Question

Let the probability of success on a Bernoulli trial be 0.16.

a. In seven Bernoulli trials, what is the probability that there will be 6 failures? (Do not round intermediate calculations. Round your final answers to 4 decimal places.)

b. In seven Bernoulli trials, what is the probability that there will be more than the expected number of failures? (Do not round intermediate calculations. Round your final answers to 4 decimal places.)

Answer 1

For Bernoulli trial, probability of X successes, P(X) = nCx p^x q^n-x,

where n is the number of trials,

p is the probability of success and

q is the probability of failure = 1 - p

a) n = 7

p = 0.16

q = 1 - 0.16 = 0.84

P(6 failures) = P(1 success)

= 7C1 x 0.16 x 0.84⁶

= 0.3935

b) n = 7

p = 0.16

q = 0.84

Expected number of failure = nq

= 7 x 0.84 = 5.88

P(more than the expected number of failures) = P(6 or 7 failures)

= P(6 failures) + P(7 failures)

= P(1 success) + P(0 successes)

= 7C1 x 0.16 x 0.84⁶ + 0.84⁷

= 0.6885