help with a tough probability

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dts5044

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an urn contains 10 red balls and 10 white balls. The balls are drawn from the urn at random, one at a time. Find the probability that the fourth white ball is the sixth ball drawn if the sampling is done:

a) with replacement
b) without replacement

In general I don't really understand when to use permutations and when to use combinations. Could someone answer this question, and if possible give a quick explanation about using permutations and combinations?
 
dts5044 said:
an urn contains 10 red balls and 10 white balls. The balls are drawn from the urn at random, one at a time. Find the probability that the fourth white ball is the sixth ball drawn if the sampling is done:

a) with replacement
b) without replacement

In general I don't really understand when to use permutations and when to use combinations. Could someone answer this question, and if possible give a quick explanation about using permutations and combinations?
Click to expand...

Please share with us your work/thoughts - so that we know where to begin to help you.
 
dts5044 said:
In general I don't really understand when to use permutations and when to use combinations.
Click to expand...
In general, combinations are for when order doesn't matter: pay me with the ten, the five, and the C-note, or pay me with the five, the C-note, and the ten; I really don't care, so long as ya pay me my hundred fifteen bucks!

On the other hand, permutations are for when order does matter: socks, and then shoes; underpants, and then pants; and not the other way 'round! :oops: :lol:

dts5044 said:
Could someone answer this question, and if possible give a quick explanation about using permutations and combinations?
Click to expand...
Why wait? There are just loads of great lessons already available online! :wink:

. . . . .Google results for "permutations combinations"

Have fun! :D

Eliz.
 
yeah I do understand that much, but somehow don't get it when it comes to problems like this.

So I need to figure out how many ways the first five balls can be drawn such that 3 are white and 2 are red
let me try with replacement:

in this case, I need a combination because I don't care what order the 3 white and 2 red balls come in, just as long as there are three white and 2 red. So 5C3 = 5!/(3!*2!) = 10

so there are 10 different ways the first five balls can be chosen that set me up for drawing the sixth ball and having it be the 4th white ball drawn.
and the probability of drawing a white ball on the sixth draw is 1/2
but the first five balls, even though there are ten different ways to go about it, have to be drawn with 3 white and 2 red. So even though I don't care about the order, the probability of drawing 3 white balls and 2 red on the first five draws is 10 * (1/2)^5.

So the probability of drawing the 4th white ball on the 6th draw is 10 * (1/2)^6!

hmmm...somehow re-hearing that bit about permutations and combinations, helped! thanks! :)

for b)

there are still 10 different ways to choose the first 5 balls in the pattern I need, but the probability of these patterns isn't (1/2)^5, it is (10/20 * 9/19 * 8/18 * 10/17 * 9/16) = 6480/186048, which I multiply by the 10 combinations to get 64800/186048 = 5400/15504. this multiplied by the probability of the sixth ball being white (7/15) is the answer.
so 5400/15504 * 7/15 = about .162538

hmmm...this seems like it could be right but I'm not sure! I can't believe how confident I was though once I knew it was a combination! Thanks so much stapel!
 
I think they call this a negative binomial probability.

\(\displaystyle C(x-1,k-1)p^{k}(1-p)^{x-k}\)

The probability the 4th white happens on the 6th draw.

\(\displaystyle C(6-1,4-1)(.5)^{4}(.5)^{2}=C(5,3)(\frac{1}{2})^{4}(\frac{1}{2})^{2}\)
 
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