red balls and white balls

cristi92b · **Joined:** Mon, 30 Jan 2012 14:51:40 UTC **Posts:** 56

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
$\bigcup_{i=1}^{5} A_{i}=\Omega$

Soroban · **Posted:** Thu, 7 Jun 2012 01:54:58 UTC

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

cristi92b · **Joined:** Mon, 30 Jan 2012 14:51:40 UTC **Posts:** 56

Thank you!!!

S.O.S. Mathematics CyberBoard

red balls and white balls

Who is online