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 Post subject: red balls and white balls
PostPosted: Wed, 6 Jun 2012 23:00:33 UTC 
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Joined: Mon, 30 Jan 2012 14:51:40 UTC
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In a box there are 5 red balls and 10 white balls. What is the probability that if you randomly select 5 balls, you will have 2 red balls and 3 white balls?

One way of finding the probability is P(A)=\frac{\binom{10}{3}*\binom{5}{2}}{\binom{15}{5}}=\frac{400}{1001}

But can you find it using conditional probabilities?

for example:

R = Red ball selected event
W = White ball selected event

and five events A_{1},A_{2},A_{3},A_{4},A_{5} so that
\bigcup_{i=1}^{5} A_{i}=\Omega


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 Post subject: Re: red balls and white balls
PostPosted: Thu, 7 Jun 2012 01:54:58 UTC 
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Joined: Mon, 19 May 2003 19:55:19 UTC
Posts: 7950
Location: Lexington, MA
Hello, cristi92b!

Quote:
In a box there are 5 red balls and 10 white balls.
What is the probability that if you randomly select 5 balls,
. . you will have 2 red balls and 3 white balls?

One way of finding the probability is: P(A)=\frac{\binom{10}{3}*\binom{5}{2}}{\binom{15}{5}}=\frac{400}{1001}

But can you find it using conditional probabilities?

For example:

R = Red ball selected event
W = White ball selected event

and five events A_{1},A_{2},A_{3},A_{4},A_{5} so that: .\bigcup_{i=1}^{5} A_{i}=\Omega

I suppose you could . . . but who would want to?

First of all, "the five events" would have ten possible outcomes.
. . rrwww,\,rwrww,\,rwwrw,\,rwwwr,\, wrrww,\,wrwrw,\, wrwwr,\,wwrrw,\,wwrwr,\,wwwrr

Then for each one. we might go through this reasoning:

Consider P(RRWWW)

\begin{array}{c}P(\text{1st R}) \:=\:\frac{5}{15} \\ \\[-3mm]

P(\text{2nd R}\,|\,\text{1st R}) \:=\:\frac{4}{14} \\ \\[-3mm]

P(\text{3rd W}\,|\,\text{1st \& 2nd R}) \:=\:\frac{10}{13} \\ \\[-3mm]

P(\text{4th W}\,|\,\text{1st \& 2nd R, 3rd W}) \:=\:\frac{9}{12} \\ \\[-3mm]

P(\text{5th W}\,|\,\text{1st \& 2nd R, 3rd \& 4th W}) \:=\:\frac{8}{11} \end{array}

Hence: .P(RRWWW) \:=\:\frac{5}{15}\cdot\frac{4}{14}\cdot\frac{10}{13}\cdot\frac{9}{12}\cdot\frac{8}{11} \;=\;\dfrac{40}{1001}


Then we go through the other nine cases
. . and we find that the probabilities are the same.

Therefore: .P(\text{2R, 3W}) \:=\:10\cdot\dfrac{40}{1001} \:=\:\dfrac{400}{1001}



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 Post subject: Re: red balls and white balls
PostPosted: Thu, 7 Jun 2012 11:33:24 UTC 
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Joined: Mon, 30 Jan 2012 14:51:40 UTC
Posts: 54
Thank you!!!


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