To find the probability that the 6th tose out of 10 tosses will be heads. Given there are exactly 2 heads out of 10 tosses. As in precending 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 tosses will be heads, given that there are exactly 2 heads out of 10 tosses, we can use the concept of conditional probability.
First, let's calculate the probability of getting exactly 2 heads out of 10 tosses. This can be calculated using the binomial probability formula:
P(X=k) = (n choose k) * p^k * (1-p)^(n-k)
Where:
- n = 10 (number of tosses)
- k = 2 (number of heads)
- p = 0.5 (probability of getting a head)
P(X=2) = (10 choose 2) * 0.5^2 * (0.5)^(10-2)
P(X=2) = 45 * 0.25 * 0.25
P(X=2) = 0.28125
Next, we need to find the probability that the 6th toss is heads, given that there are exactly 2 heads out of 10 tosses. We can use conditional probability for this:
P(E6|E2) = P(E6 ∩ E2) / P(E2)
Where:
- P(E6 ∩ E2) refers to the probability that both events E6 (6th toss is heads) and E2 (2 heads out of 10 tosses) occur
- P(E2) is the probability that event E2 (2 heads out of 10 tosses) occurs
Since the tosses are independent, the probability of getting a head on the 6th toss given that there are exactly 2 heads in 10 tosses is simply 0.5.
Therefore, the probability that the 6th toss out of 10 tosses will be heads, given there are exactly 2 heads out of 10 tosses, is 0.5.