Pikcing Balls from Urn

**Lukybear** · October 22nd 2010, 10:36 PM

An urn contains n red balls, n white balls and n blue balls. Three balls are drawn at random from the urn, one at a time, without replacement.

What is the probability that three balls are all different colours.

Answer: n/3n x n/3n-1 x n / 3n-2

**pickslides** · October 22nd 2010, 10:45 PM

This answer is pretty good. But only accounts for one of the combinations that have 3 different colours of the 27 possible outcomes.

There are 6 occasions where the balls are different colours therefore multiply your answer by 6.

**Lukybear** · October 23rd 2010, 02:30 AM

Sorry? There are six different occasions?

If its different colours, then its red, blue and white isnt it? What other possible combinations can there be?

**Plato** · October 23rd 2010, 04:33 AM

The original answer you gave is just a bit off.

$\displaystyle \frac{\binom{n}{1}\binom{n}{1}\binom{n}{1}}{\binom {3n}{3}}=\frac{2n^3}{(n)(3n-1)(3n-2)}$

**Soroban** · October 23rd 2010, 05:21 AM

Hello, Lukybear!

Let me baby-step through an explanation.

Sorry? There are six different occasions?

If it's different colours, then its red, blue and white isnt it?
What other possible combinations can there be?

What is the probability that the balls are drawn in the order Red-White-Blue?

. . $\begin{array}{cccccc} P(\text{1st is R}) &=& \frac{n}{3n} \\ \\[-3mm] P(\text{2nd is W}) &=& \frac{n}{n-1} \\ \\[-3mm] P(\text{3rd is B}) &=& \frac{n}{n-2} \end{array}$

Hence: . $P(\text{R-W-B}) \;=\;\dfrac{n}{3n} \times \dfrac{n}{3n-1} \times \dfrac{n}{3n-2}$

What is the probability that the balls are drawn in the order Blue-Red-White?

. . $\begin{array}{cccccc} P(\text{1st is B}) &=& \frac{n}{3n} \\ \\[-3mm] P(\text{2nd is R}) &=& \frac{n}{n-1} \\ \\[-3mm] P(\text{3rd is W}) &=& \frac{n}{n-2} \end{array}$

Hence: . $P(\text{B-R-W}) \;=\;\dfrac{n}{3n} \times \dfrac{n}{3n-1} \times \dfrac{n}{3n-2}$

What is the probability that the balls are drawn in the order White-Red-Blue?

. . $\begin{array}{cccccc} P(\text{1st is W}) &=& \frac{n}{3n} \\ \\[-3mm] P(\text{2nd is R}) &=& \frac{n}{n-1} \\ \\[-3mm] P(\text{3rd is B}) &=& \frac{n}{n-2} \end{array}$

Hence: . $P(\text{W-R-B}) \;=\;\dfrac{n}{3n} \times \dfrac{n}{3n-1} \times \dfrac{n}{3n-2}$

. . . etc. . . . etc.

**Lukybear** · October 23rd 2010, 11:40 PM

Ok. The answer is actually infact (nC3)^3/3nC3. But i was confused here, since i thought it they picked one and didnt replace it, then it cant be computed using combinatorics.

Namely, if the question stated, if it was picked and replaced, is the solution still (nC3)^3/3nC3?

**Plato** · October 24th 2010, 04:44 AM

Originally Posted by Lukybear

Ok. The answer is actually infact (nC3)^3/3nC3. But i was confused here, since i thought it they picked one and didnt replace it, then it cant be computed using combinatorics. Namely, if the question stated, if it was picked and replaced, is the solution still (nC3)^3/3nC3?

NO! The answer is $\displaystyle \frac{(_nC_1)^3}{_{3n}C_3}$ .

BUT the answer to the second question is $\dfrac{(3!)n^3}{(3n)^3}$

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