A box contains two coins. One is a fair coin, so that the probability of Heads i

Question

A box contains two coins. One is a fair coin, so that the probability of Heads is 0.5 and the other is biased so that the probability of Heads is 0.75.

1(d. Suppose one of the two coins is randomly chosen from the box and tossed 4 times. Find the probability that exactly 3 heads are observed Hint: Use the law of total probability. 1(e). Suppose one of the two coins is randomly chosen from the box and tossed 4 times. If exactly 3 heads are observed, use Bayes' Theorem to find the probability that the biased coin was chosen.

Explanation / Answer

Question 1d)

Here first we will compute the probability of getting 3 heads in 4 tosses in the cases of 2 coins individually. This is computed as:

P( 3 heads | fair coin ) = (4c3)*0.53*0.5 = 0.25

P( 3 heads | biased coin ) = (4c3)*0.753*0.25 = 0.421875

Also we are given that the coins are randomly selected, therefore we get:

P( fair coin ) = P( biased coin ) = 0.5

Now using the law of total probability, we get:

P( 3 heads ) = P( fair coin ) P( 3 heads | fair coin ) + P( biased coin ) P( 3 heads | biased coin )

P( 3 heads ) = 0.5*0.25 + 0.5*0.421875 = 0.3359375

Therefore 0.3359375 is the required probability here.

e) Given that 3 coins are heads out of 4 tosses, probability that it was a biased coin picked up is computed using bayes theorem as:

P( biased coin | 3 heads ) = P( biased coin ) P( 3 heads | biased coin ) / P( 3 heads )

P( biased coin | 3 heads ) = 0.5*0.421875 / 0.3359375

P( biased coin | 3 heads ) = 0.6279

Therefore 0.6279 is the required probability here.

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