Question

A bag contains $5$ red, $8$ white and $7$ 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 $20$ $(5$ red, $7$ black and $8$ white$)$

Solution(i):

No. of red or white balls in bag is $13$ $(5$ red and $8$ white$)$

Therefore, ${}^{13}{C}_1$( Selecting $1$ out of $13$ items) times out of ${}^{20}{C}_1$( Selecting $1$ out of $20$ 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)}}$$=\frac{{}^{13}{C}_1}{{}^{20}{C}_1}$$=\frac{13}{20}$

Solution(ii):

No. of non-black balls in bag is $13$ $(5$ red and $8$ white$)$

Therefore, ${}^{13}{C}_1$( Selecting $1$ out of $13$ items) times out of ${}^{20}{C}_1$( Selecting $1$ out of $20$ 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)}}$$=\frac{{}^{13}{C}_1}{{}^{20}{C}_1}$$=\frac{13}{20}$

Solution(iii):

No. of neither black nor white balls in bag is $5$ $(5$ red balls$)$

Therefore, ${}^5{C}_1$( Selecting $1$ out of $5$ items) times out of ${}^{20}{C}_1$( Selecting $1$ out of $20$ items) a neither white nor black ball is picked.

Let $E$ be the event of drawing 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)}}$$=\frac{{}^5{C}_1}{{}^{20}{C}_1}$$=\frac{5}{20}$$=\frac{1}{4}$