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

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.
It asks me to solve it using sequences of events. I have tried solving it but with no luck. Please help me
Thanks

Thanks

there are eight possible sequences for three tosses

HHH , HHT , HTH , THH , HTT , THT , TTH , TTT

look at the descriptions of the sequences , and count the relevant outcomes

To find the probability of each event, we can use the concept of conditional probability. We'll break down each part of the problem using sequences of events.

a) To find the probability that there were at least two heads, given that there was at least one head in the three tosses, we need to consider the possible outcomes. Let's denote "H" as heads and "T" as tails.

First, let's find the probability of the complementary event, which means finding the probability of having at most one head given that there was at least one head in the three tosses.

The possible outcomes with at most one head are:
1. HTT
2. THT
3. TTH
4. HHT
5. HTH
6. THH

Out of these six outcomes, only one outcome has exactly one head (HTT). So the probability of having at most one head is 1/6.

Since the probability of the complementary event is 1/6, the probability of the desired event (at least two heads) is 1 - 1/6, which is 5/6.

Therefore, the answer to part (a) is 5/6.

b) Now, let's use the answer from part (a) to find the probability that there was exactly one head given that there was at least one head in the three tosses.

To calculate this, we subtract the probability of having at least two heads (which we found to be 5/6) from the probability of having at least one head in the three tosses.

Again, let's consider the possible outcomes:
1. HTT
2. THT
3. TTH
4. HHT
5. HTH
6. THH

Out of these six outcomes, only one outcome has no heads (TTT). So the probability of having no heads is 1/6.

Since the probability of having at least one head is 1 - 1/6 (which is 5/6), and the probability of having at least two heads is 5/6, we can obtain the probability of having exactly one head by subtracting:

P(exactly one head) = P(at least one head) - P(at least two heads)
= 5/6 - 5/6
= 0

Therefore, the probability of having exactly one head given that there was at least one head in the three tosses is 0.

I hope this explanation helps you understand how the probabilities are obtained using sequences of events. Let me know if you have any further questions!