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Please solve RD Sharma class 12 chapter Probability exercise 30.7 question 29 maths textbook solution

Answers (1)

Answer:

 \frac{156}{947}

Hint:

Use baye’s theorem. 

Given:

Box

Colour

Black

White

Red

Blue

I

3

4

5

6

II

2

2

2

2

III

1

2

3

1

IV

4

3

1

5

A box is selected at random and ball is randomly drawn from the selected box. The color the ball is black. What is the probability that ball drawn is from the box III.

Solution:

Let A: Event that is black ball is selected

E1 : Event that the ball i selecting from box I

E2 : Event that the ball is selected for box II

E3 : Event that the ball is selected for box III

E4 : Event that the ball is selected for box IV

\begin{aligned} &P(E_1)=\text { Probability that ball drawn is from box I }=\frac{1}{4}\\ &P(E_2)=\text { Probability that ball drawn is from box II }=\frac{1}{4}\\ &P(E_3)=\text { Probability that ball drawn is from box III }=\frac{1}{4}\\ &P(E_4)=\text { Probability that ball drawn is from box IV }=\frac{1}{4}\\ &P\left ( \frac{A}{E_1} \right )=\frac{\text { no.of balls }}{\text { Total no.of balls in the box }}=\frac{3}{18}\\ &P\left ( \frac{A}{E_2} \right )=\frac{\text { no.of balls }}{\text { Total no.of balls in the box }}=\frac{2}{8}=\frac{1}{4}\\ &P\left ( \frac{A}{E_3} \right )=\frac{1}{7}\\ &P\left ( \frac{A}{E_4} \right )=\frac{4}{13}\\ \end{aligned}

Using Baye’s theorem we get

\begin{aligned} &P\left (\frac{E_3}{A} \right )=\frac{P(E_3).P\left ( \frac{A}{E_3} \right )}{P(E_1)\times P\left ( \frac{A}{E_1} \right )+P(E_2)\times P\left ( \frac{A}{E_2} \right )+P(E_3)\times P\left ( \frac{A}{E_3} \right )+P(E_4)\times P\left ( \frac{A}{E_4} \right )}\\ &=\frac{{\frac{1}{4}\times \frac{1}{7} }}{\frac{1}{4}\times \frac{1}{6}+\frac{1}{4}\times \frac{1}{4}+\frac{1}{4}\times \frac{1}{7}+\frac{1}{4}\times \frac{4}{13}}\\ &=\frac{\frac{1}{28}}{\frac{1}{24}+\frac{1}{16}+\frac{1}{28}+\frac{1}{13}}\\ &=\frac{156}{611+336}\\ &=\frac{156}{947} \end{aligned}

Posted by

Gurleen Kaur

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