Question about a probability problem

[reenmachine]

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

 

[mathman]

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..

 

[reenmachine]

of course! I feel a little bit stupid right now , thanks a lot!

 

[CompuChip]

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).

 

[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...)?

 

[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?

 

[mathman]

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.

 

[reenmachine]

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?

 

[reenmachine]

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

 

[mathman]

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?

I think you should describe this scenerio in more detail. What happens in game 1, what happens in game 2, etc.? What is the 16 - 7 division in reference to?