\displaystyle \textbf{Question 1: }~\text{A coin is tossed thrice and all the eight outcomes are assumed equally} \\ \text{likely. In which of the following cases are the following events }A\text{ and }B\text{ independent?} \\ \text{(i) }A=\text{ the first throw results in head, }B=\text{ the last throw results in tail} \\ \text{(ii) }A=\text{ the number of heads is odd, }B=\text{ the number of tails is odd} \\ \text{(iii) }A=\text{ the number of heads is two, }B=\text{ the last throw results in head.}
\displaystyle \text{Answer:}
\displaystyle \text{(i)  }
\displaystyle S=\{(HHH),(HHT),(HTH),(HTT),(THH),(THT),(TTH),(TTT)\}
\displaystyle P(A)=\frac{4}{8}=\frac{1}{2}
\displaystyle P(B)=\frac{4}{8}=\frac{1}{2}
\displaystyle \text{Now,}
\displaystyle P(A\cap B)=\frac{2}{8}=\frac{1}{4}
\displaystyle P(A\cap B)=P(A)P(B)
\displaystyle \text{Thus, A and B are independent events.}

\displaystyle \text{(ii)  }
\displaystyle S=\{(HHH),(HHT),(HTH),(HTT),(THH),(THT),(TTH),(TTT)\}
\displaystyle P(A)=\frac{4}{8}=\frac{1}{2}
\displaystyle P(B)=\frac{4}{8}=\frac{1}{2}
\displaystyle \text{Now,}
\displaystyle P(A\cap B)=\frac{0}{8}=0
\displaystyle P(A\cap B)\neq P(A)P(B)
\displaystyle \text{Thus, A and B are not independent events.}

\displaystyle \text{(iii)  }
\displaystyle S=\{(HHH),(HHT),(HTH),(HTT),(THH),(THT),(TTH),(TTT)\}
\displaystyle P(A)=\frac{3}{8}
\displaystyle P(B)=\frac{4}{8}=\frac{1}{2}
\displaystyle \text{Now,}
\displaystyle P(A\cap B)=\frac{2}{8}=\frac{1}{4}
\displaystyle P(A\cap B)\neq P(A)P(B)
\displaystyle \text{Thus, A and B are not independent events.}

\displaystyle \textbf{Question 2: }~\text{Prove that in throwing a pair of dice, the occurrence of the number }4\text{ on} \\ \text{the first die is independent of the occurrence of }5\text{ on the second die.}
\displaystyle \text{Answer:}
\displaystyle \text{Total number of events}=36
\displaystyle P(\text{4 on first die})=P(A)=\frac{6}{36}=\frac{1}{6}
\displaystyle P(\text{5 on second die})=P(B)=\frac{6}{36}=\frac{1}{6}
\displaystyle P(A\cap B)=\frac{1}{36}
\displaystyle P(A\cap B)=P(A)P(B)
\displaystyle \text{Thus, A and B are independent events.}

\displaystyle \textbf{Question 3: }~\text{A card is drawn from a pack of }52\text{ cards so that each card is equally likely} \\ \text{to be selected. In which of the following cases are the events }A\text{ and }B\text{ independent?} \\ \text{(i) }A=\text{ the card drawn is a king or queen, }B=\text{ the card drawn is a queen or jack} \\ \text{(ii) }A=\text{ the card drawn is black, }B=\text{ the card drawn is a king} \\ \text{(iii) }A=\text{ the card drawn is a spade, }B=\text{ the card drawn is an ace. }
\displaystyle \text{Answer:}
\displaystyle \text{(i)  }
\displaystyle P(\text{king or queen})=P(A)=\frac{8}{52}=\frac{2}{13}
\displaystyle P(\text{queen or jack})=P(B)=\frac{8}{52}=\frac{2}{13}
\displaystyle P(A\cap B)=P(\text{queen})=\frac{4}{52}=\frac{1}{13}
\displaystyle P(A\cap B)\neq P(A)P(B)
\displaystyle \text{Thus, A and B are not independent events.}

\displaystyle \text{(ii)  }
\displaystyle P(\text{black})=P(A)=\frac{26}{52}=\frac{1}{2}
\displaystyle P(\text{king})=P(B)=\frac{4}{52}=\frac{1}{13}
\displaystyle P(A\cap B)=P(\text{black king})=\frac{2}{52}=\frac{1}{26}
\displaystyle P(A\cap B)=P(A)P(B)
\displaystyle \text{Thus, A and B are independent events.}

