CBSE Class 12-science Answered

There are three boxes , first one containing 1 white , 2 red and 3 black balls; the second one containing 2 white , 3 red and 1 black ball and third one containing 3 white, 1 red and 2 black balls. A box is choosen at random and from it two balls are drawn at random. One ball is red and the other is white . what is the probability that they come from second box .

Asked by haroonrashidgkp | 05 Nov, 2018, 09:36: PM

Expert Answer

To solve this question, We have to use Bayes Theorem to get reverse probability.

Let first ball drawn is red and second ball drawn is white.

Probability of choosing a box at random = 1/3.

probabilty of drawing a red ball from box-1 = p(R|B1) = 1/3

probabilty of drawing a red ball from box-2 = p(R|B2) = 1/2

probabilty of drawing a red ball from box-3 = p(R|B3) = 1/6

probability of getting the ball from box-2 if red ball is selected is given by

begin mathsize 12px style P left parenthesis B 2 vertical line R right parenthesis space space equals space fraction numerator P left parenthesis B 2 right parenthesis cross times p left parenthesis R vertical line B 2 right parenthesis over denominator p left parenthesis B 1 right parenthesis cross times p left parenthesis R vertical line B 1 right parenthesis plus p left parenthesis B 2 right parenthesis cross times p left parenthesis R vertical line B 2 right parenthesis plus p left parenthesis B 3 right parenthesis cross times p left parenthesis R vertical line B 3 right parenthesis end fraction space equals space fraction numerator begin display style 1 third end style cross times begin display style 1 half end style over denominator begin display style 1 third end style cross times begin display style 1 third end style plus begin display style 1 third end style cross times begin display style 1 half end style plus begin display style 1 third end style cross times begin display style 1 over 6 end style end fraction equals 1 half end style

...........................(1)

Now let us consider white ball is drawn second time

probabilty of drawing a white ball from box-1 = p(W|B1) = 1/6

probabilty of drawing a white ball from box-2 = p(W|B2) = 2/5 ( note:- one ball is taken out already from box-2)

probabilty of drawing a white ball from box-3 = p(W|B3) = 1/2

probability of getting the ball from box-2 if white ball is selected is given by

begin mathsize 12px style P left parenthesis B 2 vertical line W right parenthesis space space equals space fraction numerator P left parenthesis B 2 right parenthesis cross times p left parenthesis W vertical line B 2 right parenthesis over denominator p left parenthesis B 1 right parenthesis cross times p left parenthesis W vertical line B 1 right parenthesis plus p left parenthesis B 2 right parenthesis cross times p left parenthesis W vertical line B 2 right parenthesis plus p left parenthesis B 3 right parenthesis cross times p left parenthesis W vertical line B 3 right parenthesis end fraction space equals space fraction numerator begin display style 1 third end style cross times begin display style 2 over 5 end style over denominator begin display style 1 third end style cross times begin display style 1 over 6 end style plus begin display style 1 third end style cross times begin display style 2 over 5 end style plus begin display style 1 third end style cross times begin display style 1 half end style end fraction equals 3 over 8 end style

...........................(2)

hence probability of getting the balls from box-2 if red ball selected first time and white ball second time = (1/2)×(3/8) = 3/16 ...................(3)

Now we consider the event, first selected ball is white and the second one red

probabilty of drawing a white ball from box-1 = p(W|B1) = 1/6

probabilty of drawing a white ball from box-2 = p(W|B2) = 1/3

probabilty of drawing a white ball from box-3 = p(W|B3) = 1/2

probability of getting the ball from box-2 if white ball is selected is given by

begin mathsize 12px style P left parenthesis B 2 vertical line W right parenthesis space space equals space fraction numerator P left parenthesis B 2 right parenthesis cross times p left parenthesis W vertical line B 2 right parenthesis over denominator p left parenthesis B 1 right parenthesis cross times p left parenthesis W vertical line B 1 right parenthesis plus p left parenthesis B 2 right parenthesis cross times p left parenthesis W vertical line B 2 right parenthesis plus p left parenthesis B 3 right parenthesis cross times p left parenthesis W vertical line B 3 right parenthesis end fraction space equals space fraction numerator begin display style 1 third end style cross times begin display style 1 third end style over denominator begin display style 1 third end style cross times begin display style 1 over 6 end style plus begin display style 1 third end style cross times begin display style 1 third end style plus begin display style 1 third end style cross times begin display style 1 half end style end fraction equals 1 third end style

...........................(4)

Now let us consider red ball is drawn second time

probabilty of drawing a red ball from box-1 = p(R|B1) = 1/3

probabilty of drawing a red ball from box-2 = p(R|B2) = 3/5

probabilty of drawing a red ball from box-3 = p(R|B3) = 1/6

probability of getting the ball from box-2 if red ball is selected second time is given by

begin mathsize 12px style P left parenthesis B 2 vertical line R right parenthesis space space equals space fraction numerator P left parenthesis B 2 right parenthesis cross times p left parenthesis R vertical line B 2 right parenthesis over denominator p left parenthesis B 1 right parenthesis cross times p left parenthesis R vertical line B 1 right parenthesis plus p left parenthesis B 2 right parenthesis cross times p left parenthesis R vertical line B 2 right parenthesis plus p left parenthesis B 3 right parenthesis cross times p left parenthesis R vertical line B 3 right parenthesis end fraction space equals space fraction numerator begin display style 1 third end style cross times begin display style 3 over 5 end style over denominator begin display style 1 third end style cross times begin display style 1 third end style plus begin display style 1 third end style cross times begin display style 3 over 5 end style plus begin display style 1 third end style cross times begin display style 1 over 6 end style end fraction equals 6 over 11 end style

...........................(5)

hence probability of getting the balls from box-2 if white ball is selected first time and red ball is selected second time = (1/3)×(6/11) = 2/11 ...................(6)

using (3) and (6), probability of getting two balls from box-2 if one of then red and other one is white = (3/16) + (2/11) = ( 65/176 ) = 0.37

Answered by Thiyagarajan K | 17 Nov, 2018, 12:17: PM