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

= 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) INCORRECT 0.1143

C) 0.0159

D) 0.8857

E) 0.9842

Answer 1

The probability that the train arrives on time on fewer than 4 days out of 6, using the binomial distribution, is calculated to be 0.01585. The closest answer choice is C) 0.0159.

Step-by-step explanation:

To find the probability that the train arrives on time on fewer than 4 days out of 6, we can use the binomial distribution formula, which is:

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

where:

• C(n, k) is the combination of n items taken k at a time.
• p is the probability of on-time arrival (90% or 0.9).
• n is the number of trials (6 days).
• k is the number of successful outcomes (days on which the train arrives on time).

To calculate the probability of arriving on time on fewer than 4 days, we sum the probabilities of arriving on time for k = 0, 1, 2, and 3 days:

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

After calculation, we find that:

P(X < 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 = 0.000001 + 0.000054 + 0.001215 + 0.01458 = 0.01585

Therefore, the closest option is C) 0.0159.