Need solution for RD Sharma Maths Class12 Chapter 30 Probability Exercise Fill in the Blanks Question, Question 15.

Answers (1)

Answer:  P(A)+P(B)

Given:

Ret A1B,C be pair wise independent events with P(C)>0
and P(A\cap B\cap C)=0. If P\left(\frac{A \cap B}{C}\right)=1-x, then x=…

Hint:

Using P\left(\frac{A}{B}\right)=\frac{P(A \cap B)}{P(B)}

Explanation:

Let A,B,C be pair wise independent events.
\begin{aligned} &\Rightarrow \mathrm{P}(\mathrm{A} \cap \mathrm{C})=\mathrm{P}(\mathrm{A}) \cdot \mathrm{P}(\mathrm{C}) \\ &\mathrm{P}(\mathrm{B} \cap \mathrm{C})=\mathrm{P}(\mathrm{B}) \cdot \mathrm{P}(\mathrm{C})-(*) \end{aligned}

Given
So    P(A\cap B\cap C)=0                          

\begin{aligned} &\mathrm{P}(\bar{A} \cap \overline{\mathrm{B}} \cap \mathrm{C})=0 \\ &P\left(\frac{\overline{\mathrm{A}} \overline{\mathrm{B}}}{\mathrm{C}}\right)=\frac{\mathrm{P}[(\overline{\mathrm{A}} \cap \overline{\mathrm{B}}) \cap \mathrm{C}]}{\mathrm{P}(\mathrm{C})} \\ &=\frac{\mathrm{P}(\mathrm{C})-\mathrm{P}(\mathrm{A} \cap \mathrm{C})-\mathrm{P}(\mathrm{B} \cap \mathrm{C})+\mathrm{P}(\mathrm{A} \cap \mathrm{B} \cap \mathrm{C})}{\mathrm{P}(\mathrm{C})} \\ &=\frac{\mathrm{P}(\mathrm{C})-\mathrm{P}(\mathrm{A}) \mathrm{P}(\mathrm{C})-\mathrm{P}(\mathrm{B}) \mathrm{P}(\mathrm{C})+0}{\mathrm{P}(\mathrm{C})} \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad[\because \text { Of }(*)] \end{aligned}

\begin{aligned} &=\frac{P(C) \times[1-P(A)-P(B)]}{P(C)} \\ &=1-P(A)-P(B) \\ &=1-[P(A)+P(B)]-(1) \end{aligned}

Also,

\mathrm{P}\left(\frac{\mathrm{A} \cap \mathrm{B}}{\mathrm{c}}\right)=1-\mathrm{x}-(2)

From (1) and (2),
We get x=P(A)+P(B)

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