Question about a probability problem


by reenmachine
Tags: probability
reenmachine
reenmachine is offline
#1
Feb11-13, 02:55 PM
PF Gold
reenmachine's Avatar
P: 520
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
Phys.Org News Partner Science news on Phys.org
Better thermal-imaging lens from waste sulfur
Hackathon team's GoogolPlex gives Siri extra powers
Bright points in Sun's atmosphere mark patterns deep in its interior
mathman
mathman is offline
#2
Feb11-13, 03:28 PM
Sci Advisor
P: 5,935
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
reenmachine is offline
#3
Feb11-13, 03:35 PM
PF Gold
reenmachine's Avatar
P: 520
of course!!! I feel a little bit stupid right now , thanks a lot!!!

CompuChip
CompuChip is offline
#4
Feb11-13, 03:51 PM
Sci Advisor
HW Helper
P: 4,301

Question about a probability problem


Quote Quote by reenmachine View Post
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
reenmachine is offline
#5
Feb11-13, 04:13 PM
PF Gold
reenmachine's Avatar
P: 520
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
CompuChip is offline
#6
Feb12-13, 01:58 PM
Sci Advisor
HW Helper
P: 4,301
Yes, there is an easy way of doing that, it's called the binomial coefficient:
[tex]\binom{16}{7} = \frac{16!}{7! 9!}[/tex]

Before we start writing a complete Wikipedia article here, don't you have any textbook or course material that explains this?
mathman
mathman is offline
#7
Feb12-13, 03:30 PM
Sci Advisor
P: 5,935
Quote Quote by reenmachine View Post
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
reenmachine is offline
#8
Feb13-13, 10:54 AM
PF Gold
reenmachine's Avatar
P: 520
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
reenmachine is offline
#9
Feb13-13, 10:55 AM
PF Gold
reenmachine's Avatar
P: 520
Quote Quote by CompuChip View Post
Yes, there is an easy way of doing that, it's called the binomial coefficient:
[tex]\binom{16}{7} = \frac{16!}{7! 9!}[/tex]

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
mathman is offline
#10
Feb13-13, 03:29 PM
Sci Advisor
P: 5,935
Quote Quote by reenmachine View Post
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?


Register to reply

Related Discussions
Probability question ,Random variables and probability distribution Calculus & Beyond Homework 2
Probability. Not exactly a problem, more of a question I'd like answered Precalculus Mathematics Homework 15
a simple probability problem involving the law of total probability Calculus & Beyond Homework 1
Probability question; Conditional probability and poisson distribution Precalculus Mathematics Homework 1
Probability: joint probability distribution problem? Calculus & Beyond Homework 8