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Miscellaneous 9 - II · Q95

Q.Given three identical boxes, I, II and III, each containing two coins. In box I, both coins are gold coins, in box II, both are silver coins and in the box III, there is one gold and one silver coin. A person chooses a box at random and takes out a coin. If the coin is of gold, what is the probability that the other coin in the box is also of gold?

Maharashtra MsbshseTextbookSubjectiveImportance★★★★★est
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Each box is chosen with prior 1/31/3. P(gold/Box I)=1P(\text{gold}/\text{Box I})=1 (both coins gold), P(gold/Box II)=0P(\text{gold}/\text{Box II})=0 (both silver), P(gold/Box III)=1/2P(\text{gold}/\text{Box III})=1/2 (one of each). P(gold)=13(1)+13(0)+13(12)=13+0+16=12P(\text{gold})=\frac{1}{3}(1)+\frac{1}{3}(0)+\frac{1}{3}(\frac{1}{2})=\frac{1}{3}+0+\frac{1}{6}=\frac{1}{2}. By Bayes' theorem, $P(\text{Box I}/\text{gold})=\dfrac{(1/3)(1)}{1/2}=\dfrac{1/3}{1 …

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