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) =
0.9999^{2}*0.5 + 0.1^{2}*0.5 = 0.00499900005

Clearly P(B1)P(B2) = 0.54995^{2} = 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)
= 0.9999^{2}*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) =
0.9999^{2} = P(B1 B2 |A )

Now similarly for A^{c,}

P(B1 B2 |
A^{c}) = 0.1^{2}*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}) = 0.1^{2} = P(B1 B2 |
A^{c})

**Hence proved.**

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