Question

The route used by a certain motorist in commuting to workcontains two intersections with traffic signals. The probabilitythat he must stop at the first signal is .4, the analogousprobability for the second signal is .5, and the probability thathe must stop at least one of the two signals is .6. 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) 0.3

b) 0.1

c) 0.3

Step-by-step explanation:

Lets call:

a = stop at first signal,  b = stop at second signal

The data we are given is P(a) = 0.4, and P(b)=0.5

Stoping at least at one is the event (a or b) = a ∪ b

P(a U b) = 0.6 is the other data we are given

a) Stoping at both signals is the event (a and b = a ∩ b)

The laws of probability say that:

P(a ∪ b)= P(a) + P(b) - P( a ∩ b) = 0.4 + 0.5  - P( a ∩ b) = 0.6

Then we get P( a ∩ b) = 0.3

b) The event is (a and not b) = a ∩(¬b).

P( a ∩(¬b) ) = P(a) - P( a ∩ b) = 0.1

c) The event is (a or b) without the cases in which (a and b)

P(a ∪ b) - P( a ∩ b) = 0.3

The Venn diagram can help you understand how the events are related to each other

Answer 2

Answer:

a) 0.2

b) 0.2

c) 0.5

Step-by-step explanation:

Let $S$ be the event "the car stops at the signal.

In the attached figure you can see a tree describing all possible scenarios.

For the first question the red path describes stopping at the first light but not stopping at the second. We can determine the probability of this path happening by multiplying the probabilities on the branches of the tree, thus

$P(a)=0.4\times0.5=0.2$

For the second one the blue path describes the situation

$P(b)=0.4\times 0.5=0.2$

For the las situation the sum of the two green path will give us the answer

$P(c)=0.6\times 0.5 + 0.4\times 0.5=0.3+0.2=0.5$