Question

Suppose I have two biased coins: coin #1, which lands heads with probability 0.9999, and coin #2, which lands heads with probability 0.1. I conduct an experiment as follows. First I toss a fair coin to decide which biased coin I pick (say, if it lands heads, I pick coin #1, and otherwise I pick coin #2) and then I toss the biased coin twice. Let A be the event that the biased coin #1 is chosen, B1 the event that the biased coin lands heads on the first toss, and B2 the event that the biased coin lands heads on the second toss. Show that

P(B 1∩ B 2 | A) = P(B 1 | A) P(B 2 | A) and

P(B 1 ∩ B 2 | A^c ) = P(B 1 | A^c ) P(B 2 | A^c ).

However, are B 1 and B 2 independent? Explain.

Answer 1

P(A) = 0.5, therefore P(A^c) = 0.5, equal probability of choosing both biased coins initially.

P(B1) = 0.9999*0.5 + 0.1*0.5 = 0.54995
P(B2) = P(B1) because we are tossing the same biased coin twice.

P(B1 B2) = 0.9999²*0.5 + 0.1²*0.5 = 0.00499900005

Clearly P(B1)P(B2) = 0.54995² = which is not equal to P(B1 B2). Therefore B1 and B2 are not independent

Therefore using bayes theorem, we get:
P(B1 B2 | A) =  0.9999²*0.5 / 0.5 = 0.99980001

Also P(B1 |A ) = 0.9999*0.5 / 0.5 = 0.9999
P(B2 | A) = 0.9999*0.5/0.5 = 0.9999

Therefore, we prove here that:  P(B1 |A )P(B2 | A) = 0.9999² = P(B1 B2 |A )

Now similarly for A^c,
P(B1 B2 | A^c) = 0.1²*0.5 / 0.5 = 0.01
P(B1 |A^c ) = 0.1*0.5 / 0.5 = 0.1
P(B2 | A^c) = 0.1*0.5/0.5 = 0.1

Therefore, we prove here that:  P(B1 |A^c )P(B2 | A^c) = 0.1² = P(B1 B2 | A^c)

Hence proved.