\displaystyle \text{(iii)  }
\displaystyle P(\text{spade})=P(A)=\frac{13}{52}=\frac{1}{4}
\displaystyle P(\text{ace})=P(B)=\frac{4}{52}=\frac{1}{13}
\displaystyle P(A\cap B)=P(\text{ace of spade})=\frac{1}{52}
\displaystyle P(A\cap B)=P(A)P(B)
\displaystyle \text{Thus, A and B are independent events.}

\displaystyle \textbf{Question 4: }~\text{A coin is tossed three times. Let the events }A,B\text{ and }C\text{ be defined as} \\ \text{follows: }A=\text{ first toss is head, }B=\text{ second toss is head, and }C=\text{ exactly two heads are} \\ \text{tossed in a row. Check the independence of (i) }A\text{ and }B\text{ (ii) }B\text{ and }C\text{ and (iii) }C\text{ and }A.
\displaystyle \text{Answer:}
\displaystyle \text{(i)  }
\displaystyle S=\{(HHH),(HHT),(HTH),(HTT),(THH),(THT),(TTH),(TTT)\}
\displaystyle P(A)=\frac{4}{8}=\frac{1}{2}
\displaystyle P(B)=\frac{4}{8}=\frac{1}{2}
\displaystyle P(A\cap B)=\frac{2}{8}=\frac{1}{4}=P(A)P(B)
\displaystyle \text{Thus, A and B are independent events.}

\displaystyle \text{(ii)  }
\displaystyle S=\{(HHH),(HHT),(HTH),(HTT),(THH),(THT),(TTH),(TTT)\}
\displaystyle \text{(ii) }P(C)=\frac{2}{8}=\frac{1}{4}
\displaystyle P(B)=\frac{4}{8}=\frac{1}{2}
\displaystyle P(B\cap C)=\frac{2}{8}=\frac{1}{4}\neq P(B)P(C)
\displaystyle \text{Thus, B and C are not independent events.}

\displaystyle \text{(iii)  }
\displaystyle S=\{(HHH),(HHT),(HTH),(HTT),(THH),(THT),(TTH),(TTT)\}
\displaystyle P(C)=\frac{1}{4}
\displaystyle P(A)=\frac{1}{2}
\displaystyle P(C\cap A)=\frac{1}{8}=P(C)P(A)
\displaystyle \text{Thus, A and C are independent events.}

\displaystyle \textbf{Question 5: }~\text{If }A\text{ and }B\text{ are two events such that }P(A)=\frac{1}{4},\ P(B)=\frac{1}{3} \\ \text{ and }P(A\cup B)=\frac{1}{2},\ \text{show that }A\text{ and }B\text{ are independent events.}
\displaystyle \text{Answer:}
\displaystyle P(A\cup B)=P(A)+P(B)-P(A\cap B)
\displaystyle \Rightarrow P(A\cap B)=P(A)+P(B)-P(A\cup B)
\displaystyle \Rightarrow P(A\cap B)=\frac{1}{4}+\frac{1}{3}-\frac{1}{2}
\displaystyle \Rightarrow P(A\cap B)=\frac{3+4-6}{12}
\displaystyle \Rightarrow P(A\cap B)=\frac{1}{12}=\frac{1}{4}\times\frac{1}{3}=P(A)P(B)
\displaystyle \text{Thus, A and B are independent events.}

\displaystyle \textbf{Question 6: }~\text{Given two independent events }A\text{ and }B\text{ such that }P(A)=0.3\text{ and } \\ P(B)=0.6,\ \text{find (i) }P(A\cap B)\text{ (ii) }P(\overline{A}\cap\overline{B})\text{ (iii) }P(\overline{A}\cap B) \\ \text{ (iv) }P(A\cap\overline{B}) \text{ (v) }P(A\cup B)\text{ (vi) }P(A\mid B)\text{ (vii) }P(B\mid A).
\displaystyle \text{Answer:}
\displaystyle \text{(i)  }
\displaystyle \text{Given:}
\displaystyle \text{A and B are independent events.}
\displaystyle P(A)=0.3
\displaystyle P(B)=0.6
\displaystyle P(A\cap B)=P(A)P(B)
\displaystyle =0.3\times0.6
\displaystyle =0.18

\displaystyle \text{(ii)  }
\displaystyle \text{Given:}
\displaystyle \text{A and B are independent events.}
\displaystyle P(A)=0.3
\displaystyle P(B)=0.6
\displaystyle P(A\cap \overline{B})=P(A)P(\overline{B})
\displaystyle =P(A)[1-P(B)]
\displaystyle =0.3[1-0.6]
\displaystyle =0.3\times0.4
\displaystyle =0.12

