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  • Explain Solution R.D.Sharma Class 12 Chapter 30 Probability Exercise Multiple Choice Questions Question 60 Maths Textbook Solution.

Explain Solution R.D.Sharma Class 12 Chapter 30 Probability  Exercise Multiple Choice Questions  Question 60 Maths Textbook Solution.

Answers (1)

Answer:(a)

Hint: You must know the rules of finding probability.

Given: A and B are independent such that

\begin{aligned} &0<P(A)<1 \\ &0<P(B)<1 \end{aligned}

Solution: If A and B are independent

P(A \cap B)=P(A) \times P(B)

Now,

\begin{aligned} &P(A \cap \bar{B})=P(A)-P(A \cap B) \\ &=P(A)-P(A) P(B) \\ &=P(A)\{1-P(B)\} \\ &=P(A) P(\bar{B}) \end{aligned}

Hence \mathrm{A} and \overline{\mathrm{B}} are indepemdent

Similarly,

\begin{aligned} &P(\bar{A} \cap \bar{B})=1-P(A \cup B) \\ &=1-P(A) \times P(B)-P(A \cap B) \\ &=1-P(A) \times P(B)-P(A) P(B) \\ &=1(1-P(A))-P(B)\{1-P(A)\} \\ &=\{1-P(A)\}-\{1-P(B)\} \\ &=P(\bar{A})-P(\bar{B}) \end{aligned}

\text { Hence } \overline{\mathrm{A}} \text { and } \overline{\mathrm{B}} \text { are independent }

\begin{aligned} &P\left(\frac{A}{B}\right)+P\left(\frac{A}{B}\right)=\frac{P(A \cap B)}{P(B)}+\frac{P(\bar{A} \cap B)}{P(B)} \\ &=\frac{P(A) P(B)+P(\bar{A}) P(B)}{P(B)} \\ &\text { and } 1-P(A)=P(\bar{A}) \\ &\therefore 1=P(A)+P(\bar{A}) \\ &\therefore \frac{P(A \cap B)}{P(B)}+\frac{P(\bar{A} \cap B)}{P(B)} \\ &=\frac{P(A) P(B)+P(\bar{A}) P(B)}{P(B)} \\ &=\frac{P(B)(P(A)+P(\bar{A}))}{P(B)} \\ &=1 \end{aligned}

\therefore Option \mathrm{A} is incorrect because \mathrm{P}(\mathrm{A} \cap \mathrm{B})=\mathrm{P}(A) P(B) \neq 0

Since \mathrm{P}(A)>0 and P(B)>0

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