To find P(X≥3), we need to find the probability of getting 3 or more successes in 5 trials with a success probability of 0.1.
P(X≥3) = P(X=3) + P(X=4) + P(X=5)
Using the binomial probability formula, we can calculate these probabilities individually:
P(X=k) = nCk * p^k * (1-p)^(n-k)
For k=3:
P(X=3) = 5C3 * 0.1^3 * 0.9^(5-3)
P(X=3) = 10 * 0.001 * 0.81
P(X=3) = 0.0081
For k=4:
P(X=4) = 5C4 * 0.1^4 * 0.9^(5-4)
P(X=4) = 5 * 0.0001 * 0.9
P(X=4) = 0.00045
For k=5:
P(X=5) = 5C5 * 0.1^5 * 0.9^(5-5)
P(X=5) = 1 * 0.00001 * 1
P(X=5) = 0.00001
Now, sum up these probabilities:
P(X≥3) = 0.0081 + 0.00045 + 0.00001
P(X≥3) = 0.00856
Therefore, P(X≥3) is approximately 0.0086.
Assume the random variable X has a binomial distribution with the given probability of obtaining a success. Find the following probability, given the number of trials and the probability of obtaining a success. Round your answer to four decimal places.
P(X≥3)
, n=5
, p=0.1
1 answer