\displaystyle \text{(iii)  }
\displaystyle \text{Given:}
\displaystyle \text{A and B are independent events.}
\displaystyle P(A)=0.3
\displaystyle P(B)=0.6
\displaystyle P(\overline{A}\cap B)=P(\overline{A})P(B)
\displaystyle =P(B)[1-P(A)]
\displaystyle =0.6[1-0.3]
\displaystyle =0.6\times0.7
\displaystyle =0.42

\displaystyle \text{(iv)  }
\displaystyle \text{Given:}
\displaystyle \text{A and B are independent events.}
\displaystyle P(A)=0.3
\displaystyle P(B)=0.6
\displaystyle P(\overline{A}\cap \overline{B})=P(\overline{A})P(\overline{B})
\displaystyle =[1-P(A)][1-P(B)]
\displaystyle =0.7\times0.4
\displaystyle =0.28

\displaystyle \text{(v)  }      
\displaystyle \text{Given:}
\displaystyle \text{A and B are independent events.}
\displaystyle P(A)=0.3
\displaystyle P(B)=0.6
\displaystyle P(A\cup B)=P(A)+P(B)-P(A\cap B)
\displaystyle =0.3+0.6-0.18
\displaystyle =0.9-0.18
\displaystyle =0.72

\displaystyle \text{(vi)  }      
\displaystyle \text{Given:}
\displaystyle \text{A and B are independent events.}
\displaystyle P(A)=0.3
\displaystyle P(B)=0.6
\displaystyle P\!\left(\frac{A}{B}\right)=\frac{P(A\cap B)}{P(B)}
\displaystyle =\frac{0.18}{0.6}
\displaystyle =0.3

\displaystyle \text{(vii)  }
\displaystyle \text{Given:}
\displaystyle \text{A and B are independent events.}
\displaystyle P(A)=0.3
\displaystyle P(B)=0.6
\displaystyle P\!\left(\frac{B}{A}\right)=\frac{P(A\cap B)}{P(A)}
\displaystyle =\frac{0.18}{0.3}
\displaystyle =0.6

\displaystyle \textbf{Question 7: }~\text{If }P(\overline{B})=0.65,\ P(A\cup B)=0.85,\ \text{and }A\text{ and }B \text{ are independent events,} \\ \text{then find }P(A).
\displaystyle \text{Answer:}
\displaystyle P(\overline{B})=0.65
\displaystyle \Rightarrow 1-P(B)=0.65
\displaystyle \Rightarrow P(B)=1-0.65=0.35
\displaystyle \text{Now,}
\displaystyle P(A\cup B)=P(A)+P(B)-P(A\cap B)
\displaystyle \Rightarrow P(A\cup B)=P(A)+P(B)-P(A)\times P(B)
\displaystyle \Rightarrow 0.85=P(A)+0.35-0.35\times P(A)
\displaystyle \Rightarrow 0.85-0.35=P(A)[1-0.35]
\displaystyle \Rightarrow P(A)=\frac{0.5}{0.65}=0.77

\displaystyle \textbf{Question 8: }~\text{If }A\text{ and }B\text{ are two independent events such that }\\ P(\overline{A}\cap B)=\frac{2}{15}\text{ and }P(A\cap\overline{B})=\frac{1}{6},\ \text{then find }P(B). \ [CBSE\ 2015]
\displaystyle \text{Answer:}
\displaystyle \text{Let:}
\displaystyle P(A)=x
\displaystyle P(B)=y
\displaystyle P(\overline{A}\cap B)=\frac{2}{15}
\displaystyle \Rightarrow P(\overline{A})\times P(B)=\frac{2}{15}
\displaystyle \Rightarrow (1-x)y=\frac{2}{15}\text{ ...(1)}
\displaystyle P(A\cap \overline{B})=\frac{1}{6}
\displaystyle \Rightarrow P(A)\times P(\overline{B})=\frac{1}{6}
\displaystyle \Rightarrow (1-y)x=\frac{1}{6}\text{ ...(2)}
\displaystyle \text{Subtracting (2) from (1), we get}
\displaystyle x-y=\frac{1}{30}
\displaystyle x=y+\frac{1}{30}
\displaystyle \text{Substituting the value of }x\text{ in (2), we get}
\displaystyle \left(y+\frac{1}{30}\right)(1-y)=\frac{1}{6}
\displaystyle \Rightarrow 30y^{2}-29y+4=0
\displaystyle \Rightarrow y=\frac{1}{6},\frac{4}{5}

