A spacecraft carrying two female and three male astronauts makes a trip toMars. The plan calls for a two-person detachable capsule to land at site A on the planet and a second one-person capsule to land at site B. The other two astronauts remain in orbit. Which three will board the two capsules is decided by a lottery, consisting of picking names from a box.

What is the probability that a female occupies the one-person capsule if we know that at least one member of the other capsule is female, but we are not told the order in which the astronauts were picked?

This is a problem from chapter 4 of this book. I have no clue how to approach this problem. Can anyone please help?


Define events:

\begin{align} A &= \text{At least one female in capsule } A \\ B &= \text{A female in capsule } B. \end{align}

The probability required is:

\begin{align} P(B\mid A) &= \dfrac{P(A\mid B)P(B)}{P(A)} \qquad\qquad\qquad\qquad\qquad\text{by Bayes Theorem} \\ &= \dfrac{(1-P(A^c\mid B))P(B)}{1-P(A^c)} \qquad\qquad\qquad\qquad (1)\\ \end{align}

Each of these required probabilities is easily calculated:

\begin{align} P(A^c\mid B) &= \binom{3}{2} \bigg/ \binom{4}{2} = \dfrac{1}{2} \\ P(B) &= \dfrac{2}{5} \\ P(A^c) &= \binom{3}{2} \bigg/ \binom{5}{2} = \dfrac{3}{10}. \end{align}

Substituting into $(1)$ gives:

$$P(B\mid A) = \dfrac{2}{7}.$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.