A bin contains 25 balls: 10 red, 8 yellow, and 7 blue. We draw three balls at random (without looking!) from the bin, and we will say that we "win" if our three balls represent exactly two colors. (That is, we "win" if we draw two balls of one color and another ball of a different color.) What is the probability of winning this particular game?
2006-12-28 18:01:12 · 4 answers · asked by sahsjing 7 in Science & Mathematics ➔ Mathematics
Essentially, there are two ways to do the problem. Which way do you think is simpler? Could you show your work? For discussion purpose you only need to write expression.
2006-12-28 18:23:53 · update #1
Following gjmb1960's idea,
Prob. = (1/2)[(10)(9)(8+7) + (8)(7)(10+7) + (7)(6)(10+8)]/10C3
2006-12-28 18:34:13 · update #2
Probably the easiest way to compute the probability of winning is to compute the probability of not winning and subtract that from one.
The ways NOT to win are: draw exactly one ball of each color, or draw all three balls the same color. All other draws are winners.
2006-12-28 18:09:56 · answer #1 · answered by ? 6 · 2⤊ 1⤋
I like garypopkins' idea of finding the number of ways you could lose.
The denominator will be C(25, 3) = 25x24x23/(3x2x1) = 2300 ways to choose 3 balls regardless of color.
There are 10x8x7 = 560 ways to draw one ball of each color.
There are C(10, 3) = 10x9x8/(3x2x1) = 120 ways to draw all red.
There are C(8, 3) = 8x7x6/(3x2x1) = 56 ways to draw all red.
There are C(7, 3) = 7x6x5/(3x2x1) = 35 ways to draw all blue.
Therefore the probability of winning is:
1 - (560 + 120 + 56 + 35)/2300 = 66.478%
2006-12-28 18:22:41 · answer #2 · answered by Jim Burnell 6 · 1⤊ 0⤋
This is not a problem for discussion.
The answer can be calculated with some elementar counting !
wite out all possibilities for drawing 2 the same colors
and divide this by the total
2006-12-28 18:03:36 · answer #3 · answered by gjmb1960 7 · 0⤊ 3⤋
.0016
2006-12-28 18:12:27 · answer #4 · answered by Alex 2 · 0⤊ 2⤋