Question

The route used by a certain motorist in commuting to work contains two intersections with traffic signals. The probability that he must stop at the first signal is 0.36, the analogous probability for the second signal is 0.51, and the probability that he must stop at least one of the two signals is 0.67.What is theprobability that he must stop.
a) At both signals?
b) At the first signal but not at the second one?
c) At exactly on signal?

Answer 1

Answer:

a) P(X∩Y) = 0.2

b) $P_1$ = 0.16

c) P = 0.47

Step-by-step explanation:

Let's call X the event that the motorist must stop at the first signal and Y the event that the motorist must stop at the second signal.

So, P(X) = 0.36, P(Y) = 0.51 and P(X∪Y) = 0.67

Then, the probability P(X∩Y) that the motorist must stop at both signal can be calculated as:

P(X∩Y) = P(X) + P(Y) - P(X∪Y)

P(X∩Y) = 0.36 + 0.51 - 0.67

P(X∩Y) = 0.2

On the other hand, the probability $P_1$ that he must stop at the first signal but not at the second one can be calculated as:

$P_1$ = P(X) - P(X∩Y)

$P_1$ = 0.36 - 0.2 = 0.16

At the same way, the probability $P_2$ that he must stop at the second signal but not at the first one can be calculated as:

$P_2$ = P(Y) - P(X∩Y)

$P_2$ = 0.51 - 0.2 = 0.31

So, the probability that he must stop at exactly one signal is:

$P = P_1+P_2\\P=0.16+0.31\\P=0.47$