The car wash at a particular garage has a choice of 4 different programmes. A driver chooses a programme in the garage and then, proceeds to the car wash.
The probability of any driver choosing a particular programe is
| Programme | Probability | Time taken (in mins) |
| A ( | 0.25 | 2 |
| B ( | 0.5 | 3 |
| C (Superwash) | 1/6 | 3 |
| D (Superwash and Dry) | 1/12 | 4 |
(a) the probability that the drivers of the first and second cars have chosen the same programme
(b) the probability that exactly two of the three drivers have chosen a programme which includes a dry
(c) the probability that at least one of three drivers have chosen programme C
A fourth car X joines the queue at the instant that the first car enters the car wash. Assuming that no time is lost between the time one car leaves the car wash and the programme for the next car beginning, find
(d) the probability that it will be at least 10 minutes before car X enters the car wash.
(e) the probability that it will be at least 10 minutes before car X enters the car wash, given that at least one of the previous three drivers chose programme C.
Answer:
= P(A, A) + P(B, B) + P(C, C) + P(D, D)
= (1/4 * 1/4) + (1/2 * 1/2) + (1/6 * 1/6) + (1/12 * 1/12)
= 1/16 + 1/4 + 1/36 + 1/144
= 25 / 72
(b) P(includes dry) = P(B) + P(D) = 1/2 + 1/12 = 7/12
P (does not include dry) = P(A) + P(C) = 1/4 + 1/6 = 5/12
Probability that exactly two of the three drivers have chosen a programme which includes a dry
= 7/12 * 7/12 * 5/12 * 3C2 ==> (choose 2 drivers out of 3 to have included dry)
= 245/576
(c) Probability that at least one of three drivers have chosen programme C
= 1 - Probability that none of three drivers have chosen programme C
= 1 - (5/6)3
= 91/216
(d) Probability that it will be at least 10 minutes before car X enters the car wash
= P (4 min, 4 min, 4 min) + P (4 min, 3 min, 3 min) + P (4 min, 4 min, 2 min) + P (4min, 4min, 3 min)
= (1/12)3 + 3C2 * 1/12 * 2/3 * 2/3 + 3C2 * 1/12 * 1/12 * 1/4 + 3C2 * 1/12 * 1/12 * 2/3
= 1/1728 + 1/9 + 1/192 + 1/72
= 113/864
(e) P (at least 10 minutes before car X enters the car wash | at least one of the previous three drivers chose programme C)
= P (at least 10 minutes before car X enters the car wash and at least one of the previous three drivers chose programme C) / P(at least one of the previous three drivers chose programme C)
= { P (C, C, D) + P (C, D, D) + P (B, C, D) } / {91/216}
= { (3C2 * 1/6 * 1/6 * 1/12) + (3C2 * 1/6 * 1/12 * 1/12) + ( 3! * 1/2 * 1/6 * 1/12) } / {91/216}
= {1/144 + 1/288 + 1/24} / {91/216}
= 45 / 364
