The Piggy Banks Problem - 001

Probability Level 2

Two piggy banks contains some coins. The first piggy bank contains 4 fair coins and 2 biased coins in favor of heads. The second piggy bank contains 2 fair coins and 4 biased coins in favor of tails. The biased coins follow the odds 2:1. A piggy bank is selected at random, a coin is obtained from it and is found to be biased, what is the probability that the coin is from the first piggy bank if each piggy bank has an equal chance to be selected?

2/3 1/3 1/6 5/6

Mark Elis Espiridion by Mark Elis Espiridion , 27, Philippines

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1 solution

Assume A be first piggy bank and B be the second piggy bank.

Therefore, Selecting a piggy bank where, P(A) = P(B) = 1 2 \frac{1}{2}

Let E be the biased coin. Therefore by using Baye's Theorem

P(E/A) = P(Biased coin chosen from bag A) = 2 6 \frac{2}{6} = 1 3 \frac{1}{3}

P(E/B) = P(Biased coin chosen from bag B) = 4 6 \frac{4}{6} = 2 3 \frac{2}{3}

To find: P(Coin found to be biased was chosen from bag A) = P(A\E)

P(A/E) = P ( A ) . P ( E / A ) P ( A ) . P ( E / A ) + P ( B ) . P ( E / B ) \frac{P(A) . P(E/A)}{P(A) . P(E/A) + P(B) . P(E/B)}

= ( 1 / 2 ) . ( 1 / 3 ) ( 1 / 2 ) . ( 1 / 3 ) + ( 1 / 2 ) . ( 2 / 3 ) \frac{(1/2). (1/3)}{(1/2).(1/3) + (1/2).(2/3)}

= ( 1 / 6 ) ( 1 / 6 ) + ( 2 / 6 ) \frac{(1/6)}{(1/6) + (2/6)}

= ( 1 / 6 ) ( 1 / 2 ) \frac{(1/6)}{(1/2)}

= 1 3 \frac{1}{3}

Therefore the probability is 1 3 \frac{1}{3}

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