Coloured balls are distributed in four boxes as shown in the following table:
|
Box |
Black |
White |
Red |
Blue |
|
I |
3 |
4 |
5 |
6 |
|
II |
2 |
2 |
2 |
2 |
|
III |
1 |
2 |
3 |
1 |
|
IV |
4 |
3 |
1 |
5 |
A box is selected at random and then a ball is randomly drawn from the selected box. The colour of the ball is black, what is the probability that ball drawn is from the box III?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → $0.165$ ##
Let $A, E_1, E_2, E_3$ and $E_4$ be the events as defined below:
- $A$ : a black ball is selected
- $E_1$ : box I is selected
- $E_2$ : box II is selected
- $E_3$ : box III is selected
- $E_4$ : box IV is selected
Since the boxes are chosen at random,
Therefore
$P(E_1) = P(E_2) = P(E_3) = P(E_4) = \frac{1}{4}$
Also
$P(A|E_1) = \frac{3}{18}, P(A|E_2) = \frac{2}{8}, P(A|E_3) = \frac{1}{7} \text{ and } P(A|E_4) = \frac{4}{13}$
$P(\text{box III is selected, given that the drawn ball is black}) = P(E_3|A)$. By Bayes' theorem,
$P(E_3|A) = \frac{P(E_3) \cdot P(A|E_3)}{P(E_1)P(A|E_1) + P(E_2)P(A|E_2) + P(E_3)P(A|E_3) + P(E_4)P(A|E_4)}$
$= \frac{\frac{1}{4} \times \frac{1}{7}}{\frac{1}{4} \times \frac{3}{18} + \frac{1}{4} \times \frac{1}{4} + \frac{1}{4} \times \frac{1}{7} + \frac{1}{4} \times \frac{4}{13}} = 0.165$