The probability that a student misses their bus to school is 0.4, and the probability that they missthe bus on any day is independent of whether they missed it on the previous day.Over five consecutive weekdays, what is the probability that the student(i) only misses the bus on Tuesday? (1marks)ii) misses the bus on Tuesday and on two other days? (2marks)
Question
The probability that a student misses their bus to school is 0.4, and the probability that they missthe bus on any day is independent of whether they missed it on the previous day.Over five consecutive weekdays, what is the probability that the student(i) only misses the bus on Tuesday? (1marks)ii) misses the bus on Tuesday and on two other days? (2marks)
Solution
(i) The probability that a student misses their bus on any given day is 0.4. Since the events are independent, the probability that they miss the bus on Tuesday and catch it on all other days is the product of the individual probabilities.
The probability that they catch the bus on any given day is 1 - 0.4 = 0.6.
So, the probability that they only miss the bus on Tuesday is:
0.6 (Monday) * 0.4 (Tuesday) * 0.6 (Wednesday) * 0.6 (Thursday) * 0.6 (Friday) = 0.031104.
(ii) The probability that the student misses the bus on Tuesday and two other days is a bit more complex.
First, we need to calculate the probability that they miss the bus on exactly three days. This is given by the binomial probability formula:
P(X=k) = C(n, k) * (p^k) * ((1-p)^(n-k))
where n is the number of trials (5 days), k is the number of "successes" (missed buses), p is the probability of success (0.4), and C(n, k) is the binomial coefficient "n choose k".
So, P(X=3) = C(5, 3) * (0.4^3) * (0.6^2) = 10 * 0.064 * 0.36 = 0.2304.
However, we know that one of these days is Tuesday, so we need to adjust this probability to account for the fact that there are fewer "choices" for the other two days.
There are C(4, 2) = 6 ways to choose 2 days from 4, so the final probability is 0.2304 / 6 = 0.0384.
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