A bag contains 6 red, 8 black and 4 white balls. A ball is drawn at random. What is the probability that ball drawn is not black?

Total no. of possible outcomes = 18 {6 red, 8 black, 4 white}

Let E ⟶ event of drawing black ball.

No. of favourable outcomes = 8 {8 black balls}

Probability, P(E) = (No. of favorable outcomes)/(Total no. of possible outcomes) = 8/18 = 4/9

Bar E⟶ event of not drawing black ball

P(Bar E)=1 − P(E)

= 1 − 4/9 = 5/9

Total numbers of red balls = 6

Number of black balls = 8

Number of white balls = 4

Total number of non-red balls = 6 + 8 + 4 = 18

Number of non-black balls are = 6 + 4 = 10

Probability of getting a non-black ball is = Total number of non-black balls/Total number of balls

= 10/18

= 5/9

∴ Probability of getting a non-black ball is 5/9

Total numbers of red balls = 6 

Number of black balls = 8 

Number of white balls = 4 

Total number of Non red balls = 6 + 8 + 4 = 18 

Number of non black balls are = 6 + 4 = 10

Probability of getting a non black ball is = \(\frac{Total\,number\,of\,non\,black\,balls}{Total\,number\,of\,balls}\) = \(\frac{10}{18}=\frac{5}{9}\)

Sample space, n(S) = 18 

Number of events of getting balls not black, 

n(E) = 10

∴ P(E) = \(\frac{n(E)}{n(S)}\) = \(\frac{10}{18}\) = \(\frac{5}9\)