Question

A bag contains $5$ black, $7$ red and $3$ white balls. A ball is drawn from the bag at random. Find the probability that the ball drawn is:
(i) red (ii) black or white (iii) not black

Answer 1

Total no. of balls in the bag is $15$ $(7$ red, $5$ black and $3$ white$)$

Solution(i):

No. of red balls in bag is $7$

Therefore, ${}^7{C}_1$( Selecting $1$ out of $7$ items) times out of ${}^{15}{C}_1$( Selecting $1$ out of $15$ 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)}}$$=\frac{{}^7{C}_1}{{}^{15}{C}_1}$$=\frac{7}{15}$

Solution(ii):

No. of black or white balls in bag is $8$ $(5$ black and $3$ white$)$

Therefore, ${}^8{C}_1$( Selecting $1$ out of $8$ items) times out of ${}^{15}{C}_1$( Selecting $1$ out of $15$ items) a black or white ball is picked.

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

We know that, Probability $P\left(E\right)$ $=\frac{\text{(No.of favorable outcomes)}}{\text{(Total no.of possible outcomes)}}$$=\frac{{}^8{C}_1}{{}^{15}{C}_1}$$=\frac{8}{15}$

Solution(iii):

No. of non-black balls in bag is $10$ $(7$ red and $3$ white$)$

Therefore, ${}^{10}{C}_1$( Selecting $1$ out of $10$ items) times out of ${}^{15}{C}_1$( Selecting $1$ out of $15$ items) a non-black ball is picked.

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

We know that, Probability $P\left(E\right)$ $=\frac{\text{(No.of favorable outcomes)}}{\text{(Total no.of possible outcomes)}}$$=\frac{{}^{10}{C}_1}{{}^{15}{C}_1}$$=\frac{10}{15}$$=\frac{2}{3}$