A leprechaun offers you your choice of three identical-looking pots. The leprech

Question

A leprechaun offers you your choice of three identical-looking pots. The leprechaun tells you that one of the pots contains two gold coins, one contains two silver coins, and one contains a gold and a silver coin. You are not allowed to inspect the pots or their contents before making your choice, so you choose one of the pots at random. After being given the pot, you reach in and randomly take out one of the two coins in it, without looking at the other. If the coin you took out happens to be gold, what is the probability that the other coin in the pot is also gold?

Explanation / Answer

Let P(P1) , P(P2) and P(P3) be the probability to select the pots with two gold coins, one contains two silver coins, and one contains a gold and a silver coin respectively.

Probability to select any of the pots = 1/3

Probability to select gold from the pots with 2 gold coins , P(G | P1) = 1

Probability to select gold from the pots with 2 silver coins, , P(G | P2) = 0

Probability to select gold from the pots with 1 gold coins and 1 silver coin, , P(G | P3)= 1/2

By law of total probability,

P(G) = P(P1) P(G | P1) +  P(P2) P(G | P2) + P(P3) P(G | P3)

= (1/3) * 1 + (1/3) * 0 + (1/3) * (1/2)

= 1/2

If the coin you took out happens to be gold, the probability that the other coin in the pot is also gold is equal to the probability that you selected the pots with 2 gold coins (P1). So, we need to calculate the probability,

P(P1 | G) = P(G | P1) * P(P1) / P(G) {By Bayes theorem}

= 1 * (1/3) / (1/2)

= 2/3

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