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

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.

Given:

$$P(E) = .6$$

$$P(F) = .4$$

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

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

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

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