We are in possession of two coins, one which is fairly balanced and turns up hea

Question

We are in possession of two coins, one which is fairly balanced and turns up heads with probability l/2, the other is weighted such that heads shows up with probability 3/4 and tails with probability 1/4. The two coins are identical looking and feeling so we cannot tell which is which. In order to determine which is the biased coin we toss the coin 10 times and observe the number of heads that occurred. (a) If 7 heads were observed, what is the probability that the coin flipped was the fair coin? (b) If 3 heads were observed, what is the probability that the coin flipped was the fair coin?

Explanation / Answer

A) P(7 heads on unbiased coin) = 10C7x0.57 x 0.53 = 0.1172

P(7 heads on biased coin) = 10C7x0.757 x 0.253 = 0.2503

P(fair coin | 7 heads) = P(7 heads on fair coin)/P(7 heads)

= 0.1172/(0.1172 + 0.2503)

= 0.3189

B) P(3 heads on unbiased coin) = 10C3x0.57 x 0.53 = 0.1172

P(3 heads on biased coin) = 10C3x0.257 x 0.753 = 0.0031

P(fair coin | 3 heads) = P(fair coin and 3 heads) / p(3 heads)

= 0.1172/(0.1172+ 0.0031)

= 0.9742

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