• Mar 7th 2009, 10:35 PM
azuki
A box contains 5 white balls and 3 red balls. 2 players, A and B, take turns at drawing one ball from the box at random, and balls drawn are not replaced. The play who first gets 2 red balls is the winner, and the drawing stops as soon as either player has drawn 2 balls. Player A draw first. Find the probability that
i) play A is the winner on his second draw
ii) play A is the winner, given that the winning player wins on his second draw.
• Mar 8th 2009, 12:30 AM
JeWiSh
Quote:

Originally Posted by azuki
A box contains 5 white balls and 3 red balls. 2 players, A and B, take turns at drawing one ball from the box at random, and balls drawn are not replaced. The play who first gets 2 red balls is the winner, and the drawing stops as soon as either player has drawn 2 balls. Player A draw first. Find the probability that
i) play A is the winner on his second draw
ii) play A is the winner, given that the winning player wins on his second draw.

For the first part, you just calculate in order the odds of what HAS to happen for player A to win on his second draw.
So, first he has to draw a red ball, the odds of which are 3/8. Then, in one situation Player B draws a white, which is a 5/7 chance. Lastly A has to choose another Red, which would be a 2/6 chance. $\displaystyle \frac{3}{8} * \frac{5}{7} * \frac{2}{6} = \frac{30}{336}$

However, B could have chosen a red ball on his draw, the odds of which were 2/7. Then, for B to get the last red ball would have been a 1/6 chance.$\displaystyle \frac{3}{8} * \frac{2}{7} * \frac{1}{6} = \frac{6}{336}$ So the total chance is $\displaystyle \frac{36}{336} = \frac{3}{28}$
• Mar 8th 2009, 12:33 AM
JeWiSh
The second part is somewhat similiar, but there are three different ways in which B could win on his second go, it'll take a little calculating.
• Mar 8th 2009, 04:50 AM
u2_wa
a
Quote:

Originally Posted by JeWiSh
For the first part, you just calculate in order the odds of what HAS to happen for player A to win on his second draw.
So, first he has to draw a red ball, the odds of which are 3/8. Then, in one situation Player B draws a white, which is a 5/7 chance. Lastly A has to choose another Red, which would be a 2/6 chance. $\displaystyle \frac{3}{8} * \frac{5}{7} * \frac{2}{6} = \frac{30}{336}$

However, B could have chosen a red ball on his draw, the odds of which were 2/7. Then, for B to get the last red ball would have been a 1/6 chance.$\displaystyle \frac{3}{8} * \frac{2}{7} * \frac{1}{6} = \frac{6}{336}$ So the total chance is $\displaystyle \frac{36}{336} = \frac{3}{28}$

ii) Total chances to win in second draw:
For A there is $\displaystyle \frac{3}{28}$ chance as calculated by JeWiSh.
Lets calculate for B:
1) First A draws a white ball than B draws a red ball again A draws a white ball
and B draws a red ball, prob. = $\displaystyle \frac{5}{8}*\frac{3}{7}*\frac{4}{6}*\frac{2}{5}$ = $\displaystyle \frac{1}{14}$
2) First A draws a white ball than B draws a red ball again A draws a red ball now
and B draws a red ball, prob. = $\displaystyle \frac{5}{8}*\frac{3}{7}*\frac{2}{6}*\frac{1}{5}$ = $\displaystyle \frac{1}{56}$
3) First A draws a red ball than B draws a red ball, A draws a white ball
and B draws a red ball, prob. = $\displaystyle \frac{3}{8}*\frac{2}{7}*\frac{5}{6}*\frac{1}{5}$ = $\displaystyle \frac{1}{56}$

Now add these probabilities = $\displaystyle \frac{1}{14}+\frac{1}{56}+\frac{1}{56}$ = $\displaystyle \frac{3}{28}$

prob. that A wins = $\displaystyle \frac{\frac{3}{28}}{\frac{3}{28}+\frac{3}{28}} = \frac{1}{2}$