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Box $A$ contains $2$ white, $1$ black, $3$ red balls. Box $B$ contains $3$ white, $2$ black, $4$ red balls, Box $C$ contains $4$ white, $3$ black, $2$ red balls. A die is rolled.If $1,2,$ or $3$ appears,then a ball is from Box $A$, if $4$ or $5$ appears then a ball from box $B$ will be drawn ,otherwise a ball from box $C$ will be drawn ,a ball is chosen from one of the boxes and it is found to be red. Find the probability that it is from Box $B$?

Need help in above, have tried calculating but no success.

  • 2
    Any thoughts? Seems to be entirely straightforward....just write out the cases. – 2017-01-05

1 Answers 1

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E1 dice shows 1, 2 or 3 .

E2 dice shows 4, 5.

E3 dice shows 6.

P(E1) = $\frac36$

P(E2) = $\frac26$

P(E3) = $\frac16$

A : Ball drawn red.

P(A/E1) = $\frac36$

P(A/E2) = $\frac49$

P(A/E3) = $\frac29$

By Bayes Theorem,

$P(E2/A) = \frac{\text{P(E2)P(A/E2)}}{\text{P(E1)P(A/E1) + P(E2)P(A/E2) + P(E1)P(A/E3)}}$

Put values to get the answer.