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Please solve RD Sharma class 12 chapter Probability exercise 30.6 question 1 maths textbook solution

Answers (1)

Answer:

 \frac{39}{88}

Hint:

 We use probability formula

Given:

 A bag A contains 5 white and 6 black balls. Another bag B contains 4 white and 3 black balls.

Solution:

Let

E1=A white ball is transfered bag A to bag B,then drawing the black ball
E2=A black ball is transfered bag A to bag B,then drawing the black ball

A = Ball draw is back

\begin{aligned} &\text { Probability of an event }=\frac{\text { No. of outcomes }}{ \text { Total outcomes }}\\ &P(E_1)=\frac{5}{11} \qquad ; \qquad P(E_2)=\frac{6}{11}\\ &P\left (\frac{A}{E_1} \right )=\frac{3}{8} \qquad ; \qquad P\left (\frac{A}{E_2} \right )=\frac{4}{8} \end{aligned}

Using total probabilty theorem

\begin{aligned} &\text { required probability }=P(A)=P(E_1)\times P\left (\frac{A}{E_1} \right )+ P(E_2)\times P\left (\frac{A}{E_2} \right ) \\ &P(A)=\frac{5}{11}\times \frac{3}{8}+\frac{6}{11}\times \frac{4}{8}\\ &=\frac{15}{88}+\frac{24}{88}\\ &=\frac{15+24}{88}=\frac{39}{88} \end{aligned}

Posted by

Gurleen Kaur

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