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Assume that removing a ball from bag A or bag
B is equally likely, the probability of choosing a
blue ball from bag A is 3/10 and the probability
of choosing a blue ball from bag B is 4/14 so the
probability of choosing a blue ball is 3/10 + 4/14 = 41/70.
Theoretically in 70 trials, 41 of the outcomes would give a blue
ball and 21 of these would of been drawn from bag
A . Given that the ball drawn was blue the
probability that it came from bag A is 21/41.
The only correct solution to date came from Matthew (Smithdon High School)
A man went to Monte Carlo to try and make his fortune. Is his strategy a winning one?
You and I play a game involving successive throws of a fair coin. Suppose I pick HH and you pick TH. The coin is thrown repeatedly until we see either two heads in a row (I win) or a tail followed by a head (you win). What is the probability that you win?
A gambler bets half the money in his pocket on the toss of a coin, winning an equal amount for a head and losing his money if the result is a tail. After 2n plays he has won exactly n times. Has he more money than he started with?