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  • Provide Solution For R .D. Sharma Maths Class 12 Chapter 30 Probaility Exercise 30.4 Question 8 Maths Textbook Solution.

Provide Solution For  R .D. Sharma Maths Class 12 Chapter 30 Probaility  Exercise 30.4 Question 8 Maths Textbook Solution.

Answers (1)

Answer: y=\frac{4}{5}  or  \frac{1}{6} 

Hint: Form events as an equation

Given:

            P\left ( \bar{A}\cap B \right )=\frac{2}{15}

            P\left ( A\cap \bar{B} \right )=\frac{1}{16}

P\left ( A \right ) & P\left ( B \right )are independent events

Solution: As we know, Two events are said to be independent if the product of the events are equal to their intersection. i.e. P\left ( A\cap B \right )=P\left ( A \right )P\left ( B \right )

Let,

\begin{aligned} &\mathrm{P}(\mathrm{A})=\mathrm{x} \\ &\mathrm{P}(\mathrm{B})=\mathrm{y} \\ &\mathrm{P}(\overline{\mathrm{A}} \cap \mathrm{B})=\frac{2}{15} \\ &\mathrm{P}(\overline{\mathrm{A}}) \mathrm{P}(\mathrm{B})=\frac{2}{15} \\ &(1-\mathrm{x}) \mathrm{y}=\frac{2}{15} \end{aligned}                                        ............(1)

Also

\begin{aligned} &P(A \cap \bar{B})=\frac{1}{6} \\ &P(A) P(\bar{B})=\frac{1}{6} \\ &x(1-y)=\frac{1}{6} \end{aligned}                                         ........(2)

Subtracting (1) from (2)

\begin{aligned} &y-x y=\frac{2}{15} \\ &x-x y=\frac{1}{6} \\ &y-x=\frac{2}{15}-\frac{1}{6} \\ &y-x=\frac{-3}{90} \\ &y+\frac{1}{30}=x \end{aligned}

Substitute value of x in (2)

\begin{aligned} &\left(y+\frac{1}{30}\right)(1-y)=\frac{1}{6} \\ &30 y^{2}-29 y+4=0 \end{aligned}

Solve using quadratic formula

\begin{aligned} &a=30, b=-29, c=4 \\ &\alpha=\frac{-b+\sqrt{b^{2}-4 a c}}{2 a} \\ &\quad=\frac{-(-29)+\sqrt{841-480}}{60} \\ \end{aligned}

\begin{aligned} &\alpha=\frac{29+19}{60} \\ &\alpha=\frac{48}{60} \quad \\ &\alpha=\frac{4}{5} \end{aligned}

Similarly \begin{aligned} \beta &=\frac{-b-\sqrt{b^{2}-4 a c}}{2 a} \\ \end{aligned}

                    \begin{aligned} &=\frac{29-19}{60} \\ &=\frac{10}{60} \\ &=\frac{1}{6} \\ &y=\frac{4}{5} \text { or } \frac{1}{6} \end{aligned}

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