\displaystyle \textbf{Question 9: }~\text{ }A\text{ and }B\text{ are two independent events. The probability that }A\text{ and }\\ B\text{ occur is }\frac{1}{6}\text{ and the probability that neither of them occurs is }\frac{1}{3}. \text{ Find the} \\ \text{probability of occurrence of two events.}
\displaystyle \text{Answer:}
\displaystyle P(A\cap B)=P(A)P(B)\text{ (A and B are independent events)}
\displaystyle \frac{1}{6}=P(A)P(B)
\displaystyle \Rightarrow P(A)=\frac{1}{6P(B)}\text{ ...(1)}
\displaystyle P(\overline{A}\cap \overline{B})=[1-P(A)][1-P(B)]
\displaystyle \Rightarrow \frac{1}{3}=[1-P(A)][1-P(B)]
\displaystyle \Rightarrow \frac{1}{3}=\left[1-\frac{1}{6P(B)}\right][1-P(B)]\text{ [Using (1)]}
\displaystyle \text{Let }P(B)=x
\displaystyle \Rightarrow \left(\frac{6x-1}{6x}\right)(1-x)=\frac{1}{3}
\displaystyle \Rightarrow 6x-1-6x^{2}+x=2x
\displaystyle \Rightarrow 6x^{2}-5x+1=0
\displaystyle \Rightarrow (2x-1)(3x-1)=0
\displaystyle \Rightarrow x=\frac{1}{2}\text{ or }x=\frac{1}{3}
\displaystyle \text{If }P(B)=\frac{1}{2},\text{ then }P(A)=\frac{1}{3}
\displaystyle \text{If }P(B)=\frac{1}{3},\text{ then }P(A)=\frac{1}{2}

\displaystyle \textbf{Question 10: }~\text{If }A\text{ and }B\text{ are two independent events such that } \\ P(A\cup B)=0.60 \text{ and }P(A)=0.2,\ \text{find }P(B).
\displaystyle \text{Answer:}
\displaystyle P(A\cup B)=P(A)+P(B)-P(A\cap B)
\displaystyle P(A\cup B)=P(A)+P(B)-P(A)\times P(B)\text{ [: A and B are independent events]}
\displaystyle \Rightarrow 0.6=0.2+P(B)-0.2\times P(B)
\displaystyle \Rightarrow 0.6-0.2=P(B)(1-0.2)
\displaystyle \Rightarrow P(B)=\frac{0.6-0.2}{1-0.2}
\displaystyle \Rightarrow P(B)=\frac{0.4}{0.8}
\displaystyle \Rightarrow P(B)=\frac{1}{2}
\displaystyle \Rightarrow P(B)=0.5

\displaystyle \textbf{Question 11: }~\text{A die is tossed twice. Find the probability of getting a number greater than} \\ 3 \text{ on each toss.}
\displaystyle \text{Answer:}
\displaystyle S=\{(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),(3,1), \\ (3,2),(3,3),(3,4),(3,5),(3,6),(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),(5,1),(5,2), \\ (5,3),(5,4),(5,5),(5,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6) \}
\displaystyle n(S)=36
\displaystyle E=\text{Getting a number greater than 3 on each toss}
\displaystyle =\{(44),(45),(46),(54),(55),(56),(64),(65),(66)\}
\displaystyle n(E)=9
\displaystyle P(E)=\frac{9}{36}
\displaystyle =\frac{1}{4}

\displaystyle \textbf{Question 12: }~\text{Given the probability that }A\text{ can solve a problem is }\frac{2}{3}\text{ and the} \\ \text{probability that }B\text{ can solve the same problem is }\frac{3}{5}.\ \text{Find the probability that none} \\ \text{of the two will be able to solve the problem.}
\displaystyle \text{Answer:}
\displaystyle P(\text{A solving the problem})=P(A)=\frac{2}{3}
\displaystyle P(\text{B solving the problem})=P(B)=\frac{3}{5}
\displaystyle \text{We need to find out if }P(\overline{A}\cap \overline{B})=P(\overline{A})P(\overline{B})\text{ [A and B are independent events]}
\displaystyle =[1-P(A)][1-P(B)]
\displaystyle =\left(1-\frac{2}{3}\right)\left(1-\frac{3}{5}\right)
\displaystyle =\frac{1}{3}\times\frac{2}{5}
\displaystyle =\frac{2}{15}

