find the probability exactly 3 successes in 6 trials of a binomial experiment in which the probability of success if 50%. round to the nearest tenth of a percent.

Answer 1

The probability of r successes in n trials is _nC_r(p)^r(q)^n-r, where p is the probability of success on a given trial and q = 1-p.

In this problem, n = 6, r = 3, p = q = 0.5

So, P(3 successes in 6 trials) = ₆C₃(0.5)³(0.5)³ = 20(1/64) = 20/64 = 5/16 = 0.3125 ≈ 31.3%

Answer 2

I think that the binomial formula applies here, so we can use the general formula for the binomial distribution:

P(x) = N! / [x! * (N-x)!] * p^x * (1-p)^x

where P(x) is the probability of x successes out of N trials, N is the number of trials, and p is the probability of success on a given trial. ('!' is the factorial function.)

Plugging in the appropriate values, we get

P(3) = (6! / [3! * (6-3)!] ) * .5^x * (1-.5)^x

P(3) = (720 / 36) * .125 * .125

P(3) = .3125

Answer 3

probability of 3:3 is 50%