To solve this problem, we will use the formula for the probability of k successes in n trials for a binomial experiment, which is:

P(X=k) = C(n, k) * (p^k) * ((1-p)^(n-k))

where:
- P(X=k) is the probability of k successes,
- C(n, k) is the combination of n items taken k at a time,
- p is the probability of success,
- n is the number of trials,
- k is the number of successes.

Given in the problem, we have p=0.75 (probability of success), n=5 (number of trials), and we want to find the probability of exactly 2 successes, so k=2.

First, calculate C(n, k) = C(5, 2). The formula for combination is C(n, k) = n! / [k!(n-k)!], where "!" denotes factorial. So, C(5, 2) = 5! / [2!(5-2)!] = 10.

Next, calculate p^k = (0.75)^2 = 0.5625.

Then, calculate (1-p)^(n-k) = (1-0.75)^(5-2) = (0.25)^3 = 0.015625.

Finally, substitute these values into the formula to get:

P(X=2) = C(5, 2) * (0.75)^2 * (0.25)^3 = 10 * 0.5625 * 0.015625 = 0.088.

So, the probability that the experiment results in exactly 2 successes is 0.088 or 8.8% when rounded to three decimal places.

Question

To solve this problem, we will use the formula for the probability of k successes in n trials for a binomial experiment, which is:

P(X=k) = C(n, k) * (p^k) * ((1-p)^(n-k))

where:
- P(X=k) is the probability of k successes,
- C(n, k) is the combination of n items taken k at a time,
- p is the probability of success,
- n is the number of trials,
- k is the number of successes.

Given in the problem, we have p=0.75 (probability of success), n=5 (number of trials), and we want to find the probability of exactly 2 successes, so k=2.

First, calculate C(n, k) = C(5, 2). The formula for combination is C(n, k) = n! / [k!(n-k)!], where "!" denotes factorial. So, C(5, 2) = 5! / [2!(5-2)!] = 10.

Next, calculate p^k = (0.75)^2 = 0.5625.

Then, calculate (1-p)^(n-k) = (1-0.75)^(5-2) = (0.25)^3 = 0.015625.

Finally, substitute these values into the formula to get:

P(X=2) = C(5, 2) * (0.75)^2 * (0.25)^3 = 10 * 0.5625 * 0.015625 = 0.088.

So, the probability that the experiment results in exactly 2 successes is 0.088 or 8.8% when rounded to three decimal places.

Knowee AI · Accepted Answer

To solve this problem, we will use the formula for the probability of k successes in n trials for a binomial experiment, which is:

P(X=k) = C(n, k) * (p^k) * ((1-p)^(n-k))

where:
- P(X=k) is the probability of k successes,
- C(n, k) is the combination of n items taken k at a time,
- p is the probability of success,
- n is the number of trials,
- k is the number of successes.

Given in the problem, we have p=0.75 (probability of success), n=5 (number of trials), and we want to find the probability of exactly 2 successes, so k=2.

First, calculate C(n, k) = C(5, 2). The formula for combination is C(n, k) = n! / [k!(n-k)!], where "!" denotes factorial. So, C(5, 2) = 5! / [2!(5-2)!] = 10.

Next, calculate p^k = (0.75)^2 = 0.5625.

Then, calculate (1-p)^(n-k) = (1-0.75)^(5-2) = (0.25)^3 = 0.015625.

Finally, substitute these values into the formula to get:

P(X=2) = C(5, 2) * (0.75)^2 * (0.25)^3 = 10 * 0.5625 * 0.015625 = 0.088.

So, the probability that the experiment results in exactly 2 successes is 0.088 or 8.8% when rounded to three decimal places.

A binomial experiment with probability of success =p0.75 and =n5 trials is conducted. What is the probability that the experiment results in exactly 2 successes?Do not round your intermediate computations, and round your answer to three decimal places. (If necessary, consult a list of formulas.)

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