Question

A bag contains $8$ red, $6$ white and $4$ black balls. A ball is drawn at random from the bag. Find the probability that the drawn ball is
(i) red or white (ii) not black (iii) neither white nor black

Answer 1

Total no. of balls in the bag is $18$ $(8$ red, $4$ black and $6$ white$)$

Solution(i):

No. of red or white balls in bag is $14$ $(8$ red and $6$ white$)$

Therefore, ${}^{14}{C}_1$( Selecting $1$ out of $14$ items) times out of ${}^{18}{C}_1$( Selecting $1$ out of $18$ items) a red or white ball is picked.

Let $E$ be the event of drawing a red 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)}}$$=\frac{{}^{14}{C}_1}{{}^{18}{C}_1}$$=\frac{14}{18}$$=\frac{7}{9}$

Solution(ii):

No. of non-black balls in bag is $14$ $(8$ red and $6$ white$)$

Therefore, ${}^{14}{C}_1$( Selecting $1$ out of $14$ items) times out of ${}^{18}{C}_1$( Selecting $1$ out of $18$ 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)}}$$=\frac{{}^{14}{C}_1}{{}^{18}{C}_1}$$=\frac{14}{18}$$=\frac{7}{9}$

Solution(iii):

No. of neither black nor white balls in bag is $8$ $(8$ red balls$)$

Therefore, ${}^8{C}_1$( Selecting $1$ out of $8$ items) times out of ${}^{18}{C}_1$( Selecting $1$ out of $18$ items) a neither white nor black ball is picked.

Let $E$ be the event of drawing a neither white nor black ball from bag

We know that, Probability $P\left(E\right)$ $=\frac{\text{(No.of favorable outcomes)}}{\text{(Total no.of possible outcomes)}}$$=\frac{{}^8{C}_1}{{}^{18}{C}_1}$$=\frac{8}{18}$$=\frac{4}{9}$