Question

Every Sunday morning, two children, Craig and Jill, independently try to launch their model airplanes. On each Sunday, Craig has a probability \(\frac{{\bf{1}}}{{\bf{3}}}\) of a successful launch, and Jill has the probability \(\frac{{\bf{1}}}{{\bf{5}}}\) of a successful launch. Determine the expected number of Sundays required until at least one of the two children has a successful launch.

Short answer

\(\frac{{15}}{7}\)

Step-by-step solution

Given information

The success probability of Craig is\(\frac{1}{3}\)

The success probability of Jill is\(\frac{1}{5}\)

Both launch the planes independently.

Calculate the required probability

The probability that at least one of the two children has a successful launch.

\(\begin{array}{l}{\rm{ = 1 - Pr}}\left( {{\rm{no}}\,\,{\rm{success}}} \right)\\{\rm{ = 1 - Pr}}\left( {{\rm{Craig}}\,\,{\rm{fails}}} \right){\rm{ \times Pr}}\left( {{\rm{Jill}}\,\,{\rm{fails}}} \right)\\ = 1 - \left( {1 - \frac{1}{3}} \right) \times \left( {1 - \frac{1}{5}} \right)\\ = 1 - \left( {\frac{2}{3} \times \frac{4}{5}} \right)\\ = 1 - \frac{8}{{15}}\\ = \frac{7}{{15}}\end{array}\)

Calculating the expected number of failure days

Let the probability that at least one of the two children has a successful launch follows a be a random variable denoted by X. The X follows a geometric distribution with parameter p that is \(X \sim Geo\left( {p = \frac{7}{{15}}} \right)\)

By the properties of a geometric distribution, the expectation of X is:

\(\begin{aligned}{}{\rm E}\left( X \right) &= \frac{1}{p}\\ &= \frac{1}{{\frac{7}{{15}}}}\\ &= \frac{{15}}{7}\end{aligned}\)

Therefore, the answer is\(\frac{{{\bf{15}}}}{{\bf{7}}}\).