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Chapter 5: Problem 5

Evaluate these binomial probabilities: a. \(C_{2}^{8}(.3)^{2}(.7)^{6}\) b. \(C_{0}^{4}(.05)^{0}(.95)^{4}\) c. \(C_{3}^{10}(.5)^{3}(.5)^{7}\) d. \(C_{1}^{7}(.2)^{1}(.8)^{6}\)

Short Answer

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Question: Evaluate the given binomial probabilities: a. \(C_{2}^{8}(.3)^{2}(.7)^{6}\) b. \(C_{0}^{4}(.05)^{0}(.95)^{4}\) c. \(C_{3}^{10}(.5)^{3}(.5)^{7}\) d. \(C_{1}^{7}(.2)^{1}(.8)^{6}\) Answer: a. \(P(X=2) \approx 0.3241\) b. \(P(X=0) \approx 0.8145\) c. \(P(X=3) \approx 0.1172\) d. \(P(X=1) \approx 0.3670\)

Step by step solution

01

Calculating the combinations

For each problem, we are given the number of successful trials (k) and the total number of trials (n). To find the combinations, we use the formula: \(C_{k}^{n} = \frac{n!}{k!(n-k)!}\) For each problem, calculate the combinations: a. \(C_{2}^{8} = \frac{8!}{2!(8-2)!} = 28\) b. \(C_{0}^{4} = \frac{4!}{0!(4-0)!} = 1\) c. \(C_{3}^{10} = \frac{10!}{3!(10-3)!} = 120\) d. \(C_{1}^{7} = \frac{7!}{1!(7-1)!} = 7\)
02

Evaluating the binomial probabilities

Now, plug in our calculated combinations and given probability values into the binomial probability formula and evaluate the probabilities: a. \(P(X=2) = 28(.3)^{2}(.7)^{6} \approx 0.3241\) b. \(P(X=0) = 1(.05)^{0}(.95)^{4} \approx 0.8145\) c. \(P(X=3) = 120(.5)^{3}(.5)^{7} \approx 0.1172\) d. \(P(X=1) = 7(.2)^{1}(.8)^{6} \approx 0.3670\) So, the evaluated binomial probabilities are as follows: a. \(P(X=2) \approx 0.3241\) b. \(P(X=0) \approx 0.8145\) c. \(P(X=3) \approx 0.1172\) d. \(P(X=1) \approx 0.3670\)

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