There are two traffic lights. Let $E$ be the event that you stop at the first light and $F$ be the event that you stop at the second light.


$P(E) = .6$

$P(F) = .4$

$P(E \text{ and } F) = .25$

What is the probability that you stop at only the first traffic light?

My attempt:

We want $P(E \text{ and } F^c)$

$P(E \text{ and } F^c) = P(E|F^c)P(F^c)$

The question then becomes what is $P(E|F^c)$?

Conditioning on $F$ leads to:

$P(E) = P(E|F) + P(E|F^c)$

Where $P(E|F) = \frac{P(E \text{ and } F)}{P(F)}$

$P(E) - P(E|F) = P(E|F^c)$

$P(E) - \frac{P(E \text{ and } F)}{P(F)} = P(E|F^c)$

But $P(E) = .6$ and $\frac{P(E \text{ and } F)}{P(F)}=.625$ so I get a negative number....

What did I do wrong, and what's the easiest way to solve this problem? Thanks

  • 4
    $\begingroup$ Note that $P(E\cap F)+P(E\cap F^c)=P(E)$ $\endgroup$
    – lulu
    Commented Sep 29, 2022 at 20:11

2 Answers 2

  1. $P(E\cap F^c)=P(F^c| E)P(E), \text{with}\ \ P(E)=0.6.$

  2. $\text{But}\ \ \ P(F^c|E)=1-P(F|E)=1-\frac{P(F\cap E)}{P(E)}=1-\frac{0.25}{0.6}=\frac{7}{12}.$

  3. $\text{Therefore,}\ \ \ P(E\cap F^c)=\frac{7}{12}\frac{6}{10}=\frac{7}{20}.$

In your development, $P(E) = P(E|F)\color{red}{P(F)} + P(E|F^c)\color{red}{P(F^c)}$, the terms in red are missing.


Using lulu's hint (which you can see if you imagine a Venn diagram):

$$P(E\cap F^c)=P(E)-P(E\cap F)=.6-.25=.35.$$


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