Ms. Tucker travels through two intersections with traffic lights as she drives to the market. The traffic lights operate independently. The probability that both lights will be red when she reaches them is 0.22. The probability that the first light will be red and the second light will not be red is 0.33. What is the probability that the second light will be red when she reaches it?

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The probability that the first light is red is simply the sum of the two probabilities given since they comprise of all the possible situations where the first light is red.

$$ P(F_r) = 0.22 + 0.33 = 0.55 $$

The probability that both lights are red is 0.22

$$ P(F_r)\cap P(S_r) = P(F_r)\cdot P(S_r) $$ $$ 0.22 = 0.55 \cdot P(S_r) $$ $$ P(S_r) = \frac{0.22}{0.55} $$ $$ P(S_r)= \boxed{0.40} $$

Writing down what's given:

$$P(F_r)\cdot P(S_r) = 0.22 $$ $$ P(F_r)\cdot P(S_g) = 0.33 $$

We want to find \(P(S_r)\). We can do a few algebraic manipuations.

Dividing the two equations:

$$ \frac{P(F_r)\cdot P(S_r)}{P(F_r)\cdot P(S_g)} = \frac{0.22}{0.33} $$ $$ \frac{\cancel{P(F_r)}\cdot P(S_r)}{\cancel{P(F_r)}\cdot P(S_g)} = \frac{0.22}{0.33} $$ $$ \frac{P(S_r)}{P(S_g)} = \frac{2}{3} $$

Note that the second light can only be on or off, so \(P(S_r)+P(S_g)=1\)

$$ P(S_g)=1-P(S_r) $$ $$ \frac{P(S_r)}{1-P(S_r)} = \frac{2}{3} $$ $$ 3P(S_g) = 2-2P(S_r) $$ $$ 5P(S_g) = 2 $$ $$ p(S_g) = \frac{2}{5} = \boxed{0.40}$$