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
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.