Q78 / 160 · AP EAMCET 2024 (19 May Shift 2) · Mathematics

There are 2 bags each containing 3 white and 5 black balls and 4 bags each containing 6 white and 4 black balls. If a ball drawn randomly from a bag is found to be black, then the probability that this ball is from the first set of bags is

  • Correct answer
    A
    $\frac{25}{57}$
  • B
    $\frac{25}{41}$
  • C
    $\frac{2}{5}$
  • D
    $\frac{3}{5}$

Solution

Let, $B_1=$ Bags of first group, $B_2=$ Bags of second group $P\left(B_1\right)=\frac{2}{6}=\frac{1}{3}, P\left(B_2\right)=\frac{4}{6}=\frac{2}{3}$ and let $B=$ the event that the ball is black $P\left(\frac{B}{B_1}\right)=\frac{5}{8}, P\left(\frac{B}{B_2}\right)=\frac{2}{5}$ Now, $P\left(\frac{B_1}{B}\right)=\frac{P\left(B_1\right) P\left(\frac{B}{B_1}\right)}{P\left(B_1\right) P\left(\frac{B}{B_1}\right)+P\left(B_2\right) P\left(\frac{B}{B_2}\right)}$ $=\frac{\frac{1}{3} \times \frac{5}{8}}{\frac{1}{3} \times \frac{5}{8}+\frac{2}{3} \times \frac{2}{5}}=\frac{\frac{5}{8}}{\frac{5}{8}+\frac{4}{5}}=\frac{25}{57}$

Hence, the correct answer is option: 'A'.

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