Coloured balls are distributed in four boxes as shown in the foll… - Vidyadip

Question

Coloured balls are distributed in four boxes as shown in the following table:

Box	Colour
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

Let $A, E_1, E_2, E_3$ and $E_4$ be the events as defined below :
$A :$ 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, $P(E_1) = P(E_2) = P(E_3) = P(E_4)$
Also $P(A|E_1) = \frac{3}{18}, P(A|E_2) = \frac {2}{8}, P(A|E_3) = \frac{1}{7}$ and $P(A|E_4) = \frac{4}{13}$
$P($box $III$ is selected, given that the drawn ball is black$) = P(E_3|A).$ By Bayes' theorem,
$P(E_3|A) = \frac{\mathrm{P}\left(\mathrm{E}_{3}\right) \cdot \mathrm{P}\left(\mathrm{A|E}_{3}\right)}{\mathrm{P}\left(\mathrm{E}_{1}\right) \mathrm{P}\left(\mathrm{A|E}_{1}\right)+\mathrm{P}\left(\mathrm{E}_{2}\right) \mathrm{P}\left(\mathrm{A|E}_{2}\right)+\mathrm{P}\left(\mathrm{E}_{3}\right) \mathrm{P}\left(\mathrm{A|E}_{3}\right)+\mathrm{P}\left(\mathrm{E}_{4}\right) \mathrm{P}\left(\mathrm{A|E}_{4}\right)}$
$= \frac{\frac{1}{4} \times \frac{1}{7}}{\frac{1}{4} \times \frac{3}{18}+\frac{1}{4} \times \frac{1}{4}+\frac{1}{4} \times \frac{1}{7}+\frac{1}{4} \times \frac{4}{13}} = 0.165$

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