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A box contains $12$ red and $6$ white balls. Balls are drawn from the bag one at a time without replacement. If in $6$ draws, there are at least $4$ white balls, find the probability that exactly one white ball is drawn in the next two draws.

I understand that we can use Bayes theorem so we should find $P(E|A)+P(E|B)+P(E|C)$ isn't it? where $A=$drawing $4$ white balls

$B=$drawing $5$ white balls

$C=$drawing $6$ white balls ( can be neglected, no white balls will be left for the next draws)

$E=$exactly one white ball is drawn in the next two draws

What's wrong with my solution?

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    In the first 6 draws 4 whites were drawn? Or in 6 draws 4 white remain? – 2017-02-20
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    At least four white were drawn in the first 6 draws. – 2017-02-20
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    Ok I would say the first thing wrong with your solution is you can find the probability, be you can not say for sure when the balls will be drawn.. it is likely to be a function of remaining white balls. – 2017-02-20

2 Answers 2

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What is wrong with your solution is that you write: $$P(E|A\cup B\cup C)=P(E|A)+P(E|B)+P(E|C)$$ And what you should calculate (given that $A$, $B$, and $C$ are disjoint) is: $$\begin{align} \frac{P(E\cap(A\cup B\cup C))}{P(A\cup B\cup C)}&=\frac{P(E\cap A)+P(E\cap B)+P(E\cap C)}{P(A)+P(B)+P(C)}\\ &=\frac{P(E|A)P(A)+P(E|B)P(B)+P(E|C)P(C)}{P(A)+P(B)+P(C)} \end{align}$$

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    Why did you divide by P(A)+P(B)+P(C) – 2017-02-20
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    By the definition of conditional probability $P(E|A\cup B\cup C)=\frac{P(E\cap(A\cup B\cup C))}{P(A\cup B\cup C)}$. And $P(A\cup B\cup C)=P(A)+P(B)+P(C)$ since $A,B$ and $C$ are disjoint. – 2017-02-20
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$$\begin{align}P(E|A\cup B\cup C)&=\frac{P(E,A\cup B\cup C)}{P(A\cup B\cup C)}\\&=\frac{P((E,A)\cup(E,B)\cup(E,C))}{P(A\cup B\cup C)}\\&=\frac{P(E,A)}{P(A\cup B\cup C)}+\frac{P(E,B)}{P(A\cup B\cup C)}+\frac{P(E,C)}{P(A\cup B\cup C)}\\&\neq\frac{P(E,A)}{P(A)}+\frac{P(E,B)}{P(B)}+\frac{P(E,C)}{P(C)}\\&=P(E|A)+P(E|B)+P(E|C)\end{align}$$