Question

Two balls are drawn at random from a bag containing 2 white, 3 red. 5 green and 4 black balls, one by one without, replacement. Find the probabilitythat both the balls are of different colours.

Answer 1

Out of 14 balls, two balls can be chosen in ${}^{14}{C}_2$ ways.

So, favourable number of elementary events = ${}^{14}{C}_2$

Let E be the event that all balls are of the same colour
E = {WW, RR, GO, BB}

$\therefore n\left(E\right){=}^2{C}_2{+}^3{C}_2{+}^5{C}_2{+}^4{C}_2$

$P\left(E\right)=\frac{{}^2{C}_2{+}^3{C}_2{+}^5{C}_2{+}^4{C}_2}{{}^{14}{C}_2}=\frac{40}{182}=\frac{20}{91}$

$\therefore$ Probability that both are of different colour

$P\left(\overline{E}\right)=1-P\left(E\right)$

$=1-\frac{20}{91}$

$=\frac{71}{91}=0.78$