Question

A table has drawers. Drawer 1 contains two red and five black biros, draw 2 contains four red and three black biros and drawer 3 contains on red and six black biros. A drawer is chosen at random and a biro is chosen from the drawer. Find the probability that the biro chosen is from drawer 1 if the chosen biro is black

Answer 1

Answer on Question #63074 – Math – Statistics and Probability

Question

A table has drawers. Drawer 1 contains two red and five black biros, drawer 2 contains four red and three black biros and drawer 3 contains one red and six black biros. A drawer is chosen at random and a biro is chosen from the drawer. Find the probability that the biro is chosen from drawer 1 if the chosen biro is black.

Solution

Let $A_i =$ 'a randomly chosen biro is from the $i$th drawer', where $i = 1,2,3$.

Let $B =$ 'a randomly chosen biro is black'.

If a drawer is chosen at random, then we have

$$P(A_1) = P(A_2) = P(A_3) = \frac{1}{3}$$

The probability that the biro is black given the biro was chosen from $i$th drawer ($i = 1,2,3$) will be

$$P(B|A_1) = \frac{5}{7},\ P(B|A_2) = \frac{3}{7},\ P(B|A_1) = \frac{6}{7}.$$

The law of total probability gives

$$\begin{aligned} P(B) &= \sum_{i=1}^{3} P(A_i) \cdot P(B|A_i) = \\ &= P(A_1) \cdot P(B|A_1) + P(A_2) \cdot P(B|A_2) + P(A_3) \cdot P(B|A_3) = \\ &= \frac{1}{3} \left( \frac{5}{7} + \frac{3}{7} + \frac{6}{7} \right) = \frac{2}{3} \end{aligned}$$

We know that the event $B$ has occurred, and we want to calculate the conditional probability of the event $A_1$.

By Bayes' theorem, we have

$$P(A_1|B) = \frac{P(A_1) \cdot P(B|A_1)}{P(B)} = \frac{\frac{1}{3} \times \frac{5}{7}}{\frac{2}{3}} = \frac{5}{14}$$

Answer: $\frac{5}{14}$

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