The route used by a certain motorist in commuting to work contains two

Rena Giron

Rena Giron

Answered question

2021-11-17

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 .4, the analogous probability for the second signal is .5, and the probability that he must stop at at least one of the two signals is .7. What is the probability that he must stop
a. At both signals?
b. At the first signal but not at the second one?
c. At exactly one signal?

Answer & Explanation

Kathleen Ashton

Kathleen Ashton

Beginner2021-11-18Added 15 answers

Step 1
Let A be the event that the person stops at first signal.
P(A)=0.4
Let B be the event that the person stops at second signal.
P(B)=0.5
And the probability that he must stop at at least one of the two signals is 0.7
P(A∪B)=0.7
Step 2
a) The probability that person stops at both signals is P(A∩B).
P(A∩B)=P(A)+P(B)−P(A∪B)
=0.4+0.5−0.7
=0.2
The probability that person stops at both signals is equal to 0.2.
Step 3
b) The probability of person stop at the first signal but not at the second one is P(A∩Bc).
P(A∩Bc)=P(A)−P(A∩B)
=0.4−0.2
=0.2
The probability of person stop at the first signal but not at the second one is equal to 0.2.
Step 4
c) The probability of person stops at exactly one signal is P(A∩Bc)+P(Ac∩B).
P(A∩Bc)+P(Ac∩B)=[P(A)−P(A∩B)]+[P(B)−P(A∩B)]
=[0.4−0.2]+[0.5−0.2]
=0.2+0.3
=0.5
The probability of person stops at exactly one signal is equal to 0.5
Mr Solver

Mr Solver

Skilled2023-06-14Added 147 answers

Answer:
a. The probability that the motorist must stop at both signals is 0.2.
b. The probability that the motorist must stop at the first signal but not at the second one is 0.2.
c. The probability that the motorist must stop at exactly one signal is 0.5.
Explanation:
a. To find the probability that the motorist must stop at both signals, we can use the concept of independent events. The probability of stopping at the first signal is given as P(stop at first signal)=0.4, and the probability of stopping at the second signal is given as P(stop at second signal)=0.5. Since these events occur at different intersections, we can assume they are independent. Therefore, the probability of stopping at both signals is the product of the individual probabilities:
P(stop at both signals)=P(stop at first signal)×P(stop at second signal)=0.4×0.5=0.2.
b. To find the probability that the motorist must stop at the first signal but not at the second one, we can subtract the probability of stopping at both signals from the probability of stopping at the first signal:
P(stop at first signal but not at second signal)=P(stop at first signal)−P(stop at both signals)=0.4−0.2=0.2.
c. To find the probability that the motorist must stop at exactly one signal, we need to consider two cases: stopping at the first signal but not at the second, or stopping at the second signal but not at the first.
The probability of stopping at the first signal but not at the second is already calculated as 0.2 in part b. Similarly, the probability of stopping at the second signal but not at the first can be calculated as:
P(stop at second signal but not at first signal)=P(stop at second signal)−P(stop at both signals)=0.5−0.2=0.3.
Since these two cases are mutually exclusive (the motorist cannot stop at both signals), we can add their probabilities to find the overall probability:
P(stop at exactly one signal)=P(stop at first signal but not at second signal)+P(stop at second signal but not at first signal)=0.2+0.3=0.5.
Therefore, the probability that the motorist must stop at exactly one signal is 0.5
Nick Camelot

