Question

There are two traffic lights on the route used by a certain individual to go from home to work. Let E denote the event that the individual must stop at the first light, and define the event F in a similar manner for the second light. Suppose that P(E) = 0.4, P(F) = 0.2 and P(E ∩ F) = 0.15.(a) What is the probability that the individual must stop at at least one light; that is, what is the probability of the event P(E ∪ F)?(b) What is the probability that the individual needn't stop at either light?(c) What is the probability that the individual must stop at exactly one of the two lights?(d) What is the probability that the individual must stop just at the first light? (Hint: How is the probability of this event related to P(E) and P(E ∩ F)? A Venn diagram might help.)

Answer 1

Answer with Step-by-step explanation:

We are given that

E=Event denote the event that the individual must stop at the first light

F=Event denote the event that the individual must stop at the second light

P(E)=0.4

P(F)=0.2

$P(E\cap F)=0.15$

a.We have to find the probability that the individual must stop at atleast one light.

We know that

$P(E\cup F)=P(E)+P(F)-P(E\cap F)$

Substitute the value in the given formula then, we get

$P(E\cup F)=0.4+0.2-0.15=0.45$

$P(E\cup F)=0.45$

b.$P(E\cup F)'=1-P(E\cup F)$

$P(E\cup F)'=1-0.45=0.55$

c.We have to find the probability that the individual must stop at exactly one of the two lights.

P(must stop at exactly one of the two lights)=$P(E\cup F)-P(E\cap F)=0.45-0.15=0.3$

P(must stop at exactly one of the two lights)=0.3

d.We have to find the probability that the individual must stop just at the first light.

P(must stop juts at the first light)=$P(E)-P(E\cap F)$

P(must stop juts at the first light)=0.4-0.15=0.25