Q77 / 160 · AP EAMCET 2023 (15 May Shift 1) · Mathematics

Bag $B_1$ contains 4 white and 2 black balls. Bag $B_2$ contains 3 white and 4 black balls. A bag is chosen at random and a ball is drawn from it at random, then the probability that the ball drawn is white, is

  • A
    $\frac{1}{42}$
  • B
    $\frac{42}{32}$
  • C
    $\frac{33}{42}$
  • Correct answer
    D
    $\frac{23}{42}$

Solution

Let $P\left(B_1\right), P\left(B_2\right)$ and $P(W)$ are the probabilty of selecting bag 1 , bag 2 and white bag respectively. $$ P\left(B_1\right)=\frac{1}{2}=P\left(B_2\right) $$ $$ \begin{aligned} & P\left(\frac{W}{B_1}\right)=\frac{4}{6}=\frac{2}{3} \\ & P\left(\frac{W}{B_2}\right)=\frac{3}{7} \end{aligned} $$ Now, from the total probability theorem: $$ \begin{aligned} & P(W)=P\left(\frac{W}{B_1}\right) P\left(B_1\right)+\mathrm{P}\left(\frac{W}{B_2}\right) P\left(B_2\right) \\ & =\frac{1}{2} \times \frac{2}{3}+\frac{1}{2} \times \frac{3}{7}=\frac{23}{42} \end{aligned} $$

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

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