A lvl H2 Maths: Probability

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 (Wash)	0.25	2
B (Wash and Dry)	0.5	3
C (Superwash)	1/6	3
D (Superwash and Dry)	1/12	4

At a particular instant, there are 3 cars queueing to use the car wash. Find

(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.

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Answer:

(a) Probability that drivers of first and second car chose the same programe
= 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

A lvl H2 Maths: Probability

Question from http://www.sgforums.com/forums/2297/topics/314258

1) Two boxes are labelled A and B. box A contain 3 blue discs and 2 white discs and box B contain 2 blue discs and 3 white discs. A random sample of 2 discs is selected without replacement from each box. Find the probability that at most one of the discs drawn is white.

2) Two friends Jane and Mary often go their shopping on sat afternoon. If Mary goes shopping, the probability that Jane also goes is 0.96. If Jane goes, the probability that Mary goes is 0.8. The probability that neither goes is 0.07. Find the probability that both go shopping.

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Answer:

1)

P(none white) = 3/5 * 2/4 * 2/5 * 1/4 = 12/400 = 3/100

P(1 white)
= P(1 white from box A and no white from box B) + P( no white from box A and 1 white from box B)
= ( 2/5 *3/4 *2! ) * (2/5*1/4) + (3/5 * 2/4) * (3/5 * 2/4 * 2!)
= 3/50 + 9/50 = 12/50

Summing up gives 27/100

2)

P(at least one of them go) = P(J goes) + P(M goes) - P(both J and M goes) {double counted}

1 - P(none go) = P(both J and M goes)/0.96 + P(both J and M goes)/0.8 - P(both J and M goes)

1-0.07 = P(both J and M goes)[1/0.96+1/0.8-1]

P(both J and M goes) = 0.72

(I'm not very sure of Question 2, will review it again later... If you have any ideas, please email me. Thanks!)