Maths probability 的問題

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

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2.
Denote:
H: A head is tossed
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

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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
=