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&gt;2)P(X&gt;2), n=6n=6, p=0.7p=0.7Answer

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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&amp;gt;2)P(X&amp;gt;2), n=6n=6, p=0.7p=0.7Answer

Eliza Beth13 · Accepted Answer

To solve the problem, we are given the random variable X, which has a binomial distribution with parameters n and p. Here, n represents the number of trials, and p represents the probability of obtaining a success in each trial. We need to find the probability P(X &amp;gt; 2) when n = 6 and p = 0.7.The probability mass function (PMF) of a binomial distribution is given by the formula:P(X=k)=(nk)·pk·(1−p)n−kwhere (nk) represents the binomial coefficient, which is the number of ways to choose k successes out of n trials.To find P(X &amp;gt; 2), we need to calculate the probability of X being greater than 2, which includes the probabilities of X being 3, 4, 5, and 6. We can calculate each probability separately and then sum them up.P(X&amp;gt;2)=P(X=3)+P(X=4)+P(X=5)+P(X=6)Let&#039;s calculate each term step by step:P(X=3)=(63)·(0.7)3·(1−0.7)6−3P(X=3)=(63)·(0.7)3·(0.3)3To calculate the binomial coefficient (63), we use the formula:(63)=6!3!·(6−3)!(63)=6!3!·3!=6·5·43·2·1=20Substituting the values into the formula, we have:P(X=3)=20·(0.7)3·(0.3)3Calculating this expression gives:P(X=3)≈0.3087Next, we calculate the probability for X = 4:P(X=4)=(64)·(0.7)4·(1−0.7)6−4P(X=4)=(64)·(0.7)4·(0.3)2Using the binomial coefficient formula, we find:(64)=6!4!·(6−4)!=6·52·1=15Substituting the values into the formula, we have:P(X=4)=15·(0.7)4·(0.3)2Calculating this expression gives:P(X=4)≈0.1852Next, we calculate the probability for X = 5:P(X=5)=(65)·(0.7)5·(1−0.7)6−5P(X=5)=(65)·(0.7)5·(0.3)1Using the binomial coefficient formula, we find:(65)=6!5!·(6−5)!=61=6Substituting the values into the formula, we have:P(X=5)=6·(0.7)5·(0.3)1Calculating this expression gives:P(X=5)≈0.0882Finally, we calculate the probability for X = 6:P(X=6)=(66)·(0.7)6·(1−0.7)6−6P(X=6)=(66)·(0.7)6·(0.3)0Using the binomial coefficient formula, we find:(66)=6!6!·(6−6)!=11=1Substituting the values into the formula, we have:P(X=6)=1·(0.7)6·(0.3)0Calculating this expression gives:P(X=6)≈0.1176Now, we can find the probability P(X &amp;gt; 2) by summing up the probabilities we calculated:P(X&amp;gt;2)=P(X=3)+P(X=4)+P(X=5)+P(X=6)P(X&amp;gt;2)≈0.3087+0.1852+0.0882+0.1176Calculating this expression gives:P(X&amp;gt;2)≈0.6997Therefore, the probability of X being greater than 2, given n = 6 and p = 0.7, is approximately 0.6997. Rounded to four decimal places, the answer is 0.6997.

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Assume the random variable X has a binomial

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