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 precend8ng segment continue to addume all tosses are independent and that each toss has same fixed prob 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 in 10 tosses, we can use the concept of conditional probability.

Let A be the event that the 6th toss is heads, and let B be the event that there are exactly 2 heads in 10 tosses.

We want to find P(A|B), which is the conditional probability of A given B.

We know that P(A and B) = P(A|B) * P(B), where P(A and B) is the probability of both A and B occurring.

We also know that P(B) = the probability of getting exactly 2 heads in 10 tosses, which can be calculated using the binomial distribution formula:
P(B) = (10 choose 2) * (1/2)^2 * (1/2)^(10-2)

We can also calculate P(A and B), which is the probability of getting exactly 2 heads in 10 tosses and the 6th toss being heads:
P(A and B) = (10 choose 2) * (1/2)^2 * (1/2)^(10-2) * (1/2)

Now we can find P(A|B) using the formula:
P(A|B) = P(A and B) / P(B)

Let's now calculate these probabilities using the given information.