Jason is tossing a fair coin. He tosses the coin ten times and it lands on heads eight times. If Jason tosses the coin an eleventh time, what is the probability it lands on heads?
P(heads) is always 1/2 for a fair coin.
To calculate the probability of the coin landing on heads in the eleventh toss, we need to make some assumptions. Since the coin is fair, we can assume that each toss is independent and has an equal probability of landing on heads or tails, which is 1/2 or 0.5.
Since Jason has already tossed the coin ten times and it has landed on heads eight times, we can consider this a binomial distribution problem. The probability of getting exactly k successes in n independent Bernoulli trials, where the probability of success on each trial is p, can be calculated using the binomial probability formula:
P(X = k) = (n choose k) * p^k * (1 - p)^(n - k)
In this case, we want to find the probability of getting heads on the eleventh toss, so n = 11, k = 8, and p = 0.5.
Using this information, we can calculate the probability as follows:
P(X = 8) = (11 choose 8) * (0.5^8) * (0.5^(11 - 8))
= (11! / (8! * 3!)) * 0.5^8 * 0.5^3
= (11 * 10 * 9) / (3 * 2 * 1) * 0.5^11
= (11 * 10 * 9) / (3 * 2 * 1) * 0.5^11
= 165 / 8 * 0.5^11
= 165 / 2048
≈ 0.0801
Therefore, the probability that the coin lands on heads in the eleventh toss is approximately 0.0801 or 8.01%.