# Maths probability 的問題

1.

Box A Box B

Total no. of bulbs 8 14

No. of defective bulbs 2 6

A box is chosen at random and then a bulb is drawn

If it is known that the bulb drawn is defective, find the probability that the box chosen is B.

Ans: 12/19

2. Calvin and Katherine take turns tossing a coin. The one who first gets a head will win the game. Suppose Katherine tosses the coin first.

Find the probability that Katherine wins the game.

Ans: 2/3

3. Probability of John, May, Jason wins the game are 0.8, 0.7, 0.6 respectively. Find the probablity that one of them wins the game.

Why the steps are like this?

0.8 + (1-0.8)(0.7) + (1-0.8)(1-0.7)(0.6)

Thanks so much!!

Rating

1.

Denote:

A: Box A is chosen

B: Box B is chosen

d: The bulb is defective

P(A and d)

= P(A) x P(d from Box A)

= (1/2) x (2/8)

= 1/8

P(B and d)

= P(B) x P(d from Box B)

= (1/2) x (6/14)

= 3/14

P(d)

= P(A and d) + P(B and d)

= (1/8) + (3/14)

= 19/56

P(B | d)

= P(B and d) / P(d)

= (3/14) / (19/56)

= (3/14) x (56/19)

= 12/19

=====

2.

Denote:

T: A tail is tossed

Katherine tosses in the 1st, 3rd, 5th ...... (2n - 1)th terms

Katherine wins if:

She tosses H in the 1st turn. Prob. = 1/2

Or: T in the first 2 turns, but she tosses H in the 3rd turn = (1/2)2(1/2)

Or: T in the first 4 turns, but she tosses H in the 5th turn = (1/2)4(1/2)

.......

P(Katherine wins)

= (1/2) + (1/2)2(1/2) + (1/2)4(1/2) + (1/2)6(1/2) + .........

= (1/2) + (1/4)(1/2) + (1/4)2(1/2) + (1/4)3(1/2) + ......

It is the sum of a G.P. (or G.S)

with first term a = 1/2 and common ratio r =

Hence, P(Katherine wins)

= (1/2) + (1/4)(1/2) + (1/4)2(1/2) + (1/4)3(1/2) + ......

= a/(1 - r)

= (1/2)/[1 - (1/4)]

= (1/2)/(3/4)

= (1/2) x (4/3)

= 2/3

=====

3.

Assume that John plays the game first. If John loses, May plays the game. If May also loses, Jason plays the game.

P(John wins the game)

= 0.8

P(May wins the game)

= P(John loses and then May wins)

= P(John loses) x P(May wins)

= (1 - 0.8) x 0.7

P(Jason wins the game)

= P(John loses, May loses, and Jason wins)

= P(John loses) x P(May loses) x P(Jason wins)

= (1 - 0.8) x (1 - 0.7) x 0.6

P(one of them wins the game)

= P(John wins the game) + P(May wins the game) + P(Jason wins the game)

= 0.8 + (1 - 0.8) x 0.7 + (1 - 0.8) x (1 - 0.7) x 0.6

= 0.8 + 0.14 + 0.036

= 0.976

=