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
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.