# Question about a probability problem

by reenmachine
Tags: probability
 PF Gold P: 521 Hi all , here's hoping someone could help me with this problem. :) The problem is basically as follow: You have 7 balls in your pocket , 3 are whites and 4 are blacks.If you take out 3 balls out of your pocket , what are the odds that you'll have 2 white balls and 1 black ball in your hand ? Now I have no mathematic education at all so here's how I proceeded: 3/7 * (1/2 + 1/3) * (1/5 + 2/5 + 3/5) 3/7 * 5/6 * 5/5 3/7 * 5/6 = 5/14 5/14 = 35,7% = 36% That was my final answer and I got it wrong , the answer happened to be 34%. How could my method of calculating the odds produce an answer that was so close yet wrong? Help would be greatly appreciated :D thanks in advance
 Sci Advisor P: 6,080 There are three ways to get the desired result. wwb, wbw, bww. P(wwb) = (3/7)(2/6)(4/5) P(wbw) = (3/7)(4/6)(2/5) P(bww) = (4/7)(3/6)(2/5) Each term = 4/35. Sum = 12/35 = .342857..
 PF Gold P: 521 of course!!! I feel a little bit stupid right now , thanks a lot!!!
HW Helper
P: 4,300

 Quote by reenmachine How could my method of calculating the odds produce an answer that was so close yet wrong?
That doesn't really mean anything, especially in statistics. If you are on the right track, you are usually off by a multiplicative factor instead of a small absolute percentage.

For example, if you'd done the P(wwb) correctly but forgot to take into account the other two possibilities you would be off by exactly a factor of 3. That you get 36% instead of 34% (after rounding both numbers to integer percentages) is probably just a co-incidence (which is quite likely if you mess about with various combinations of 3, 5, 6 and 7 in a fraction).
 PF Gold P: 521 Another question related to this problem , finding all the ways to get there (wwb,wbw,bww) is pretty easy in this case , but suppose the problem would require me to find all the possible ways for a larger sample like for example: I have 7 white balls and 16 black balls in a bag (each ball indicate a potential victory for each team) and I want to calculate the odds of winning in a Best-of-7 sport series for the white team (of course each time a ball is picked she doesn't return in the bag) is there a quicker way to do it than trying to write all the ways down (like wwwwbbb , wwwbwbb etc...)?
 Sci Advisor HW Helper P: 4,300 Yes, there is an easy way of doing that, it's called the binomial coefficient: $$\binom{16}{7} = \frac{16!}{7! 9!}$$ Before we start writing a complete Wikipedia article here, don't you have any textbook or course material that explains this?
P: 6,080
 Quote by reenmachine Another question related to this problem , finding all the ways to get there (wwb,wbw,bww) is pretty easy in this case , but suppose the problem would require me to find all the possible ways for a larger sample like for example: I have 7 white balls and 16 black balls in a bag (each ball indicate a potential victory for each team) and I want to calculate the odds of winning in a Best-of-7 sport series for the white team (of course each time a ball is picked she doesn't return in the bag) is there a quicker way to do it than trying to write all the ways down (like wwwwbbb , wwwbwbb etc...)?
Be careful - you are confusing two different problems. Picking a ball and not returning is different from best of seven for sports. The latter is analogous to picking a ball and then returning it before picking the next ball.
PF Gold
P: 521
 Be careful - you are confusing two different problems. Picking a ball and not returning is different from best of seven for sports. The latter is analogous to picking a ball and then returning it before picking the next ball. Y 07:58 PM
Suppose you have 2 fantasy teams and 23 judges who votes on the series , then the result comes out 16 and 7 and we want to decide who wins game 1 , game 2 and so on until one team wins 4 out of 7 games , wouldn't my technic to pick a ball and not return it to the bag be the way to go? If so , that's why I asked how to calculate the number of possible ways wbbwbwb wwwwbbb etc...

Or is what you are saying is that I should always keep all the balls in the bag before each game , making it a 16/7 odds for every game and ultimately the series outcome?
PF Gold
P: 521
 Quote by CompuChip Yes, there is an easy way of doing that, it's called the binomial coefficient: $$\binom{16}{7} = \frac{16!}{7! 9!}$$ Before we start writing a complete Wikipedia article here, don't you have any textbook or course material that explains this?
I'm not going to school , but I'll try to find the explanation online and self-teach it to myself if possible.

thanks