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  • Need solution for RD sharma maths class 12 chapter 30 probablity exercise Very short answer question, question 18

Need solution for RD sharma maths class 12 chapter 30 probablity exercise Very short answer question, question 18

Answers (1)

The value of p is \frac{5}{12} or \frac{1}{3}

Hint:

If A and B are independent then,

P(A\cap B)=P(A).P(B)  and

P(exactly one of the two events occurs) = P(A)+P(B) - 2P(A \cap B)

Given:

A and B are independent events such that

P(A) = p

P(B) = 2p

P(exactly one of A and B occurs) = \frac{5}{9}

Explanation:

It is given that

P(exactly one of A and B occurs) = \frac{5}{9}
\begin{aligned} &\Rightarrow \frac{5}{9}=P(A)+P(B)-2 P(A \cap B) \\ &=p+2 p-2 \cdot P(A) \cdot P(B) \\ \end{aligned}                                                         [ since,PA ∩B=PAPB] 

⇒\begin{aligned} &\Rightarrow \frac{5}{9}=p+2 p-2 \cdot p \cdot 2 p \\ &\Rightarrow \frac{5}{9}=3 p-4 p^{2} \\ &\Rightarrow 27 p-36 p^{2}=5 \\ &\Rightarrow 36 p^{2}-27 p+5=0 \\ \end{aligned}

Using quadratic formula, we have

\begin{aligned} p &=-\frac{(-27) \pm \sqrt{(-27)^{2}-4(36) 5}}{2 \times 36} \\ \end{aligned}

\begin{aligned} &=\frac{27 \pm \sqrt{729-720}}{72} \\ &=\frac{27 \pm \sqrt{9}}{72} \\ &=\frac{27 \pm 3}{72} \\ \end{aligned}

\begin{aligned} &\Rightarrow p=\frac{27+3}{72} \text { or } \quad p=\frac{27-3}{72} \\ &\Rightarrow p=\frac{30}{72} \quad \text { or } p=\frac{24}{72}\\ &\Rightarrow p=\frac{5}{12} \quad \text { or } p=\frac{1}{3} \end{aligned}

Hence, the value of p is  \frac{5}{12} or \frac{1}{3}

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