Find the probability that the ball is :

A bag contains 6 red balls, 8 white balls, 5 green balls, and 3 black balls. One ball is drawn at random from the bag. Find the probability that the ball is :
(i) white
(ii) red or black
(iii) not green
(iv) neither white nor black.

Solution:

In a bag,
Number of red balls = 6
Number of white balls = 8
Number of green balls = 5
and number of black balls = 3
Total number of balls in the bag
= 6 + 8 + 5 + 3 = 22
(i) Probability of white balls
P(E) = $\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }$
= $\\ \frac { 8 }{ 22 }$
= $\\ \frac { 4 }{ 11 }$
(ii) Probability of red or black balls
P(E) = $\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }$
= $\\ \frac { 6+3 }{ 22 }$
= $\\ \frac { 9 }{ 22 }$
(iii) Probability of not green balls i.e. having red, white and black balls.
P(E) = $\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }$
= $\\ \frac { 6+8+3 }{ 22 }$
= $\\ \frac { 17 }{ 22 }$
(iv) Probability of neither white nor black balls i.e. red and green balls
P(E) = $\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }$
= $\\ \frac { 6+5 }{ 22 }$
= $\\ \frac { 11 }{ 22 }$
= $\\ \frac { 1 }{ 2 }$