Coloured balls are distributed in four boxes | vedclass

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(A) Let $A$ be the event that a black ball is selected,and $E_1, E_2, E_3, E_4$ be the events that boxes $I, II, III, IV$ are selected respectively.Si

Vedclass · Accepted Answer

$0.165$

Vedclass · Answer

(A) Let $A$ be the event that a black ball is selected,and $E_1, E_2, E_3, E_4$ be the events that boxes $I, II, III, IV$ are selected respectively.Since the boxes are chosen at random,$P(E_1) = P(E_2) = P(E_3) = P(E_4) = \frac{1}{4}$.The probabilities of drawing a black ball from each box are:$P(A|E_1) = \frac{3}{3+4+5+6} = \frac{3}{18} = \frac{1}{6}$$P(A|E_2) = \frac{2}{2+2+2+2} = \frac{2}{8} = \frac{1}{4}$$P(A|E_3) = \frac{1}{1+2+3+1} = \frac{1}{7}$$P(A|E_4) = \frac{4}{4+3+1+5} = \frac{4}{13}$Using Bayes' theorem,the probability that the ball is from box $III$ given it is black is:$P(E_3|A) = \frac{P(E_3)P(A|E_3)}{\sum_{i=1}^{4} P(E_i)P(A|E_i)}$$P(E_3|A) = \frac{\frac{1}{4} 	imes \frac{1}{7}}{\frac{1}{4}(\frac{1}{6} + \frac{1}{4} + \frac{1}{7} + \frac{4}{13})} = \frac{\frac{1}{7}}{\frac{1}{6} + \frac{1}{4} + \frac{1}{7} + \frac{4}{13}}$Calculating the denominator: $\frac{1}{6} + \frac{1}{4} + \frac{1}{7} + \frac{4}{13} = \frac{364 + 546 + 312 + 672}{2184} = \frac{1894}{2184} \approx 0.8672$$P(E_3|A) = \frac{1/7}{1894/2184} = \frac{312}{1894} \approx 0.16475 \approx 0.165$.

Box	Black	White	Red	Blue
$I$	$3$	$4$	$5$	$6$
$II$	$2$	$2$	$2$	$2$
$III$	$1$	$2$	$3$	$1$
$IV$	$4$	$3$	$1$	$5$

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