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  • There are two bags A and B. Bag A contains 3 white and 4 red balls whereas bag B contains 4 white and 3 red balls. Three balls are drawn at random (without replacement) from one of the bags and are found to be two white and one red. Find the probabilit

There are two bags A and B. Bag A contains 3 white and 4 red balls whereas bag B contains 4 white and 3 red balls. Three balls are drawn at random (without replacement) from one of the bags and are found to be two white and one red. Find the probability that these were drawn from bag B.

 

 

 

 
 
 
 
 

Answers (1)

Let E1: Selecting bag A, E2: Selecting bag B and A: getting 2 white and 1 red balls out of 3 drawn (without replacement)

Here,    \mathrm{P(E_1) = P(E_2)=\frac{1}{2}\quad P(A|E_1) = \frac{^3C_2\times ^4C_1}{^7C_3}}

                                                                                    \mathrm{=\frac{12}{35}}

\mathrm{P(A|E_2) = \frac{^4C_2\times ^3C_1}{^7C_3} = \frac{18}{35}}

Now by Baye's Theorem \mathrm{P(E_2|A) = \frac{P(E_2)P(A|E_2)}{P(E_1)P(A|E_1) + P(E_2)P(A|E_2)}}

\mathrm{P(E_2|A) = \frac{\frac{1}{2}\times \frac{18}{25}}{\frac{1}{2}\times \frac{12}{25}\times \frac{1}{2}\times \frac{18}{35}} = \frac{18}{30} \ or \ \frac{3}{5}}

Posted by

Ravindra Pindel

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