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?

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

Answer 1

To find the probability that the train arrived on time in fewer than 4 days, we can use the binomial distribution and calculate the probability using the formula. We get 0.1143.

Step-by-step explanation:

To find the probability that the train arrived on time in fewer than 4 days, we can use the binomial distribution. The binomial distribution is used when there are only two possible outcomes for each trial, in this case, the train either arrives on time or it doesn't.

We can calculate the probability using the formula:

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

where Y is the number of days on which the train arrives on time.

To find the probability for each individual case, we use the formula P(Y = k) = C(n, k) * p^k * q^(n-k), where C(n, k) represents the combination of n items taken k at a time, p is the probability of success (in this case, the train arriving on time), and q is the probability of failure (1 - p).

So, the probability that the train arrived on time in fewer than 4 days is:

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

= 0.1143.

We find that the probability that the train arrived on time in fewer than 4 days is approximately 0.1143.

Therefore, the correct answer is B) 0.1143.