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Need solution for RD Sharma maths class 12 chapter Probability exercise 30.7 question 31

Answers (1)

Answer:

 \frac{18}{133}

Hint:

 Use Baye’s theorem.

Given:

 Bag A contain 3 red, 5 black, white bag contain 4 red and 4 black. Two ball are transferred at random from bag A to bag B and then a ball B drawn from bag B at random..

Solution:

2 ball drawn from bag A could be both red probability =

\frac{3C_2}{8C_2}

2 one red and one black probability =

\frac{3\times 5}{8C_2}

3 both black probability =

\frac{8C_2}{5C_2}

The no of ball in bag B in each case would be

1. 6 red 4 black = probability pick red =

 \frac{6}{10}

2. 5 red 5 black = probability pick red =

\frac{5}{10}

3. 4 red 6 black = probability pick red =

\frac{4}{10}

Probability of two red ball transfer under the condition that red ball found

Using Baye’s theorem we get

=\frac{\frac{3C_2}{8C_2}\times \frac{6}{10}}{\frac{3C_2}{8C_2}\times \frac{6}{10}+\frac{3\times 5}{8C_2}\times \frac{5}{10}+\frac{3C_2}{5C_2}\times \frac{4}{10}}\\

=\frac{18}{133}

Posted by

Gurleen Kaur

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