\displaystyle \textbf{Question 13: }~\text{An unbiased die is tossed twice. Find the probability of getting }4,5,\text{ or } \\ 6\text{ on the first toss and }1,2,3\text{ or }4\text{ on the second toss.}
\displaystyle \text{Answer:}
\displaystyle P(A)=P(\text{4, 5 or 6 on first toss})=\frac{18}{36}=\frac{1}{2}
\displaystyle P(B)=P(\text{1, 2, 3 or 4 on second toss})=\frac{24}{36}=\frac{2}{3}
\displaystyle \text{It is clear that A and B are independent events.}
\displaystyle \Rightarrow P(A\cap B)=P(A)P(B)
\displaystyle \Rightarrow P(A\cap B)=\frac{1}{2}\times\frac{2}{3}
\displaystyle \therefore P(A\cap B)=\frac{1}{3}

\displaystyle \textbf{Question 14: }~\text{A bag contains }3\text{ red and }2\text{ black balls. One ball is drawn from it at random.} \\ \text{Its colour is noted and then it is put back in the bag. A second draw is made and the same} \\ \text{procedure is repeated. Find the probability of drawing (i) two red balls, (ii) two black balls,} \\ \text{(iii) first red and second black ball.}
\displaystyle \text{Answer:}
\displaystyle \text{Total balls}=3+2=5
\displaystyle P(\text{Red})=\frac{3}{5},\ \ P(\text{Black})=\frac{2}{5}
\displaystyle \text{Since the ball is replaced after each draw, the draws are independent.}
\displaystyle \text{(i) Probability of two red balls}
\displaystyle P(RR)=\frac{3}{5}\times\frac{3}{5}=\frac{9}{25}
\displaystyle \text{(ii) Probability of two black balls}
\displaystyle P(BB)=\frac{2}{5}\times\frac{2}{5}=\frac{4}{25}
\displaystyle \text{(iii) Probability of first red and second black}
\displaystyle P(RB)=\frac{3}{5}\times\frac{2}{5}=\frac{6}{25}

\displaystyle \textbf{Question 15: }~\text{Three cards are drawn with replacement from a well shuffled pack of cards.} \\ \text{Find the probability that the cards drawn are king, queen and jack.}
\displaystyle \text{Answer:}
\displaystyle P(\text{king})=\frac{4}{52}
\displaystyle P(\text{queen})=\frac{4}{52}
\displaystyle P(\text{jack})=\frac{4}{52}
\displaystyle \text{These cards can be drawn in }{}^{3}P_{3}\text{ ways.}
\displaystyle P(\text{king, queen and jack})=\frac{4}{52}\times\frac{4}{52}\times\frac{4}{52}\times{}^{3}P_{3}
\displaystyle =\frac{6}{2197}

\displaystyle \textbf{Question 16: }~\text{An article manufactured by a company consists of two parts }X\text{ and }Y.\\ \text{In } \text{the process of manufacture of the part }X,\ 9\text{ out of }100\text{ parts may be defective.} \\ \text{Similarly, }5\text{ out of }100\text{ are likely to be defective in the manufacture of the part }Y.\\\ \text{Calculate the probability that the assembled product will not be defective.}
\displaystyle \text{Answer:}
\displaystyle \text{Let:}
\displaystyle A=\text{Particle X is defective}
\displaystyle B=\text{Particle Y is defective}
\displaystyle \therefore P(A)=\frac{9}{100}
\displaystyle P(B)=\frac{5}{100}
\displaystyle \text{Required probability}=P(\overline{A}\cap \overline{B})
\displaystyle =P(\overline{A})\times P(\overline{B})
\displaystyle =[1-P(A)]\times[1-P(B)]
\displaystyle =\left[1-\frac{9}{100}\right]\times\left[1-\frac{5}{100}\right]
\displaystyle =\frac{91}{100}\times\frac{95}{100}
\displaystyle =0.91\times0.95
\displaystyle =0.8645

\displaystyle \textbf{Question 17: }~\text{The probability that }A\text{ hits a target is }\frac{1}{3}\text{ and the probability} \\ \text{that }B\text{ hits it, is }\frac{2}{5}.\ \text{What is the probability that the target will be hit, if each one of } \\ A\text{ and }B\text{ shoots at the target?}
\displaystyle \text{Answer:}
\displaystyle P(A)=P(\text{A hits target})=\frac{1}{3}
\displaystyle P(B)=P(\text{B hits target})=\frac{2}{5}
\displaystyle \text{Now,}
\displaystyle P(A\cup B)=P(\text{target will be hit by either A or B})
\displaystyle \Rightarrow P(A\cup B)=P(A)+P(B)-P(A\cap B)
\displaystyle \Rightarrow P(A\cup B)=P(A)+P(B)-P(A)P(B)\text{ [A and B are independent]}
\displaystyle \Rightarrow P(A\cup B)=\frac{1}{3}+\frac{2}{5}-\frac{1}{3}\times\frac{2}{5}
\displaystyle \Rightarrow P(A\cup B)=\frac{5+6-2}{15}
\displaystyle \Rightarrow P(A\cup B)=\frac{9}{15}
\displaystyle \Rightarrow P(A\cup B)=\frac{3}{5}

