15:I[8216,["972","static/chunks/972-b28da0de92681928.js","47","static/chunks/47-a7f06e9eda1ac756.js","115","static/chunks/app/%5Bexam%5D/questions/%5Byear%5D/%5Bsubject%5D/%5Bchapter%5D/%5BquestionId%5D/page-7b6f0a12b9cc0206.js"],"BackButton"] 16:I[1553,["972","static/chunks/972-b28da0de92681928.js","47","static/chunks/47-a7f06e9eda1ac756.js","115","static/chunks/app/%5Bexam%5D/questions/%5Byear%5D/%5Bsubject%5D/%5Bchapter%5D/%5BquestionId%5D/page-7b6f0a12b9cc0206.js"],"QuestionViewer"] 17:T4c1,

We have, $$P(A \cap B)=\frac{1}{12}$$

Since, $$A$$ and $$B$$ are two independent events.

$$\therefore \quad P(A \cap B)=P(A) P(B)=\frac{1}{12}$$ ..... (i)

$$\begin{array}{l} \text { Also, } & P\left(A^{\prime} \cap B^{\prime}\right)=\frac{1}{2} \\ \Rightarrow & P(\overline{A \cup B})=\frac{1}{2} \\ \Rightarrow & 1-P(A \cup B)=\frac{1}{2} \end{array}$$

$$\begin{aligned} \Rightarrow & \quad P(A \cup B)=\frac{1}{2} \\ \Rightarrow & \quad P(A)+P(B)-P(A \cap B)=\frac{1}{2} \end{aligned}$$

$$\Rightarrow \quad P(A)+P(B)-\frac{1}{12}=\frac{1}{2} \quad \text { [Using Eq. (i)] }$$

$$\Rightarrow \quad P(A)+P(B)=\frac{7}{12}$$ ..... (ii)

Now, from Eqs. (i) and (ii), we get

$$P(A)\left\{\frac{7}{12}-P(A)\right\}=\frac{1}{12}$$

$$\begin{array}{lr} \Rightarrow & \frac{7}{12} P(A)-P(A)^2=\frac{1}{12} \\ \Rightarrow & P(A)^2-\frac{7}{12} P(A)+\frac{1}{12}=0 \\ \Rightarrow & 12 P(A)^2-7 P(A)+1=0 \\ \Rightarrow & (3 P(A)-1)(4 P(A)-1)=0 \\ \Rightarrow & P(A)=\frac{1}{3} \text { or } P(A)=\frac{1}{4} \end{array}$$

$$\therefore$$ When $$P(A)=\frac{1}{3}$$, then $$P(B)=\frac{1}{4}$$

and when $$P(A)=\frac{1}{4}$$, then $$P(B)=\frac{1}{3}$$

Probability

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