Three urns A, B and C contain 7 red, 5 black; 5 red, 7 black and 6 red, 6 black balls, respectively. One of…

Three urns A, B and C contain 7 red, 5 black; 5 red, 7 black and 6 red, 6 black balls, respectively. One of the urn is selected at random and a ball is drawn from it. If the ball drawn is black, then the probability that it is drawn from urn $\mathrm{A}$ is :
  1. $\frac{5}{18}$
  2. $\frac{5}{16}$
  3. $\frac{4}{17}$
  4. $\frac{7}{18}$

Solution

$\begin{array}{ccc}\text { A } & \text { B } & \text { C } \\ 7 \mathrm{R}, 5 \mathrm{~B} & 5 \mathrm{R}, 7 \mathrm{~B} & 6 \mathrm{R}, 6 \mathrm{~B}\end{array}$
$\begin{aligned} & \mathrm{P}(\mathrm{B})=\frac{1}{3} \cdot \frac{5}{12}+\frac{1}{3} \cdot \frac{7}{12}+\frac{1}{3} \cdot \frac{6}{12} \\ & \text { required probability }=\frac{\frac{1}{3} \cdot \frac{5}{12}}{\frac{1}{3} \cdot\left[\frac{5}{12}+\frac{7}{12}+\frac{6}{12}\right]}=\frac{5}{18}\end{aligned}$

Asked in: JEE Main 2024 (04 Apr Shift 1)

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