\displaystyle \textbf{Question 18: }~\text{An anti-aircraft gun can take a maximum of }4\text{ shots at an enemy plane} \\ \text{moving away from it. The probabilities of hitting the plane at the first, second, third} \\ \text{and fourth shot are }0.4,0.3,0.2\text{ and }0.1\text{ respectively. What is the probability that} \\ \text{the gun hits the plane?}
\displaystyle \text{Answer:}
\displaystyle P(\text{gun hits the plane})=1-P(\text{gun does not hit the plane})
\displaystyle \Rightarrow P(A)=1-P(\overline{A})
\displaystyle \text{Now,}
\displaystyle \Rightarrow P(\overline{A})=(1-0.4)(1-0.3)(1-0.2)(1-0.1)
\displaystyle =0.6\times0.7\times0.8\times0.9
\displaystyle =0.3024
\displaystyle \therefore P(A)=1-0.3024
\displaystyle =0.6976

\displaystyle \textbf{Question 19: }~\text{The odds against a certain event are }5\text{ to }2\text{ and the odds in favour} \\ \text{of another event, independent to the former are }6\text{ to }5.\ \text{Find the probability} \\ \text{that (i) at least one of the events will occur, and (ii) none of the events will occur.}
\displaystyle \text{Answer:}
\displaystyle \text{The odds against event }A\text{ are }5\text{ to }2.
\displaystyle P(A)=\frac{2}{5+2}=\frac{2}{7}
\displaystyle \text{The odds in favour of event }B\text{ are }6\text{ to }5.
\displaystyle P(B)=\frac{6}{6+5}=\frac{6}{11}
\displaystyle \text{(i) }P(\text{at least one event occurs})
\displaystyle =P(A\cup B)
\displaystyle =P(A)+P(B)-P(A\cap B)
\displaystyle =P(A)+P(B)-P(A)\times P(B)
\displaystyle =\frac{2}{7}+\frac{6}{11}-\frac{2}{7}\times\frac{6}{11}
\displaystyle =\frac{22+42}{77}-\frac{12}{77}
\displaystyle =\frac{22+42-12}{77}=\frac{52}{77}
\displaystyle \therefore P(A\cup B)=\frac{52}{77}
\displaystyle \text{(ii) }P(\text{none of the events occurs})
\displaystyle =1-P(A\cup B)
\displaystyle =1-\frac{52}{77}
\displaystyle =\frac{25}{77}

\displaystyle \textbf{Question 20: }~\text{A die is thrown thrice. Find the probability of getting an odd number} \\ \text{at least once.}
\displaystyle \text{Answer:}
\displaystyle P(\text{getting an odd number in one throw})=\frac{1}{2}
\displaystyle \text{Here, getting an odd number in three throws refers to 3 independent events.}
\displaystyle P(A)=P(B)=P(C)=\frac{1}{2}
\displaystyle P(A\cup B\cup C)=P(A)+P(B)+P(C)-[P(A\cap B)+P(B\cap C)+P(C\cap A)]+P(A\cap B\cap C)
\displaystyle =\frac{1}{2}+\frac{1}{2}+\frac{1}{2}-\left[\frac{1}{2}\times\frac{1}{2}+\frac{1}{2}\times\frac{1}{2}+\frac{1}{2}\times\frac{1}{2}\right]+\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}
\displaystyle =\frac{3}{2}-\frac{3}{4}+\frac{1}{8}
\displaystyle =\frac{12-6+1}{8}
\displaystyle =\frac{7}{8}

