Question

A bag contains $3$ red balls and $5$ black balls. A ball is drawn at random from the bag. What is the probability that the ball drawn is:
(i) red (ii) black

Answer 1

Solution(i):

Total no. of balls in the bag is $8$ $(3$ red and $5$ black$)$

No. of red balls in bag is $3$

Therefore, ${}^3{C}_1$( Selecting $1$ out of $3$ items) times out of ${}^8{C}_1$( Selecting $1$ out of $8$ items) a red ball is picked.

Let $E$ be the event of drawing a red ball from bag

We know that, Probability $P\left(E\right)$ $=\frac{\text{(No.of favorable outcomes)}}{\text{(Total no.of possible outcomes)}}$$=\frac{{}^3{C}_1}{{}^8{C}_1}$$=\frac{3}{8}$

Therefore,Probability that the ball drawn is red $=\frac{3}{8}$

Solution(iI):

Total no. of balls in the bag is $8$ $(3$ red and $5$ black$)$

No. of black balls in bag is $5$

Therefore, ${}^5{C}_1$( Selecting $1$ out of $5$ items) times out of ${}^8{C}_1$( Selecting $1$ out of $8$ items) a black ball is picked.

Let $E$ be the event of drawing a black ball from bag

We know that, Probability $P\left(E\right)$ $=\frac{\text{(No.of favorable outcomes)}}{\text{(Total no.of possible outcomes)}}$$=\frac{{}^5{C}_1}{{}^8{C}_1}$$=\frac{5}{8}$

Therefore,Probability that the ball drawn is black $=\frac{5}{8}$