Bellow said:
In a game, two players take turns at drawing, and immediately replacing, a marble from a bag containing two green and three red marbles. The game is won by player A drawing a green marble, or player B drawing a red marble. Player A draws first.
Find the probability that:
(i) A wins on her first draw;
(ii)B wins on her first draw;
(iii)A wins in less than four of her turns;
(iv)A wins eventually. (whats this mean?)
i) P(A wins first draw) = 2/5 (2 green marbles in a bag of 5)
ii) For B to win A must first draw a red (i.e. not win) then B must draw a red:
P(B wins first draw) = (3/5)(3/5) = 9/25
iii) P(A wins in less than four turns) = P(A wins 1st turn) + P(A wins 2nd turn) + P(A wins 3rd turn)
= (2/5) +
[(3/5)(2/5)](2/5) +
[(3/5)(2/5)(3/5)(2/5)](2/5) (notice the pattern here)
= 1622/3125
iv) P (A wins eventually) means A could win on her 1st turn, her 2nd turn, ..., her nth turn (n --> ∞ ). You have to add all these cases together. Looking at the pattern above A has to lose once and B has to lose once for A to win on her 1st tern. A must loose (n-1) times and B must loose (n-1) times for A to win on her nth turn. This sets up a geometric series:
P(A wins eventually) = 2/5 + 2/5(6/25) + 2/5(6/25)<sup>2</sup> + ...
= 2/5[1/(1 - 6/25)]
= 2/5(25/19)
= 10/19
