Question

According to New Jersey Transit, the 8:00 a.m. weekday train from Princeton to New York City has a 90% chance of arriving on time on a randomly selected day. Suppose this claim is true. Choose 6 days at random. Let Y = the number of days on which the train arrives on time. What is the probability that the train arrived on time in fewer than 4 days? Show steps

A) 0.0984
B) 0.1143
C) 0.0159
D) 0.8857
E) 0.9842

Answer 1

Using the binomial probability formula for 6 days with a probability of 0.9 for an on-time arrival, we calculated the sum of probabilities for the train arriving on time in fewer than 4 days. The result is 0.0984, corresponding to Answer A.

Step-by-step explanation:

To calculate the probability that the train arrived on time in fewer than 4 days out of 6, we use the binomial probability formula:

P(Y = k) = C(n, k) * (p)^k * (1-p)^(n-k)

Where:

• C(n, k) is the number of combinations of n items taken k at a time.
• p is the probability of a success on any given trial (the train arriving on time).
• n is the number of trials (days in this case).
• k is the number of successful outcomes (days the train arrives on time).

Since the probability p given is 90% or 0.9, and the number of trials n is 6, we need to find the sum of probabilities for k = 0, 1, 2, and 3:

P(Y < 4) = P(Y = 0) + P(Y = 1) + P(Y = 2) + P(Y = 3)

Calculating each term:

1. P(Y = 0) = C(6, 0) * (0.9)^0 * (0.1)^6
2. P(Y = 1) = C(6, 1) * (0.9)^1 * (0.1)^5
3. P(Y = 2) = C(6, 2) * (0.9)^2 * (0.1)^4
4. P(Y = 3) = C(6, 3) * (0.9)^3 * (0.1)^3

Adding these probabilities together gives us the total probability for P(Y < 4).

After plugging in the values and calculating, we find that the sum of these probabilities is approximately 0.0984, which corresponds to Answer A.