Final answer:
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:
- P(Y = 0) = C(6, 0) * (0.9)^0 * (0.1)^6
- P(Y = 1) = C(6, 1) * (0.9)^1 * (0.1)^5
- P(Y = 2) = C(6, 2) * (0.9)^2 * (0.1)^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.