Question

# In the binomial probability distribution, let the number of trials be n = 3, and let...

In the binomial probability distribution, let the number of trials be n = 3, and let the probability of success be p = 0.3742. Use a calculator to compute the following.

(a) The probability of two successes. (Round your answer to three decimal places.)

(b) The probability of three successes. (Round your answer to three decimal places.)

(c) The probability of two or three successes. (Round your answer to three decimal places.)

Solution: Let X be the random variable following Binomial distribution, i.e., X ~ Bin(3,0.3742)

thus, the pmf of X is written as f(x) = (3Cx)*(0.3742)x(1-0.3742)3-x, x=0,1,2,3

where 3Cx = 3!/((3-x)! x!)

(a) The probability of two successes = P(X=2) = (3C2)*(0.3742)2(1-0.3742)3-2 =  0.263

(b) The probability of three successes = P(X=3) = (3C3)*(0.3742)3(1-0.3742)3-3 =  0.052

(c) The probability of two or three successes = P(X=2) + P(X=3) = 0.263 + 0.052 = 0.315

(all the answers are rounded to three decimal places)

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