determining probability

Question

a game involving six cards A,B,C,D,E and F is to be played.the cards carry respective digits 4,7,8,2 and 6. no cards has more than one digit. instructions for the game are as follows:

pay £50 for one game.

shuffle all cards, digits facing down. select three cards.

if the number is more than 483 but not more than 724, get your money back and a reward of £250.

1. determine the probability of winning one game.

2. Dominic paid £100 for two consecutive games. determine the probability of: a, loosing only the first game . b, loosing only one game . c, taking home a reward of £500.

Answer 1

P(win) = P(487) + P(6XX) + P(724)

= 1/6*1/5*1/4 + 1/6 + 1/6*1/5*1/4

= 1/120 + 1/6 + 1/120

= 1/120 + 20/120 + 1/120

= 22/120

= 11/60

= .1833333....


P(losing first game and winning second) = 49/60 * 11/60 = 539/3600 = .149722222....

P(losing only one game) = 11/60*49/60 + 11/60*49/60 = 2 * 11/60 * 49/60 = 2 * 539/3600 = 0.2994444444.....

P(500 reward) = P(winning both games) = 11/60 * 11/60 = 121/3600 = .033611111....