Assume the geometric distribution applies. Use the given probability of success p to find the indicated probability. Find $P(5)$ when $p=0.10$
\[
P(5)=\square \text { (Rgund to five decimal places as needed.) }
\]
Answer
So, the probability that the first success occurs on the 5th trial, given that the probability of success on each trial is 0.10, is approximately \(\boxed{0.06561}\).
Step 1 ：We are given a problem that involves the geometric distribution. The geometric distribution is a probability distribution that describes the number of trials needed to get the first success in repeated Bernoulli trials. The probability mass function of the geometric distribution is given by: \(P(X=k) = (1-p)^{k-1}p\), where \(P(X=k)\) is the probability that the first success occurs on the \(k\)th trial, \(p\) is the probability of success on each trial, and \(k\) is the number of trials.
Step 2 ：In this case, we are asked to find \(P(5)\) when \(p=0.10\). This means we want to find the probability that the first success occurs on the 5th trial, given that the probability of success on each trial is 0.10.
Step 3 ：We can substitute \(k=5\) and \(p=0.10\) into the formula to find the answer.
Step 4 ：By substituting the given values into the formula, we get \(P_5 = 0.06561\).
Step 5 ：So, the probability that the first success occurs on the 5th trial, given that the probability of success on each trial is 0.10, is approximately \(\boxed{0.06561}\).
