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There are two traffic lights on Fred’s route from home to work. Over the years Fred has tracked when he has to stop at these lights. He determined that 45% of the time he has to stop at the first light; 30% of the time he has to stop at the second light, and 20% of the time he has to stop at both lights.Let A be the event that Fred has to stop at the first light. Let B be the event that Fred has to stop at the second light. What is the probability that Fred has to stop at one or more lights on his way to work?0.6150.550.650.95

Question

There are two traffic lights on Fred’s route from home to work. Over the years Fred has tracked when he has to stop at these lights. He determined that 45% of the time he has to stop at the first light; 30% of the time he has to stop at the second light, and 20% of the time he has to stop at both lights.Let A be the event that Fred has to stop at the first light. Let B be the event that Fred has to stop at the second light. What is the probability that Fred has to stop at one or more lights on his way to work?0.6150.550.650.95

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Solution

To solve this problem, we need to find the probability of either event A (Fred stops at the first light) or event B (Fred stops at the second light) happening.

The formula for the probability of either of two events happening is P(A U B) = P(A) + P(B) - P(A ∩ B).

Here, P(A) is the probability that Fred has to stop at the first light, which is 0.45. P(B) is the probability that Fred has to stop at the second light, which is 0.30. P(A ∩ B) is the probability that Fred has to stop at both lights, which is 0.20.

Substituting these values into the formula gives us:

P(A U B) = 0.45 + 0.30 - 0.20 = 0.55.

So, the probability that Fred has to stop at one or more lights on his way to work is 0.55 or 55%. Therefore, the correct answer is 0.55.

This problem has been solved

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