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All of Adam's friends either jog or go to the gym, or possibly both. He knows 70% of friends jog and 35% gym. What percentage of Adam's friends go to the gym, but do not jog?

My solution:

$$P(\text{gym}\mid\text{no jog}) = \dfrac{P(\text{gym} \, \cap \, \text{no jog})}{P(\text{no jog})}$$

I would like verify a few things.

  1. Is the probability of 'no jog' actually just $1-0.7=0.3$, or is it not this simple?
  2. To calculate $P(\text{gym} \, \cap \, \text{no jog})$, must I draw a Venn diagram and work it out? Is there a better way?
joriki
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Trogdor
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2 Answers2

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Yes, is seems a bit simple. You can draw a Venn-digram. But if you have only two variables I find it easier to make a contingency table. From the given information we can obtain the following table.

$$\begin{array}{|c|c|c|} \hline & G & \overline G & \\ \hline J & & &0.7\\ \hline \overline J & & \color{blue}0 & 0.3\\ \hline & 0.35 & 0.65& 1\\ \hline \end{array}$$

$G$: Go to the gym,$\overline G$: Not go to the gym, $J$: Go jogging, $\overline J$: Not go jogging

Because of the information that all of Adam's friends either jog or go to the gym we can insert the blue $0$.

Now it is simple to evaluate the missing values in the table, especially $P(\overline J\cap G)$. This is the probability what is asked for.

callculus42
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"All of Adam's friends either jog or go to the gym, or possibly both" imply: $$P(J\cup G)=P(J)+P(G)-P(J\cap G)=1 \iff \\ P(J\cap G)=0.7+0.35-1=0.05.$$ Then: $$P(G)=P(G\cap J)+P(G\cap J^C) \iff 0.35=0.05+P(G\cap J^C) \iff P(G\cap J^C)=0.3.$$

farruhota
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