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Provide solution for RD Sharma Maths Class12 Chapter 30 Probability Exercise Fill in the Blanks Question, Question 2.

Answers (1)

Answer:   \frac{5}{12},\frac{1}{3}

Given:

If A and B are independent events such-that

PA=p,

P(B)=2pand

P( Exactly one ofA_1B=\frac{5}{9}) then

p=?

Hint:

P(A \cap A B)=P(A) P(B) if A and B are independent.

Explanation:

As A and B ane independent events.
So P(A \cap A B)=P(A) P(B)       - (1)

Now
P( Exadly one of A and B occurs )= \frac{5}{9}
⇒P( only A)+P( onlyB)= \frac{5}{9}

\begin{aligned} &\Rightarrow[\mathrm{P}(\mathrm{A})-\mathrm{P}(\mathrm{A} \cap \mathrm{B})]+[\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A} \cap \mathrm{B})]=\frac{5}{9} \\ &\Rightarrow[\mathrm{P}(\mathrm{A})-\mathrm{P}(\mathrm{A}) \mathrm{P}(\mathrm{B})]+[\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A}) \mathrm{P}(\mathrm{B}))]=\frac{5}{9} \\ &\Rightarrow\left[\mathrm{P}(\mathrm{A}) \times(1-\mathrm{P}(\mathrm{B}))+\mathrm{P}(\mathrm{B}) \times[1-\mathrm{P}(\mathrm{A})]=\frac{5}{9}\right. \\ &\Rightarrow \mathrm{p}(1-2 \mathrm{p})+2 \mathrm{p}(1-\mathrm{p})=\frac{5}{9} \\ &\Rightarrow \mathrm{p}-2 \mathrm{p}^{2}+2 \mathrm{p}-2 \mathrm{p}^{2}=\frac{5}{9} \\ &\Rightarrow 3 \mathrm{p}-4 \mathrm{p}^{2}=\frac{5}{9} \\ &\Rightarrow 27 \mathrm{p}-36 \mathrm{p}^{2}=5 \\ &\Rightarrow 36 \mathrm{p}^{2}-27 \mathrm{p}+5=0 \end{aligned}

So using quadratic formula, \mu.

\begin{aligned} &p=\frac{-(-27) \pm \sqrt{(-27)^{2}-4 \times 36 \times 5}}{2 \times 36} \\ &=\frac{27 \pm \sqrt{729-720}}{72} \\ &=\frac{27 \pm \sqrt{9}}{72} \\ &=\frac{27 \pm 3}{72} \end{aligned}

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

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