\displaystyle \textbf{Question 21: }~\text{Two balls are drawn at random with replacement from a box containing } \\ 10\text{ black and }8\text{ red balls. Find the probability that (i) both balls are red, (ii) first ball is} \\ \text{black and second is red, (iii) one of them is black and other is red.}
\displaystyle \text{Answer:}
\displaystyle \text{Total balls}=10\text{ black}+8\text{ red}=18\text{ balls}
\displaystyle P(\text{first red ball})=\frac{8}{18}
\displaystyle P(\text{second red ball})=\frac{8}{18}
\displaystyle P(\text{first ball is black})=\frac{10}{18}
\displaystyle P(\text{second ball is black})=\frac{10}{18}
\displaystyle \text{(i) }P(\text{two red balls})=\frac{8}{18}\times\frac{8}{18}
\displaystyle =\frac{16}{81}
\displaystyle \text{(ii) }P(\text{first ball is black and second is red})=\frac{10}{18}\times\frac{8}{18}
\displaystyle =\frac{20}{81}
\displaystyle \text{(iii) }P(\text{one of them is black and the other is red})=P(\text{first ball is red and second is black})+P(\text{first ball is black and second is red})
\displaystyle =\frac{8}{18}\times\frac{10}{18}+\frac{10}{18}\times\frac{8}{18}
\displaystyle =\frac{20}{81}+\frac{20}{81}
\displaystyle =\frac{40}{81}

\displaystyle \textbf{Question 22: }~\text{An urn contains }4\text{ red and }7\text{ black balls. Two balls are drawn at} \\ \text{random with replacement. Find the probability of getting (i) }2\text{ red balls,} \\ \text{(ii) }2\text{ black balls, }\text{(iii) one red and one black ball. \ [CBSE\ 2007]}
\displaystyle \text{Answer:}
\displaystyle \text{(i)  }
\displaystyle \text{Total balls}=4\text{ red}+7\text{ blue}=11\text{ balls}
\displaystyle P(\text{2 red balls})=P(\text{first ball is red})\times P(\text{second ball is red})
\displaystyle =\frac{4}{11}\times\frac{4}{11}
\displaystyle =\frac{16}{121}

\displaystyle \text{(ii)  }
\displaystyle \text{Total balls}=4\text{ red}+7\text{ blue}=11\text{ balls}
\displaystyle P(\text{2 blue balls})=P(\text{first ball is blue})\times P(\text{second ball is blue})
\displaystyle =\frac{7}{11}\times\frac{7}{11}
\displaystyle =\frac{49}{121}

\displaystyle \text{(iii)  }
\displaystyle \text{Total balls}=4\text{ red}+7\text{ blue}=11\text{ balls}
\displaystyle P(\text{one red and one blue})=P(\text{first red and second blue})+P(\text{first blue and second red})
\displaystyle =\frac{4}{11}\times\frac{7}{11}+\frac{7}{11}\times\frac{4}{11}
\displaystyle =\frac{28}{121}+\frac{28}{121}
\displaystyle =\frac{56}{121}
\displaystyle \text{Disclaimer: In the question, instead of black balls it should be blue balls.}

\displaystyle \textbf{Question 23: }~\text{The probabilities of two students }A\text{ and }B\text{ coming to the school in time} \\ \text{are }\frac{3}{7}\text{ and }\frac{5}{7}\text{ respectively. Assuming that the events, }A\text{ coming in time' and}B\text{ coming in} \\ \text{time are independent, find the probability of only one of them coming to the school in} \\ \text{time. Write at least one advantage of coming to school in time. \ [CBSE\ 2013]}
\displaystyle \text{Answer:}
\displaystyle P(\text{A coming in time})=\frac{3}{7}
\displaystyle P(\text{A not coming in time})=1-\frac{3}{7}=\frac{4}{7}
\displaystyle P(\text{B coming in time})=\frac{5}{7}
\displaystyle P(\text{B not coming in time})=1-\frac{5}{7}=\frac{2}{7}
\displaystyle P(\text{only one of A and B coming in time})=P(A)P(\overline{B})+P(\overline{A})P(B)
\displaystyle =\frac{3}{7}\times\frac{2}{7}+\frac{4}{7}\times\frac{5}{7}
\displaystyle =\frac{6}{49}+\frac{20}{49}
\displaystyle =\frac{26}{49}

