Question

Suppose that there is a 1 in 60 chance of injury on a single skydiving attempt.

A friend claims there is a 100% chance of injury if a skydiver jumps 60 times. Assume that the results of repeated jumps are mutually independent.

(a) What is the probability that 60 jumps will be completed without an injury?

(b) What is the probability that at least one injury will occur in 60 jumps?

(c) What is the maximum number of jumps, n, the skydiver can make if the probability is at least 0.70 that all n jumps will be completed without injury?

Answer 1

(a)

Let X be the number of jumps out of 60 to get an injury.

X ~ Binomial(n = 60, p = 1/60)

Probability that 60 jumps will be completed without an injury = P(X = 0)

= 0.3647923

(b)

Probability that at least one injury will occur in 60 jumps = P(X 1)

= 1 - P(X = 0)

= 1 - 0.3647923

= 0.6352077

(c)

X ~ Binomial(n, p = 1/60)

P(X = 0) 0.70

Taking natural log, we get

n ln(59/60) ln(0.70)

n * -0.01680712   -0.3566749

n 0.3566749 / 0.01680712

n 21.22

Thus, the maximum number of jumps is 21, such that the probability is at least 0.70 when all n jumps will be completed without injury.