Answer:
To find the probability of no more than 2 successes in 5 trials of a binomial experiment in which the probability of success in any one trial is 18%, we can use the formula for the probability mass function of the binomial distribution:
P(X = k) = (n choose k) * p^k * (1 - p)^(n-k)
Where:
X is the number of successes in the binomial experiment
n is the number of trials
p is the probability of success in any one trial
k is the number of successes we are interested in
So, in this case, we want to find the probability of X being equal to 0, 1, or 2 successes. We can do this by summing the probabilities of each of these events:
P(X = 0) + P(X = 1) + P(X = 2)
Plugging in the values from the problem, we get:
P(X = 0) = (5 choose 0) * (0.18)^0 * (1 - 0.18)^(5-0) = 0.3199
P(X = 1) = (5 choose 1) * (0.18)^1 * (1 - 0.18)^(5-1) = 0.4199
P(X = 2) = (5 choose 2) * (0.18)^2 * (1 - 0.18)^(5-2) = 0.2082
So the probability of no more than 2 successes in 5 trials of a binomial experiment in which the probability of success in any one trial is 18% is:
0.3199 + 0.4199 + 0.2082 = 0.948
Explanation: