danlew wrote:
The probability that a train arrives late at a particular station is 25%. There are four trains scheduled to arrive at the station today. What is the probability that at least three of them will arrive later than the scheduled time?
A) \(\frac{1}{256}\)
B) \(\frac{5}{256}\)
C) \(\frac{13}{256}\)
D) \(\frac{20}{256}\)
E) \(\frac{255}{256}\)
We are given that the probability that a train arrives late is 1/4.
We need to determine the probability that at least 3 trains arrive late. That is, we need to determine the probability that exactly 3 trains arrive late or that all 4 trains arrive late.
Let’s start with 3 late trains and 1 on-time train (L = late and T = on time):
P(L-L-L-T) = 1/4 x 1/4 x 1/4 x 3/4 = 3/256
Since we can organize the 3 Ls and 1 T in 4!/3! = 4 ways, the total probability is actually 3/256 x 4 = 12/256.
Next we can determine the probability that all 4 trains arrive late:
P(L-L-L-L) = 1/4 x 1/4 x 1/4 x 1/4 = 1/256
Thus, the probability of at least 3 trains arriving late is 12/256 + 1/256 = 13/256.
Answer: C
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