Suppose you are given a bag containing n unbiased coins. You are told that n-1 o

Question

Suppose you are given a bag containing n unbiased coins. You are told that n-1 of these coins are normal, with heads on one side and tails on the other, wheras one coin is a fake, with heads on both sides.
a. Suppose you reach into the bag, pick out a coin at random, flip it, and get a head. What is the (conditional) probability that the coin you chose is the fake coin?
b. Suppose you contine flipping the coin for a total of k times after picking it and see k heads. Now what is the conditional probability that you picked the fake coin?
c. Suppose you wanted to decide whether the chosen coin was fake by flipping it k times. The decision procedure returns fake if all k flips come up with heads; otherwise it returns normal. What is the (unconditional) probability that this procedure makes an error?

Explanation / Answer

The conditional probability of A given B is defined as the quotient of the joint probability of A and B, and the marginal probability of B P(A|B) = P(A n B)/P(B) (a) probability of random coin being head P(B) = (n-1)/n * (1/2) + 1/n*1 = (3n - 1)/2n. probability that the random coin flipped head is a fake coin = P(A n B)= 1/n * 1 Hence , conditional probability that the coin you chose is the fake coin after you pick out a coin at random, flip it, and get a head = P(A|B) = P(A n B) / P(B) = (1/n) / [(3n-1) / 2n ] = 2/(3n-1). (b)probability that you picked the fake coin and you contine flipping the coin for a total of k times after picking it and see k heads = P(A n B) = 1/n *1 =1/n probability that after flipping a coin for k times and see k heads = P(B) = [(n-1)/n ]*(1/2)^k + 1/n * 1 = (n + 2^k -1)/(n*2^k) conditional probability that you picked the fake coin = (1/n) / [(n + 2^k -1)/(n*2^k)] = 2^k/ (n + 2^k -1) (c)error is the probability that a normal coin be chosen and it gives heads all the k - times is [(n-1)/n ] * (1/2)^k = (n-1)/(2^k * n)

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