Question

A bag contains 5 black, 4 green and 6 red balls. If a ball is drawn at random from the bag, find the probability that it will be:
i) red
ii) black
iii) green
[3 MARKS]

Answer 1

Each subpart: 1 Mark each

Total number of balls in the bag
= 5(B) + 4(G) + 6(R) = 15 balls
Thus, n(S) = 15
i) Event = {Red ball}, i.e., n(E) = 6
$\therefore P\left(E\right)=\frac{n\left(E\right)}{n\left(S\right)}=\frac{6}{15}=\frac{2}{5}$

ii) Event = {Black ball}, i.e., n(E) = 5
$\therefore P\left(E\right)=\frac{n\left(E\right)}{n\left(S\right)}=\frac{5}{15}=\frac{1}{3}$

iii) Event = {Green ball}, i.e., n(E) = 4
$\therefore P\left(E\right)=\frac{n\left(E\right)}{n\left(S\right)}=\frac{4}{15}$