A biased coin whose chance of heads is 0.4 is tossed five times in a row. What is the probability that heads is tossed exactly 2 times?
20.2%
23.04%
36.8%
40.1%
45.6%
To find the probability of getting exactly 2 heads when tossing a biased coin with a 0.4 chance of heads five times in a row, we can use the binomial probability formula.
The formula for the probability of getting exactly k successes in n independent trials, each with a probability of success p, is:
P(X = k) = C(n, k) * p^k * (1 - p)^(n - k)
Where:
- P(X = k) is the probability of getting exactly k successes
- C(n, k) is the number of combinations of n items taken k at a time
- p^k is the probability of k successes
- (1 - p)^(n - k) is the probability of (n - k) failures
In this case, we want to find the probability of getting exactly 2 heads when tossing the coin 5 times, so k = 2, n = 5, and p = 0.4.
Using the formula:
P(X = 2) = C(5, 2) * 0.4^2 * (1 - 0.4)^(5 - 2)
= (5! / (2! * (5 - 2)!)) * 0.4^2 * (0.6)^3
= (5! / (2! * 3!)) * 0.4^2 * (0.6)^3
= (5 * 4 * 3! / (2! * 3!)) * 0.4^2 * (0.6)^3
= (5 * 4 / 2) * 0.4^2 * 0.6^3
= 10 * 0.16 * 0.216
= 0.3456
Therefore, the probability of getting exactly 2 heads is 0.3456, which is approximately 34.56%.
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