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, whereas one coin is a fake, with heads on both sides. . 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?. Suppose you continue 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? . 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 heads; otherwise it returns normal. What is the (unconditional) probability that this procedure makes an error?

a) [2/n+1]

b) 2ᵏ/[2ᵏ + (n-1)]

Explanation:

(n-1) coins are fair and only 1 is a fake coin with two heads.

This means there are a total of n coins in the bag.

Let the probability of getting a head be P(h)

Probability of picking a fake coin = P(f) = (1/n)

Probability of getting head on a fake coin = P(h|f) = 1

Probability of picking a fake coin and getting a head = P(f n h) = (1/n) × 1 = (1/n)

Probability of picking the right (original) coin = P(r) = (n-1)/n

Probability of getting a head on the right coin = P(h|r) = (1/2)

Probability of picking a real coin and getting a head = P(h n r) = [(n-1)/n] × (1/2) = [(n-1)/2n]

a) The required conditional probability is the probability that the coin is fake, given that it turns up head

P(f|h) = P(f n h)/P(h)

P(f n h) = (1/n)

But, we do not have P(h). We can obtain it through the relation,

P(h) = P(h n r) + P(h n f)

= P(r) P(h|r) + P(f) P(h|f)

= {[(n-1)/n] × [1/2] + [1/n] [1]

= [(n-1)/2n] + [1/n]

= (n+1)/2n

P(f|h) = P(f n h)/P(h)

= (1/n) ÷ [(n+1)/2n] = [2/n+1]

b) The second required probability

Probability that it's a fake coin, given that we get k heads.

P(f|k heads) = P(f n k heads) ÷ P(k heads)

P(f n k heads) = (1/n) × 1ᵏ = (1/n)

But, we do not have P(k heads). We can obtain it through the relation,

P(k heads) = P(r n k heads) + P(f n k heads)

P(r n k heads) = [(n-1)/n] × [1/2)ᵏ

= [(n-1)/2ᵏn]

P(f n k heads) = (1/n)

P(k heads) = P(r n k heads) + P(f n k heads)

= [(n-1)/2ᵏn] + [1/n]

P(f|k heads) = P(f n k heads) ÷ P(k heads)

= [1/n] ÷ {[(n-1)/2ᵏn] + [1/n]}

On simplifying,

We obtain

P(f|k heads) = 2ᵏ/[2ᵏ + (n-1)]

Note that, Bayes rule is the guiding principle in all the conditional probability evaluations we have done above.

Hope this Helps!!!

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