A bag contains 4 red, 5 black and 6 white balls

Solution:

GIVEN: A bag contains 4 red, 5 black and 6white balls and a ball is drawn at random

TO FIND: Probability of getting a

(i) white ball

(ii) red ball

(iii) not black ball

(iv) red or white

Total number of balls 

(i) Total number white balls are 6

We know that PROBABILITY = 

Hence probability of getting white a ball is 

(ii) Total number of red are 4

We know that PROBABILITY = 

Hence probability of getting red a ball is equal to 

(iii) Total number of black balls are 5

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.

$P(E)+P(\bar{E})=1$

$\frac{1}{3}+P(\bar{E})=1$

$P(\bar{E})=1-\frac{1}{3}$

$P(\bar{E})=\frac{2}{3}$

Hence the probability of getting non black ball is $P(\bar{E})=\frac{2}{3}$

(iv) Total number of red or white balls is $4+6=10$

We know that PROBABILITY $=\frac{\text { Number of favourable event }}{\text { Total number of event }}$

Hence probability of getting white or red ball $\frac{10}{15}=\frac{2}{3}$