For n repeated independent trials, with constant probability of success p for all trials, find the probability of exactly x successes. Round your answer to four decimal places

n = 10, p = 1/7, x = 3

A. 0.1189

B. 0.1546

C. 0.1308

D. 0.1427

Plug the variables in.

For 10 repeated independent trials, with constant probability of success 1/7 for all trials, find the probability of exactly 3 successes. Round your answer to four decimal places

first find probablility of 3 successes,

then find probability of 7 failures.

To find the probability of exactly x successes in n repeated independent trials, we can use the binomial probability formula:

P(x) = (nCx) * p^x * (1-p)^(n-x)

where n is the number of trials, p is the probability of success in one trial, x is the number of successes, and (nCx) represents the number of combinations of n items taken x at a time.

In this case, n = 10, p = 1/7, and x = 3.

To calculate (nCx), we use the combination formula:

(nCx) = n! / (x! * (n-x)!)

Let's calculate each component step by step:

1. Calculate (nCx):
(nCx) = 10! / (3! * (10-3)!)
= 10! / (3! * 7!)
= (10 * 9 * 8) / (3 * 2 * 1)
= 120 / 6
= 20

2. Calculate p^x * (1-p)^(n-x):
p^x = (1/7)^3 = 1/343
(1-p)^(n-x) = (1 - 1/7)^(10-3) = (6/7)^7 ≈ 0.3802

3. Multiply (nCx) by p^x and (1-p)^(n-x):
20 * 1/343 * 0.3802 ≈ 0.0022

Therefore, the probability of exactly 3 successes in 10 repeated independent trials, with a constant probability of success of 1/7, is approximately 0.0022.

Since none of the provided options match this result, there may be a mistake in the calculation or the answer choices are incorrect.