A biased coin lands heads with probabilty 2/3. The coin is tossed 3 times

Question

a) Given that there was at least one head in the three tosses, what is the probability that there were at least two heads?
b) use your answer in a) to find the probability that there was exactly one head, give that there was at least one head in the three tosses.

Can someone please help me solve it using sequences of events?

Thanks

Answer 1

a) Let A be the event that there is at least one head in the three tosses, and B be the event that there are at least two heads.
We want to compute P(B|A), the probability of B given A. By the definition of conditional probability, we have P(B|A) = P(A∩B) / P(A).

First, let's compute the probabilities of the complementary events A' and B', which are "no heads in the three tosses" and "fewer than two heads", respectively.

For A': the only way to get no heads is to have the coin land tails three times. This happens with probability (1/3)^3 = 1/27. Since P(A) + P(A') = 1, we get P(A) = 26/27.

For B': there can be either no heads or exactly one head. We already computed the probability of no heads, which is 1/27. To get exactly one head, there are three scenarios: H-T-T, T-H-T, or T-T-H. The probability of each of these scenarios is (2/3)*(1/3)*(1/3) = 2/27. In total, P(B') = P(no heads) + P(exactly one head) = 1/27 + 3*(2/27) = 1/3. Since P(B) + P(B') = 1, we get P(B) = 2/3.

Now we can compute the intersection of A and B, A∩B. We know that B implies at least two heads, so A∩B is just the event "at least two heads". We know that P(at least two heads) = P(B) = 2/3.

Finally, we can compute the conditional probability P(B|A) = P(A∩B) / P(A) = (2/3) / (26/27) = (2/3)*(27/26) = 9/13.

b) Let C be the event of exactly one head. To find P(C|A), notice that C is the complement of B given A, since if there is at least one head and there are not at least two heads, then there must be exactly one head. Thus, P(C|A) = 1 - P(B|A) = 1 - 9/13 = 4/13.