Question

An urn contains 5 red, 3 black and 2 white balls. If three balls are chosen at random then what is the probability that balls will be of different colours?

• A. $\frac{1}{2}$
• B. $\frac{1}{4}$
• C. $\frac{3}{11}$
• D. $\frac{4}{15}$

Answer 1

Answer: (B)

$\frac{1}{4}$

To get $3$ balls of different colours, we need to choose $1$ ball of each of the three given colours.

So, probability to get $3$ balls of different colours $=\frac{Choosing1red,1black,1whiteballs}{Choosingany3fromall10balls}=\frac{{5_C}_1\times {3_C}_1\times {2_C}_1}{{{10}_C}_3}$

$=\frac{5\times 3\times 2}{\frac{10!}{3!(10-3)!}}$ Since ${n_C}_1=n$ and ${n_C}_r=\frac{n!}{r!(n-r)!}$

$=\frac{5\times 3\times 2}{\frac{10!}{3!7!}}=\frac{5\times 3\times 2\times 3!\times 7!}{10!}=\frac{5\times 3\times 2\times 3\times 2\times 7!}{10\times 9\times 8\times 7!}=\frac{1}{4}$