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

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) = 3 18 , P(A|E_2) = 2 8 , P(A|E_3) = 1 7 and P(A|E_4) = 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) = P (E_ 3 ) 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 ) = 1 4 1 7 1 4 3 18 +1 4 1 4 +1 4 1 7 +1 4 4 13 = 0.165

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

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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 $III?$

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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,
$P(E_3|A) = \frac{\mathrm{P}\left(\mathrm{E}_{3}\right) \cdot \mathrm{P}\left(\mathrm{A|E}_{3}\right)}{\mathrm{P}\left(\mathrm{E}_{1}\right) \mathrm{P}\left(\mathrm{A|E}_{1}\right)+\mathrm{P}\left(\mathrm{E}_{2}\right) \mathrm{P}\left(\mathrm{A|E}_{2}\right)+\mathrm{P}\left(\mathrm{E}_{3}\right) \mathrm{P}\left(\mathrm{A|E}_{3}\right)+\mathrm{P}\left(\mathrm{E}_{4}\right) \mathrm{P}\left(\mathrm{A|E}_{4}\right)}$
= $\frac{\frac{1}{4} \times \frac{1}{7}}{\frac{1}{4} \times \frac{3}{18}+\frac{1}{4} \times \frac{1}{4}+\frac{1}{4} \times \frac{1}{7}+\frac{1}{4} \times \frac{4}{13}} = 0.165$