A bag contains $3$ red balls and $5$ black balls. if A ball is drawn at random from the bag. Then What is the probability that the ball drawn is (i) red (ii) not red

Given that,

A bag contains $3$ red and $5$ black balls.

Therefore, the total no. of possible outcomes $=8$ ($3$ red $+5$ black)

(i)Now Let E = event of getting red ball.

So, No. of favorable outcomes $=3$ (as there are $3$ red)

We know that Probability, P(E) = Number of favorable outcomes/ Total number of outcomes

P(E) $=3/8$

(ii)  Let
$(\overset{\scriptscriptstyle\rightharpoonup}{E})$ event of getting no red ball.

Therefore, From the previous question we already have P(E) $=3/8$

$P(\overset{\scriptscriptstyle\rightharpoonup}{E})+P(\overset{\scriptscriptstyle\rightharpoonup}{E})=1$

$P(\overset{\scriptscriptstyle\rightharpoonup}{E})=1-P(E)$

$P(\overset{\scriptscriptstyle\rightharpoonup}{E})=1-\frac{3}{8}$

$P(\overset{\scriptscriptstyle\rightharpoonup}{E})=\frac{5}{8}$