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JEE Main & Advanced Mathematics Probability Question Bank Conditional probability bayes theorem

  • question_answer
    Bag A contains 4 green and 3 red balls and bag B contains 4 red and 3 green balls. One bag is taken at random and a ball is drawn and noted it is green. The probability that it comes bag B       [DCE 2005]

    A)                 \[\frac{2}{7}\]       

    B)                 \[\frac{2}{3}\]

    C)                 \[\frac{3}{7}\]       

    D)                 \[\frac{1}{3}\]

    Correct Answer: C

    Solution :

                      It is based on Baye?s theorem.            Probability of picked bag A \[P(A)=\frac{1}{2}\]                    Probability of picked bag B \[P(B)=\frac{1}{2}\]                    Probability of green ball picked from bag A                                  \[=P(A).P\left( \frac{G}{A} \right)\]\[=\frac{1}{2}\times \frac{4}{7}=\frac{2}{7}\]                    Probability of green ball picked from bag B                                 \[=P(B).P\left( \frac{G}{B} \right)\]\[=\frac{1}{2}\times \frac{3}{7}=\frac{3}{14}\]                    Total probability of green ball = \[\frac{2}{7}+\frac{3}{14}=\frac{1}{2}\]                    Probability of fact that green ball is drawn from bag B                                  \[P\left( \frac{G}{B} \right)=\frac{P(B)P\left( \frac{G}{B} \right)}{P(A)P\left( \frac{G}{A} \right)+P(B)P\left( \frac{G}{B} \right)}=\frac{\frac{1}{2}\times \frac{3}{7}}{\frac{1}{2}\times \frac{4}{7}+\frac{1}{2}\times \frac{3}{7}}=\frac{3}{7}\].


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