Question

Assume that a procedure yields a binomial distribution with a trial repeated n = 9 times. Use either the binomial probability formula (or technology) to find the probability of k = 6 successes given the probability p = 0.53 of success on a single trial. (Report answer accurate to 4 decimal places.) P(X = k) = C

Answer 1

n = 9

no of success = k = 6

P (success) = 0.53

P(x = 6)

$P(x) =\frac{N!}{x! (N-x)!} \pi ^{x} (1-\pi)^{N-x}$

$P(6) =\frac{9!}{6! (9-6)!} 0.53 ^{6} (1-0.53)^{9-6}$

$P(6) =\frac{9!}{6! (3)!} 0.53 ^{6} (0.47)^{3}$

P(6) = 0.1933

ANS : Probability of 6 success = 0.1933