }6,12,18,\ldots,198.
\displaystyle n(A)=33\qquad(\because\ 198=6+(n-1)\times6\Rightarrow n=33).
\displaystyle \text{Let }B\text{ be the event that the chosen integer is divisible by }8.
\displaystyle \text{Multiples of }8\text{ are }8,16,24,\ldots,200.
\displaystyle n(B)=25\qquad(\because\ 200=8+(n-1)\times8\Rightarrow n=25).
\displaystyle \text{Multiples of both }6\text{ and }8\text{ are multiples of }24.
\displaystyle \text{They are }24,48,\ldots,192.
\displaystyle n(A\cap B)=8\qquad(\because\ 192=24+(n-1)\times24\Rightarrow n=8).
\displaystyle P(A\cup B)=P(A)+P(B)-P(A\cap B).
\displaystyle =\frac{n(A)}{n(S)}+\frac{n(B)}{n(S)}-\frac{n(A\cap B)}{n(S)}.
\displaystyle =\frac{33}{200}+\frac{25}{200}-\frac{8}{200}=\frac{50}{200}=\frac{1}{4}.
\displaystyle \therefore \text{The required probability is }\frac{1}{4}.
\displaystyle \\

\displaystyle \textbf{Question 24: }\text{Find the probability of getting }2\text{ or }3\text{ tails when a coin is tossed}
\displaystyle \text{four times.}
\displaystyle \text{Answer:}
\displaystyle \text{When a coin is tossed four times, the total number of possible outcomes}=2^4=16.
\displaystyle \therefore n(S)=16.
\displaystyle \text{Let }A\text{ be the event of getting exactly }2\text{ tails.}
\displaystyle A=\{TTHH,\ THTH,\ THHT,\ HTTH,\ HTHT,\ HHTT\}.
\displaystyle \therefore n(A)=6.
\displaystyle P(A)=\frac{n(A)}{n(S)}=\frac{6}{16}=\frac{3}{8}.
\displaystyle \text{Let }B\text{ be the event of getting exactly }3\text{ tails.}
\displaystyle B=\{TTTH,\ TTHT,\ THTT,\ HTTT\}.
\displaystyle \therefore n(B)=4.
\displaystyle P(B)=\frac{n(B)}{n(S)}=\frac{4}{16}=\frac{1}{4}.
\displaystyle \text{Since }A\text{ and }B\text{ are mutually exclusive events,}
\displaystyle P(A\cup B)=P(A)+P(B).
\displaystyle =\frac{3}{8}+\frac{1}{4}=\frac{5}{8}.
\displaystyle \therefore \text{The required probability is }\frac{5}{8}.
\displaystyle \\

\displaystyle \textbf{Question 25: }\text{Suppose an integer from }1\text{ through }1000\text{ is chosen at random. Find the}
\displaystyle \text{probability that it is a multiple of }2\text{ or a multiple of }9.
\displaystyle \text{Answer:}
\displaystyle \text{Total number of possible outcomes}=1000.
\displaystyle \therefore n(S)=1000.
\displaystyle \text{Let }A\text{ be the event that the chosen integer is divisible by }2.
\displaystyle \text{Multiples of }2\text{ are }2,4,6,\ldots,1000.
\displaystyle n(A)=500\qquad(\because\ 1000=2+(n-1)\times2\Rightarrow n=500).
\displaystyle \text{Let }B\text{ be the event that the chosen integer is divisible by }9.
\displaystyle \text{Multiples of }9\text{ are }9,18,27,\ldots,999.
\displaystyle n(B)=111\qquad(\because\ 999=9+(n-1)\times9\Rightarrow n=111).
\displaystyle \text{Multiples of both }2\text{ and }9\text{ are multiples of }18.
\displaystyle \text{They are }18,36,\ldots,990.
\displaystyle n(A\cap B)=55\qquad(\because\ 990=18+(n-1)\times18\Rightarrow n=55).
\displaystyle n(A\cup B)=500+111-55=556.
\displaystyle P(A\cup B)=\frac{556}{1000}=\frac{139}{250}=0.556.
\displaystyle \therefore \text{The required probability is }\frac{139}{250}.
\displaystyle \\

