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
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
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?
. . $\displaystyle \begin{array}{cccccc}
P(\text{1st is R}) &=& \frac{n}{3n} \\ \\[-3mm]
P(\text{2nd is W}) &=& \frac{n}{n-1} \\ \\[-3mm]
P(\text{3rd is B}) &=& \frac{n}{n-2} \end{array}$
Hence: .$\displaystyle 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?
. . $\displaystyle \begin{array}{cccccc}
P(\text{1st is B}) &=& \frac{n}{3n} \\ \\[-3mm]
P(\text{2nd is R}) &=& \frac{n}{n-1} \\ \\[-3mm]
P(\text{3rd is W}) &=& \frac{n}{n-2} \end{array}$
Hence: .$\displaystyle 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?
. . $\displaystyle \begin{array}{cccccc}
P(\text{1st is W}) &=& \frac{n}{3n} \\ \\[-3mm]
P(\text{2nd is R}) &=& \frac{n}{n-1} \\ \\[-3mm]
P(\text{3rd is B}) &=& \frac{n}{n-2} \end{array}$
Hence: .$\displaystyle P(\text{W-R-B}) \;=\;\dfrac{n}{3n} \times \dfrac{n}{3n-1} \times \dfrac{n}{3n-2} $
. . . etc. . . . etc.
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?