\displaystyle \textbf{Question 24: }~\text{Two dice are thrown together and the total score is noted. The events } E, \\ F\text{ and }G\text{ are "a total }4\text{", "a total of }9\text{ or more", and "a total divisible by }5\text{", respectively.} \\ \text{Calculate }P(E),P(F)\text{ and }P(G)\text{ and decide which pairs of events, if any, are independent.}
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle S=\{(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),(3,1), \\ (3,2),(3,3),(3,4),(3,5),(3,6),(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),(5,1),(5,2), \\ (5,3),(5,4),(5,5),(5,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)\}
\displaystyle n(S)=36
\displaystyle E=\text{a total of }4=\{(1,3),(2,2),(3,1)\}\text{ i.e. }n(E)=3
\displaystyle F=\text{a total of }9\text{ or more}=\{(3,6),(4,5),(4,6),(5,4),(5,5),(5,6),(6,3),(6,4),(6,5),(6,6)\}\text{ i.e. }n(F)=10
\displaystyle G=\text{a total divisible by }5=\{(1,4),(2,3),(3,2),(4,1),(4,6),(5,5),(6,4)\}\text{ i.e. }n(G)=7
\displaystyle \text{Now,}
\displaystyle P(E)=\frac{3}{36}=\frac{1}{12}
\displaystyle P(F)=\frac{10}{36}=\frac{5}{18}
\displaystyle P(G)=\frac{7}{36}
\displaystyle \text{Also,}
\displaystyle E\cap F=\phi,\;E\cap G=\phi\text{ and }F\cap G=\{(4,6),(5,5),(6,4)\}\text{ i.e. }n(F\cap G)=3
\displaystyle \text{Since }P(F)\times P(G)=\frac{5}{18}\times\frac{7}{36}=\frac{35}{648}\neq\frac{3}{36}
\displaystyle \text{i.e. }P(F)\times P(G)\neq P(F\cap G)
\displaystyle \text{So, no pair is independent.}

\displaystyle \textbf{Question 25: }~\text{Let }A\text{ and }B\text{ be two independent events such that} \\ P(A)=p_1 \text{ and } P(B)=p_2.\ \text{Describe in words the events whose probabilities are:} \\ \text{(i) }p_1p_2\text{ (ii) }(1-p_1)p_2\text{ (iii) }1-(1-p_1)(1-p_2)\text{ (iv) }p_1+p_2=2p_1p_2. 
\displaystyle \text{Answer:}
\displaystyle \text{(i)  }
\displaystyle \text{As, }p_{1}p_{2}=P(A)\times P(B)
\displaystyle \text{And, A and B are independent events.}
\displaystyle \text{i.e. }P(A)\times P(B)=P(A\cap B)
\displaystyle \text{So, }P(A\cap B)=p_{1}p_{2}
\displaystyle \text{Hence, }p_{1}p_{2}=P(\text{A and B occur})

\displaystyle \text{(ii)  }
\displaystyle \text{As, }(1-p_{1})p_{2}=[1-P(A)]\times P(B)=P(\overline{A})\times P(B)
\displaystyle \text{And, A and B are independent events.}
\displaystyle \text{i.e. }P(\overline{A})\times P(B)=P(\overline{A}\cap B)
\displaystyle \text{So, }P(\overline{A}\cap B)=(1-p_{1})p_{2}
\displaystyle \text{Hence, }(1-p_{1})p_{2}=P(\text{A does not occur, but B occurs})

\displaystyle \text{(iii)  }
\displaystyle \text{As, }1-(1-p_{1})(1-p_{2})=1-[1-P(A)]\times[1-P(B)]
\displaystyle =1-P(\overline{A})\times P(\overline{B})
\displaystyle \text{And, A and B are independent events.}
\displaystyle \text{i.e. }P(\overline{A})\times P(\overline{B})=P(\overline{A}\cap \overline{B})
\displaystyle \Rightarrow 1-(1-p_{1})(1-p_{2})=1-P(\overline{A}\cap \overline{B})=P(A\cup B)
\displaystyle \text{So, }P(A\cup B)=1-(1-p_{1})(1-p_{2})
\displaystyle \text{Hence, }1-(1-p_{1})(1-p_{2})=P(\text{at least one of A and B occurs})

\displaystyle \text{(iv)  }
\displaystyle \text{As, }p_{1}+p_{2}-2p_{1}p_{2}=(p_{1}-p_{1}p_{2})+(p_{2}-p_{1}p_{2})
\displaystyle =[P(A)-P(A)\times P(B)]+[P(B)-P(A)\times P(B)]
\displaystyle \text{And, A and B are independent events.}
\displaystyle \text{i.e. }P(A)\times P(B)=P(A\cap B)
\displaystyle \Rightarrow p_{1}+p_{2}-2p_{1}p_{2}=[P(A)-P(A\cap B)]+[P(B)-P(A\cap B)]
\displaystyle =P(\text{only A})+P(\text{only B})
\displaystyle \text{So, }P(\text{only A})+P(\text{only B})=p_{1}+p_{2}-2p_{1}p_{2}
\displaystyle \text{Hence, }p_{1}+p_{2}-2p_{1}p_{2}=P(\text{Exactly one of A and B occurs})

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