To find the probability that the 6 th Rose out of 10 tosses will be heads. Given there are exactly 2 heads out of 10 tosses. As in precendingg segment continue to assume all tosses are independent and that each toss has same fixed probability of heads

Answer 1

To find the probability that the 6th toss out of 10 will be heads given that there are exactly 2 heads out of 10 tosses, we need to use the concept of conditional probability.

Let event A be the event that the 6th toss is heads.

Let event B be the event that there are exactly 2 heads out of 10 tosses.

We want to find P(A|B), the probability of event A occurring given that event B has occurred.

By conditional probability formula, we have:

P(A|B) = P(A and B) / P(B)

First, let's calculate P(B), the probability that there are exactly 2 heads out of 10 tosses using the binomial probability formula:

P(B) = (10 choose 2) * (0.5)^2 * (0.5)^(10-2) = 45 * 0.25 * 0.25 ≈ 0.2813

Now, let's calculate P(A and B), the probability that the 6th toss is heads and there are exactly 2 heads out of 10 tosses. Since we want the 6th toss to be heads and there are 2 heads overall, we have 1 head occurring before the 6th toss. This can be calculated using the binomial probability formula:

P(A and B) = (5 choose 1) * (0.5)^1 * (0.5)^(5-1) * (5 choose 1) * (0.5)^1 * (0.5)^(10-6) = 5 * 0.5 * 0.5 * 5 * 0.5 * 0.5 ≈ 0.1563

Finally, we can calculate P(A|B):

P(A|B) = P(A and B) / P(B) = 0.1563 / 0.2813 ≈ 0.556

Therefore, the probability that the 6th toss out of 10 will be heads given that there are exactly 2 heads out of 10 tosses is approximately 0.556.