# A bag contains 5 red, 8 white and 7 black balls. A ball is drawn at random form the bag.

Question:

A bag contains 5 red, 8 white and 7 black balls. A ball is drawn at random form the bag. Find the probability that the drawn ball is

(i) red or white

(ii) not black

(iii) neither white nor black.

Solution:

GIVEN: A bag contains 5 red, 7 black and 8 white balls and a ball is drawn at random

TO FIND: Probability of getting a

(i) red or white ball

(ii) not black ball

(iii) neither white nor black

Total number of balls

(i) Total number red and white balls are

We know that PROBABILITY =

Hence probability of getting red or white ball

(ii) Total number of black balls are 7

We know that PROBABILITY =

Hence probability of getting black ball

We know that sum of probability of occurrence of an event and probability of non occurrence of an event is 1.

$P(\mathrm{E})+P(\overline{\mathrm{E}})=1$

$\frac{7}{20}+\mathrm{P}(\overline{\mathrm{E}})=1$

$\mathrm{P}(\overline{\mathrm{E}})=1-\frac{7}{20}$

$\mathrm{P}(\overline{\mathrm{E}})=\frac{20-7}{20}$

$\mathrm{P}(\overline{\mathrm{E}})=\frac{13}{20}$

Hence the probability of getting a non black ball is $P(\bar{E})=\frac{13}{20}$

(iii) Total number of neither red nor black balls i.e. red ball is 5

We know that PROBABILITY =

Hence probability of getting neither white nor black ball