Total no. of possible outcomes = 8 {3 red, 5 black}

(i) Let E ⟶ event of drawing red ball.

No. favourable outcomes = 1 {1 ace card}

P(E) = (No.of favorable outcomes)/(Total no.of possible outcomes) = 3/8

(ii) Let E ⟶ event of drawing black ball.

No. favourable outcomes = 5 {5 black balls}

P(E) = 5/8

Given: A bag contains 3 red, and 5 black balls. A ball is drawn at random

Required to find: Probability of getting a

(i) red ball

(ii) white ball

Total number of balls 3 + 5 = 8

(i) Total number red balls are 3

We know that, Probability = Number of favourable outcomes/ Total number of outcomes

Thus, the probability of drawing a red ball = 3/8

(ii) Total number of black ball are 5

We know that, Probability = Number of favourable outcomes/ Total number of outcomes

Thus, the probability of drawing a black ball = 5/8

A bag contains 3 red balls, 5 black balls. 

Totally there are 8 balls. 

∴ n(S) = 8 

i) Possibility that the red ball drawn, n(E) = 3 

∴ Probability, P(E) = \(\frac{n(E)}{n(S)} = \frac{3}{8}\)

(ii) Possibility that the 1 black ball drwn is n(F) = 5 

∴ Probability, P(F) = \(\frac{n(F)}{n(S)} = \frac{5}{8}\)

Total no. of possible outcomes = 8 {3 red, 5 black}

(i) E ⟶ event of getting red ball.

No. of favourable outcomes = 3 {3 red}

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

P(E) = 3/8

(ii) Bar E ⟶ event of getting no red ball.

()+P(Bar E)=1

P(̅)=1−P(Bar E)

= 1 − 3/8 = 5/8