A box contains two biased coins. One coin shows heads with probability 2/3, and tails with probability 1/3. The other coin shows heads with probability 1/3, and tails with probability 2/3.
a. Suppose you pick a coin at random from the box and flip it ten times and obtain 7 heads and 3 tails. What is the probability you picked the coin that is biased towards heads?
b. Suppose you pick a coin at random from the box and flip it 100 times and obtain 52 heads and 48 tails. What is the probability you picked the coin that is biased towards heads? (Note: you should get the same result as in part (a)!)

where event A₁= obtaining the coin that is biased towards heads , A₂= obtaining the coin that is biased towards tails B= obtaining 7 heads and 3 tails when flipped the coin

(P(B)= obtaining the result with the first coin + obtaining the result with the second coin )

then

P(A₁/B)=P(B∩A₁)/[P(B∩A₁)+P(B∩A₂)] = 1/(1+ P(B∩A₂)/P(B∩A₁))

then

P(B∩A₁) = probability of obtaining 7 heads and 3 tails when flipped the coin biased towards heads

P(B∩A₂) = probability of obtaining 7 heads and 3 tails when flipped the coin biased towards tails

and P(B∩A₁) and P(B∩A₂) follow a binomial distribution

P(B∩A₁) = C(10,7)* p₁⁷ *(1-p₁)³

P(B∩A₂) = C(10,7)* p₂⁷ *(1-p₂)³

where C(10,7) denote the combinations of 7 heads in 10 flips and p denote the probabilities of obtaining heads in one toss. Thus

P(B∩A₂)/P(B∩A₁) = (p₁/p₂)⁷*[(1-p₁)/(1-p₂)]³

then replacing values

P(B∩A₂)/P(B∩A₁) = [(2/3)/(1/3)]⁷*[(1/3)/(2/3)]³ = 2⁷/2³ = 2⁴ = 16

then the probability that we want to obtain is

P(A₁/B)= 1/(1+ P(B∩A₂)/P(B∩A₁)) = 1/(1+16) = 1/17 (5.88%)

P(A₁/B)= 1/17 (5.88%)

therefore the probability you picked the coin that is biased towards heads is 1/17 (5.88%)

for b) since the ratio of probabilities P(B∩A₂)/P(B∩A₁) depend only the difference of heads and tails , the result obtained will be the same:

P(B∩A₂)/P(B∩A₁) = 2⁵²/2⁴⁸ = 2⁴ = 16

and P(A₁/B)= 1/17 (5.88%)