Nick Camelot

Skilled2023-06-14Added 164 answers

Let's denote the events as follows:
A: The motorist must stop at the first signal.
B: The motorist must stop at the second signal.
We are given the following probabilities:
P(A)=0.4
P(B)=0.5
P(A∪B)=0.7
a. To find the probability that the motorist must stop at both signals (A and B), we can use the formula for the intersection of two events:
P(A∩B)=P(A)·P(B|A)
Since the events are independent (stopping at one signal does not affect the probability of stopping at the other signal), we have:
P(B|A)=P(B)=0.5
Substituting the given values, we can calculate:
P(A∩B)=0.4·0.5=0.2
Therefore, the probability that the motorist must stop at both signals is 0.2.
b. To find the probability that the motorist must stop at the first signal (A) but not at the second signal (B), we can subtract the probability of stopping at both signals from the probability of stopping at the first signal:
P(A∩¬B)=P(A)−P(A∩B)
Using the values we know:
P(A∩¬B)=0.4−0.2=0.2
Therefore, the probability that the motorist must stop at the first signal but not at the second one is 0.2.
c. To find the probability that the motorist must stop at exactly one signal, we can use the principle of inclusion-exclusion:
P((A∩¬B)∪(¬A∩B))=P(A∩¬B)+P(¬A∩B)
We have already calculated P(A∩¬B)=0.2. Now, to find P(¬A∩B), we can use the fact that P(A∪B)=P(A)+P(B)−P(A∩B). Rearranging this equation, we get:
P(A∩B)=P(A)+P(B)−P(A∪B)
Substituting the given values, we find:
P(A∩B)=0.4+0.5−0.7=0.2
Therefore, P(¬A∩B)=P(A∩B)=0.2.
Now we can calculate the probability of stopping at exactly one signal:
P((A∩¬B)∪(¬A∩B))=P(A∩¬B)+P(¬A∩B)=0.2+0.2=0.4
Therefore, the probability that the motorist must stop at exactly one signal is 0.4.
madeleinejames20

madeleinejames20

Skilled2023-06-14Added 165 answers

Step 1:
a. To find the probability that the motorist must stop at both signals, we can multiply the probabilities of stopping at each signal. Let's denote the event of stopping at the first signal as A and the event of stopping at the second signal as B. We need to find P(A∩B).
Using the multiplication rule of probability, we have:
P(A∩B)=P(A)·P(B|A)
Given that the probability of stopping at the first signal is 0.4, we have P(A)=0.4.
The probability of stopping at the second signal given that the motorist has already stopped at the first signal is denoted by P(B|A). However, we are not given this conditional probability directly.
To find P(B|A), we can use the fact that the probability of stopping at at least one of the two signals is 0.7. We can represent this event as A∪B, where ∪ denotes the union of events.
The probability of the union of two events is given by the addition rule of probability:
P(A∪B)=P(A)+P(B)−P(A∩B)
Substituting the given probabilities, we have:
0.7=0.4+0.5−P(A∩B)
Simplifying the equation, we find:
P(A∩B)=0.4+0.5−0.7
Therefore, the probability that the motorist must stop at both signals is 0.2.
Step 2:
b. To find the probability that the motorist must stop at the first signal but not at the second signal, we need to find P(A∩¬B), where ¬B represents the complement of event B (not stopping at the second signal).
Using the multiplication rule, we have:
P(A∩¬B)=P(A)·P(¬B|A)
To find P(¬B|A), we can use the fact that the probability of stopping at at least one of the two signals is 0.7. We can represent this event as A∪B, as mentioned earlier.
The probability of not stopping at the second signal given that the motorist has already stopped at the first signal is denoted by P(¬B|A).
Using the complement rule of probability, we have:
P(¬B|A)=1−P(B|A)
Substituting the given probability P(B|A)=0.5, we find:
P(¬B|A)=1−0.5=0.5
Therefore, the probability that the motorist must stop at the first signal but not at the second signal is:
P(A∩¬B)=0.4·0.5=0.2
Step 3:
c. To find the probability that the motorist must stop at exactly one signal, we can use the subtraction rule of probability. Let's denote the event of stopping at exactly one signal as A⊕B, where ⊕ represents the exclusive OR (XOR) operation between events.
The probability of A⊕B is given by:
P(A⊕B)=P(A∪B)−P(A∩B)
We already know that P(A∪B)=0.7 and P(A∩B)=0.2. Substituting these values, we find:
P(A⊕B)=0.7−0.2=0.5
Therefore, the probability that the motorist must stop at exactly one signal is 0.5.

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