marbles

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May 9th 2006, 04:53 PM
swoopesjr01

marbles

A bag contains 4 red marbles and 2 blue marbles. You draw a marble at random, without replacement, until the first blue marble is drawn. Let X = the number of draws.

a) If you repeated this experiment a very large number of times, on average how many draws would you make before a blue marble was drawn?
May 9th 2006, 06:48 PM
ThePerfectHacker

Quote:

Originally Posted by swoopesjr01

A bag contains 4 red marbles and 2 blue marbles. You draw a marble at random, without replacement, until the first blue marble is drawn. Let X = the number of draws.

a) If you repeated this experiment a very large number of times, on average how many draws would you make before a blue marble was drawn?

Not sure about this one. I guess that one which is most likely to happen.

Probability
On 1st]=2/6=1/3=.3333
On 2nd]=(4/6)(2/5)=4/15=.2666
On 3rd]=(4/6)(3/5)(2/4)=2/10=.2000
On 4th]=(4/6)(3/5)(2/4)(2/3)=2/15=.1333
On 5th]=(4/6)(3/5)(2/4)(1/3)(2/2)=1/15=.0666

It seems like on the first try to me.
May 9th 2006, 08:16 PM
swoopesjr01

the answer is µ = 2.334..im not really sure i understand what you did in your reply
May 9th 2006, 09:23 PM
JakeD

Quote:

Originally Posted by ThePerfectHacker

Not sure about this one. I guess that one which is most likely to happen.

Probability
On 1st]=2/6=1/3=.3333
On 2nd]=(4/6)(2/5)=4/15=.2666
On 3rd]=(4/6)(3/5)(2/4)=2/10=.2000
On 4th]=(4/6)(3/5)(2/4)(2/3)=2/15=.1333
On 5th]=(4/6)(3/5)(2/4)(1/3)(2/2)=1/15=.0666

It seems like on the first try to me.

The average number of draws is the mean $\sum_{x=1}^{5} xf(x)$ where $f(x)$ are PH's probabilities. The mean is 2.33 as swoopes says.

Why don't quotes within quotes work? Using the quote button on PH's post loses his quote of swoopes. :confused:
May 9th 2006, 10:33 PM
CaptainBlack

Quote:

Originally Posted by JakeD

Quote:

Originally Posted by PH

Originally Posted by ThePerfectHacker
Not sure about this one. I guess that one which is most likely to happen.

Probability
On 1st]=2/6=1/3=.3333
On 2nd]=(4/6)(2/5)=4/15=.2666
On 3rd]=(4/6)(3/5)(2/4)=2/10=.2000
On 4th]=(4/6)(3/5)(2/4)(2/3)=2/15=.1333
On 5th]=(4/6)(3/5)(2/4)(1/3)(2/2)=1/15=.0666

It seems like on the first try to me.

The average number of draws is the mean $\sum_{x=1}^{5} xf(x)$ where $f(x)$ are PH's probabilities. The mean is 2.33 as swoopes says.

Why don't quotes within quotes work? Using the quote button on PH's post loses his quote of swoopes. :confused:

They do, you just have to include them yourself :confused:

RonL
May 9th 2006, 10:36 PM
CaptainBlack

Quote:

Originally Posted by CaptainBlack

Originally Posted by JakeD
Quote:
Originally Posted by PH

Originally Posted by ThePerfectHacker
Not sure about this one. I guess that one which is most likely to happen.

Probability
On 1st]=2/6=1/3=.3333
On 2nd]=(4/6)(2/5)=4/15=.2666
On 3rd]=(4/6)(3/5)(2/4)=2/10=.2000
On 4th]=(4/6)(3/5)(2/4)(2/3)=2/15=.1333
On 5th]=(4/6)(3/5)(2/4)(1/3)(2/2)=1/15=.0666

It seems like on the first try to me.

The average number of draws is the mean where are PH's probabilities. The mean is 2.33 as swoopes says.

Why don't quotes within quotes work? Using the quote button on PH's post loses his quote of swoopes.

They do, you just have to include them yourself :confused:

RonL

But you may have some trouble includong the TeX (unless you have moderator privaleges :cool: )

RonL
May 11th 2006, 11:28 AM
swoopesjr01

what would be the probability of obtaining the same blue marble on the 6th draw if it was done with replacement?