If exactly one broken pen is chosen, what is the probability that the girl chose it?

Question

A box of colouring pens contains five broken pens and nine working pens. A pen is required for each of the two children, a girl and a boy. If exactly one broken pen is chosen, what is the probability that the girl chose it?

My attempt: $(5/14 \cdot 5/13)+(9/14 \cdot 5/13)$, so $90/182$

Is this correct? If not why?

Many thanks for your help.

Cheers.

Answer 1

Without any calculation the requested probability is 0.5

If you want to calculate it, observe that the probability to get exactly one broken pen is

$$P(X=1)= \frac{\binom{5}{1}\binom{9}{1}}{\binom{14}{2}}+\frac{\binom{5}{1}\binom{9}{1}}{\binom{14}{2}} =A+A $$

The conditional requested probability is evidently

$$\frac{A}{A+A}=0.5$$

Answer 2

We are asked for the conditional probability that the girl selects the broken pen given that exactly one of the two selected pens is broken. To find it, we must divide the probability that exactly one pen is broken and it is selected by the girl by the probability that exactly one pen is broken.

There are two possibilities if exactly one of the selected pens is broken:

• The boy selects one of the nine working pens and the girl selects one of the broken pens when two of the fourteen pens are selected, which occurs with probability $$\frac{\dbinom{9}{1}\dbinom{5}{1}}{\dbinom{14}{2}}$$
• The boy selects one of the five broken pens and the girl selects one of the nine working pens when two of the fourteen pens are selected, which occurs with probability $$\frac{\dbinom{5}{1}\dbinom{9}{1}}{\dbinom{14}{2}}$$

Since these two possibilities are mutually exclusive and exhaustive, the probability that exactly one pen is broken is $$\frac{\dbinom{9}{1}\dbinom{5}{1}}{\dbinom{14}{2}} + \frac{\dbinom{5}{1}\dbinom{9}{1}}{\dbinom{14}{2}}$$ The probability that exactly one pen is broken and the girl selects it is $$\frac{\dbinom{9}{1}\dbinom{5}{1}}{\dbinom{14}{2}}$$ Hence, \begin{align*} & \Pr(\text{girl selects broken pen given exactly one pen is broken})\\ & \qquad = \frac{\Pr(\text{exactly one pen is broken and it is selected by the girl})}{\Pr(\text{exactly one pen is broken})}\\ & \qquad = \frac{\frac{\dbinom{9}{1}\dbinom{5}{1}}{\dbinom{14}{2}}}{\frac{\dbinom{9}{1}\dbinom{5}{1}}{\dbinom{14}{2}} + \frac{\dbinom{5}{1}\dbinom{9}{1}}{\dbinom{14}{2}}}\\ & \qquad = \frac{1}{2} \end{align*} as we should expect since it is equally likely that the boy or the girl would select the broken pen if exactly one of the two selected pens is broken.

If you would prefer to take the order of selection into account, there are four ways exactly one broken pen could be selected.

• The boy selects a working pen, then the girl selects a broken pen, which occurs with probability $$\frac{9}{14} \cdot \frac{5}{13}$$
• The boy selects a broken pen, then the girl selects a working pen, which occurs with probability $$\frac{5}{14} \cdot \frac{9}{13}$$
• The girl selects a working pen, then the boy selects a broken pen, which occurs with probability $$\frac{9}{14} \cdot \frac{5}{13}$$
• The girl selects a broken pen, then the boy selects a working pen, which occurs with probability $$\frac{5}{14} \cdot \frac{9}{13}$$

The first and fourth possibilities are favourable (to the boy, anyway). Hence, \begin{align*} & \Pr(\text{girl selects broken pen given exactly one pen is broken})\\ & \qquad = \frac{\Pr(\text{exactly one pen is broken and it is selected by the girl})}{\Pr(\text{exactly one pen is broken})}\\ & \qquad = \frac{\dfrac{9}{14} \cdot \dfrac{5}{13} + \dfrac{5}{14} \cdot \dfrac{9}{13}}{\dfrac{9}{14} \cdot \dfrac{5}{13} + \dfrac{5}{14} \cdot \dfrac{9}{13} + \dfrac{9}{14} \cdot \dfrac{5}{13} + \dfrac{5}{14} \cdot \dfrac{9}{13}}\\ & \qquad = \frac{1}{2} \end{align*}