Question

You pass through a particular stoplight on the way to school each morning. You have a 35% chance of catching the light red on any given day. In a three-day period:

a. What is the probability that you will catch the light red twice?
b. What is the probability that you catch the light red all three days?
c. what is the probability that you don’t catch the light red any of the three days?

Answer 1

To solve this problem, we can use the binomial distribution, which is a probability distribution that describes the number of successes in a fixed number of independent trials.

For each day, we have two possible outcomes: catching the light red (success) or not catching the light red (failure). The probability of success is 0.35 and the probability of failure is 0.65.

a. The probability of catching the light red twice in a three-day period can be calculated as follows:

P(catching red light twice) = (3 choose 2) * (0.35)^2 * (0.65)^1

where (3 choose 2) = 3! / (2! * (3 - 2)!) = 3 is the number of ways to choose 2 out of 3 days.

P(catching red light twice) = 0.44175

Therefore, the probability of catching the light red twice in a three-day period is 0.44175 or approximately 0.44.

b. The probability of catching the light red all three days can be calculated as follows:

P(catching red light all three days) = (0.35)^3

P(catching red light all three days) = 0.042875

Therefore, the probability of catching the light red all three days is 0.042875 or approximately 0.04.

c. The probability of not catching the light red any of the three days can be calculated as follows:

P(not catching red light any of the three days) = (0.65)^3

P(not catching red light any of the three days) = 0.274625

Therefore, the probability of not catching the light red any of the three days is 0.274625 or approximately 0.27.