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Probability

Question
CBSEENMA12033749

Bag I contains 3 red and 4 black balls and bag II contains 4 red and 5 black balls. One ball is transferred from bag I to bag II, then a ball is drawn from bag II, the ball so drawn is found to be red in colour. Find the probability that the transferred ball is black.

Solution

Bag I contains 3 red and 4 black balls.
Bag II contain 4 red and 5 black balls.
Let E1 : Event that a red ball is drawn from bag I
E: Event that a black ball is drawn from bag I
therefore space space space space space straight P left parenthesis straight E subscript 1 right parenthesis space equals space 3 over 7 comma space space straight P left parenthesis straight E subscript 2 right parenthesis space equals space 4 over 7

After transferring a red ball from bag I to bag II, the bag II will have 5 red and 5 black balls.
Let A be the event of drawing red ball
         therefore space space space space straight P left parenthesis straight A space left enclose straight E subscript 1 end enclose right parenthesis space equals space 5 over 10 space equals 1 half
Further when a black ball is transferred from bag I to bag II, it will contain 4 red and 6 black balls.
straight P space open parentheses straight A space left enclose straight E subscript 2 end enclose close parentheses space equals space 4 over 10 space equals space 2 over 5
Required probability  = straight P left parenthesis straight E subscript 2 space left enclose space straight A end enclose right parenthesis space equals space fraction numerator straight P left parenthesis straight E subscript 2 right parenthesis thin space straight P left parenthesis straight A space left enclose space straight E subscript 2 end enclose right parenthesis over denominator straight P left parenthesis straight E subscript 1 right parenthesis space straight P left parenthesis straight A space left enclose straight E subscript 1 end enclose right parenthesis space plus space straight P left parenthesis straight E subscript 2 right parenthesis thin space straight P left parenthesis straight A space left enclose straight E subscript 2 end enclose right parenthesis end fraction
space equals space fraction numerator begin display style 4 over 7 end style cross times begin display style 2 over 5 end style over denominator begin display style 3 over 7 end style cross times begin display style 1 half end style plus begin display style 4 over 7 end style cross times begin display style 2 over 5 end style end fraction space equals space fraction numerator begin display style 8 over 35 end style over denominator begin display style 13 over 14 end style plus begin display style 8 over 35 end style end fraction space equals space fraction numerator begin display style 8 over 35 end style over denominator begin display style fraction numerator 15 plus 16 over denominator 70 end fraction end style end fraction space equals 8 over 35 cross times 70 over 31 equals 16 over 31.

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