Assume that a procedure yields a binomial distribution with a trial repeated $n=8$ times. Use either the binomial probability formula (or technology) to find the probability of $k=8$ successes given the probability $p=63 \%$ of success on a single trial.
(Report answer accurate to 4 decimal places.)
\[
P(X=k)=
\]
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Answer
\(\boxed{P(X=8) = 0.0631}\) is the final answer, which means the probability of getting 8 successes in 8 trials, given a 63% chance of success on each trial, is approximately 0.0631 (or 6.31% when expressed as a percentage).
Step 1 ：Given that the number of trials \(n=8\), the number of successes \(k=8\), and the probability of success on a single trial \(p=0.63\).
Step 2 ：The binomial probability formula is given by: \[P(X=k) = C(n, k) * (p^k) * ((1-p)^(n-k))\]
Step 3 ：Substitute the given values into the formula: \[P(X=8) = C(8, 8) * (0.63^8) * ((1-0.63)^(8-8))\]
Step 4 ：The combination \(C(8, 8)\) is 1 (since there's only one way to choose all 8 items from a set of 8), and \((1-0.63)^(8-8)\) is also 1 (since any number to the power of 0 is 1). So, the formula simplifies to: \[P(X=8) = 1 * (0.63^8) * 1 = 0.63^8\]
Step 5 ：Calculating this gives: \[P(X=8) = 0.0631\]
Step 6 ：\(\boxed{P(X=8) = 0.0631}\) is the final answer, which means the probability of getting 8 successes in 8 trials, given a 63% chance of success on each trial, is approximately 0.0631 (or 6.31% when expressed as a percentage).
