Answer:
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!!!
