Problem 5.1 Easy Free 1 variation

Transfer Between Urns Then Draw

Urn $A$ contains $100$ green and $400$ red balls, and urn $B$ contains $400$ green and $100$ red balls. A random sample of $300$ balls from urn $A$ is transferred into urn $B$ . A ball is then drawn uniformly at random from the resulting $800$ balls in urn $B$ . Find the probability that this ball is red.

Work it out, then compare with the answer.

SolutionShow the worked solution

Condition on where the drawn ball came from, not on the random make-up of the transferred sample.

The drawn ball is one of the $300$ transferred balls with probability $\frac{300}{800}=\frac38$ . Otherwise it is one of $B$ ’s original $500$ balls, with probability $\frac58$ .

Given that it is a transferred ball, it is uniform among the $300$ transferred balls. Those balls are a uniformly random sample from $A$ . By symmetry, every ball of $A$ is equally likely to be the one drawn, so it is red with probability $\frac{400}{500}=\frac45$ .
Given that it is an original ball of $B$ , it is uniform among those $500$ balls, so it is red with probability $\frac{100}{500}=\frac15$ .

By the law of total probability,

\mathbb{P}(\text{red}) = \frac38\cdot\frac45+\frac58\cdot\frac15=\frac{12}{40}+\frac{5}{40}=\frac{17}{40}.

\boxed{\dfrac{17}{40}=0.425}

Alternative. Let $X$ be the number of red balls transferred, so $\mathbb{E}[X]=300\cdot\frac45=240$ . Given $X$ , the draw is red with probability $\frac{100+X}{800}$ . Hence

\mathbb{P}(\text{red})=\mathbb{E}\!\left[\frac{100+X}{800}\right]=\frac{340}{800}=\frac{17}{40}.

Key idea. A transferred ball carries its source urn’s colour proportion, so the distribution of the transfer never has to be computed.

Variations

The same idea with a twist. Each solution gives only the changed step.

Variation 5.1a

A bowl $A$ holds four red cherries and five purple cherries; a bowl $B$ holds six red cherries and three purple cherries. One cherry is transferred uniformly at random from bowl $A$ into bowl $B$ , and then a cherry is drawn uniformly at random from bowl $B$ and eaten. Find the probability that the eaten cherry is red.

Work it out, then compare with the answer.

Solution

After the transfer, bowl $B$ holds $10$ cherries, and only one of them came from $A$ .

With probability $\frac1{10}$ the eaten cherry is the transferred one. That cherry is uniform over bowl $A$ , so it is red with probability $\frac49$ .
With probability $\frac9{10}$ the eaten cherry is an original resident of $B$ , which is red with probability $\frac69$ .

By the law of total probability,

\mathbb{P}(\text{red})=\frac1{10}\cdot\frac49+\frac9{10}\cdot\frac69=\frac{58}{90}=\frac{29}{45}.

\boxed{\dfrac{29}{45}}

Spotted an error? Email us