Assuming a procedure results in a binomial distribution with&nbsp;n=7&nbsp;trials and a probability of success of&nbsp;p=0.90. Use a binomial probability table to find the probability that the number of successes x is exactly three.P(3)=?&nbsp;(round to three decimal places as needed)

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Assuming a procedure results in a binomial distribution with&amp;nbsp;n=7&amp;nbsp;trials and a probability of success of&amp;nbsp;p=0.90. Use a binomial probability table to find the probability that the number of successes x is exactly three.P(3)=?&amp;nbsp;(round to three decimal places as needed)

Nathaniel Kramer · Accepted Answer

Step 1&amp;nbsp;Obtain the probability that the number of successes x is exactly 3.&amp;nbsp;The probability that the number of successes x is exactly 3 is obtained below as follows:&amp;nbsp;Let X denotes the random variable which follows binomial distribution with the probability of success 0.90 with the number of trails equals 7.N&amp;nbsp;That is,&amp;nbsp;n=7,p=0.90,q=0.10(=1−0.90).&amp;nbsp;The probability distribution is given by,&amp;nbsp;P(X=x)=(beg∈{array}{c}n∖xend{array})px(1−p)n−x; here&amp;nbsp;x=0,1,2,⋯,n&amp;nbsp;for&amp;nbsp;0≤p≤1&amp;nbsp;Where n is the number of trials and p is the probability of success for each trial.&amp;nbsp;Step 2&amp;nbsp;The required probability is,&amp;nbsp;P(X=3)&amp;nbsp;From the “Binomial probability table”, with the number of trails as&amp;nbsp;X=7&amp;nbsp;with the probability of success as 0.90 the probability value is 0.003.&amp;nbsp;Therefore,&amp;nbsp;P(X=3)=0.003&amp;nbsp;The probability that the number of successes x is exactly 3 is 0.003.

Assume that a procedure yields a binomial distribution with n=7

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