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Which urn was it

ProbabilityMedium

Urn A holds $3$ red and $2$ black balls. Urn B holds $1$ red and $4$ black balls. A fair coin picks one urn, and then a ball is drawn at random from that urn. The ball is red. What is the probability that it came from urn A? Give a fraction in lowest terms.

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Hint

Use Bayes' rule: compare the joint probabilities of red from each urn.

Solution

The probability of red from urn A is $\frac{1}{2} \times \frac{3}{5} = \frac{3}{10}$ .
The probability of red from urn B is $\frac{1}{2} \times \frac{1}{5} = \frac{1}{10}$ .
The total probability of red is $\frac{3}{10} + \frac{1}{10} = \frac{4}{10}$ .
Dividing, the probability of urn A given red is $(\frac{3}{10}) \div (\frac{4}{10})$ , which equals $\frac{3}{4}$ .

Answer: $\frac{3}{4}$

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