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=7
, p=0.3
To find P(X<3) when n=7 and p=0.3, we need to calculate the probabilities of X=0, X=1, and X=2 and then sum them up.
P(X=0) = (7 choose 0)(0.3)^0(0.7)^(7-0) = 1 * 1 * 0.7^7 = 0.0824
P(X=1) = (7 choose 1)(0.3)^1(0.7)^(7-1) = 7 * 0.3 * 0.7^6 = 0.2304
P(X=2) = (7 choose 2)(0.3)^2(0.7)^(7-2) = 21 * 0.09 * 0.7^5 = 0.2668
Now, add these probabilities to get P(X<3):
P(X<3) = P(X=0) + P(X=1) + P(X=2) = 0.0824 + 0.2304 + 0.2668 = 0.5796
Therefore, P(X<3) when n=7 and p=0.3 is 0.5796.