\displaystyle \textbf{Question 26: }\text{In a large metropolitan area, the probabilities are }0.87,\ 0.36\text{ and }0.30
\displaystyle \text{that a family owns a colour television, a black-and-white television, or both kinds}
\displaystyle \text{respectively. Find the probability that a family owns either one or both kinds of sets.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }C\text{ be the event that a family owns a colour television.}
\displaystyle \text{Let }B\text{ be the event that a family owns a black-and-white television.}
\displaystyle \text{Given }P(C)=0.87,\ P(B)=0.36\text{ and }P(C\cap B)=0.30.
\displaystyle \text{Required probability}=P(B\cup C).
\displaystyle P(B\cup C)=P(B)+P(C)-P(B\cap C).
\displaystyle =0.36+0.87-0.30=0.93.
\displaystyle \therefore \text{The required probability is }0.93.
\displaystyle \\

\displaystyle \textbf{Question 27: }\text{If }A\text{ and }B\text{ are mutually exclusive events such that }P(A)=0.35
\displaystyle \text{and }P(B)=0.45,\text{ find:}
\displaystyle \text{(i) }P(A\cup B)\qquad\text{(ii) }P(A\cap B)\qquad\text{(iii) }P(A\cap\overline{B})
\displaystyle \text{(iv) }P(\overline{A}\cap\overline{B}).
\displaystyle \text{Answer:}
\displaystyle \text{Since }A\text{ and }B\text{ are mutually exclusive events,}
\displaystyle P(A\cap B)=0.
\displaystyle \text{(i) }P(A\cup B)=P(A)+P(B)-P(A\cap B).
\displaystyle =0.35+0.45-0=0.80.
\displaystyle \text{(ii) }P(A\cap B)=0.
\displaystyle \text{(iii) }P(A\cap\overline{B})=P(A)-P(A\cap B).
\displaystyle =0.35-0=0.35.
\displaystyle \text{(iv) }P(\overline{A}\cap\overline{B})=1-P(A\cup B).
\displaystyle =1-0.80=0.20.
\displaystyle \\

\displaystyle \textbf{Question 28: }\text{A sample space consists of }9\text{ elementary events }E_1,E_2,\ldots,E_9,
\displaystyle \text{whose probabilities are }P(E_1)=P(E_2)=0.08,\ P(E_3)=P(E_4)=0.10,
\displaystyle P(E_6)=P(E_7)=0.20\text{ and }P(E_8)=P(E_9)=0.07.
\displaystyle \text{Suppose }A=\{E_1,E_5,E_8\}\text{ and }B=\{E_2,E_5,E_8,E_9\}.
\displaystyle \text{(i) Compute }P(A),\ P(B)\text{ and }P(A\cap B).
\displaystyle \text{(ii) Using the addition law of probability, find }P(A\cup B).
\displaystyle \text{(iii) List the elementary events in }A\cup B\text{ and calculate }P(A\cup B)\text{ directly.}
\displaystyle \text{(iv) Calculate }P(\overline{B})\text{ from }P(B)\text{ and also directly from its elementary events.}
\displaystyle \text{Answer:}
\displaystyle \text{Since the sum of the probabilities of all elementary events is }1,
\displaystyle P(E_5)=1-[2(0.08)+2(0.10)+2(0.20)+2(0.07)].
\displaystyle =1-0.90=0.10.
\displaystyle \text{(i) }P(A)=P(E_1)+P(E_5)+P(E_8).
\displaystyle =0.08+0.10+0.07=0.25.
\displaystyle P(B)=P(E_2)+P(E_5)+P(E_8)+P(E_9).
\displaystyle =0.08+0.10+0.07+0.07=0.32.
\displaystyle A\cap B=\{E_5,E_8\}.
\displaystyle P(A\cap B)=P(E_5)+P(E_8)=0.10+0.07=0.17.
\displaystyle \text{(ii) By the addition law of probability,}
\displaystyle P(A\cup B)=P(A)+P(B)-P(A\cap B).
\displaystyle =0.25+0.32-0.17=0.40.
\displaystyle \text{(iii) }A\cup B=\{E_1,E_2,E_5,E_8,E_9\}.
\displaystyle P(A\cup B)=P(E_1)+P(E_2)+P(E_5)+P(E_8)+P(E_9).
\displaystyle =0.08+0.08+0.10+0.07+0.07=0.40.
\displaystyle \text{(iv) }P(\overline{B})=1-P(B)=1-0.32=0.68.
\displaystyle \overline{B}=\{E_1,E_3,E_4,E_6,E_7\}.
\displaystyle P(\overline{B})=P(E_1)+P(E_3)+P(E_4)+P(E_6)+P(E_7).
\displaystyle =0.08+0.10+0.10+0.20+0.20=0.68.
\displaystyle \\

 

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