Question

Three balls are drawn at random from a collection of $7$ white, $12$ green and $4$ red balls. The probability that each is different colour is

• A. $\frac{48}{253}$
• B. $\frac{64}{253}$
• C. $\frac{23}{253}$
• D. $\frac{56}{253}$

Answer 1

The correct option is A $\frac{48}{253}$

The number of ways drawing one white, one green and one red ball $={}^7{C}_1\times {}^{12}{C}_1\times {}^4{C}_1$

$=7\times 12\times 4$

$=336$

Total number of balls $=7+13+4=23$

The probability of drawing one white, one green and one red ball $=\frac{336}{{}^{23}{C}_3}$

$=\frac{336}{\frac{23\times 22\times 21}{3\times 2\times 1}}$

$=\frac{336}{1771}$

$=\frac{48}{253}$