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

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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