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.
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
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