Question

A bag contains 5 white balls, 7 red balls, 4 black balls and 2 blue balls. One ball is drawn at random from the bag. Find the probability that the balls drawn is (i) White or blue (ii) red or black (iii) not white (iv) neither white nor black.

Answer 1

Number of white balls in a bag= 5
Number of red balls in a bag= 7
Number of black balls in a bag= 4
Number of blue balls in a bag= 2

Total Number of balls in a bag= 5 +7+4+2= 18
Total number of favourable outcomes= 18

i) Number of white  & blue balls in a bag= 5+2= 7
Number of favourable outcomes= 7

Probability = Number of favourable outcomes/ Total number of outcomes
P(white or blue) = 7/18

ii)  Number of red  & black balls in a bag= 7+4= 11
Number of favourable outcomes= 11

Probability = Number of favourable outcomes/ Total number of outcomes
P(red or black) = 11/18

iii) Number of not white balls in a bag=
Total number of balls in a bag -  number of white balls = 18 -5 = 13
Number of favourable outcomes= 13

Probability = Number of favourable outcomes/ Total number of outcomes
P(not white) = 13 /18

iv) Number of balls neither white nor black in a bag= Total number of balls in a bag -  number of white & black balls = 18 -(5+4) = 18 - 9= 9
Number of favourable outcomes= 9

Probability = Number of favourable outcomes/ Total number of outcomes
P(not white) = 9 /18 = ½.

HOPE THIS WILL HELP YOU..

Answer 2

total no of balls = total no of outcomes = 5+7+4+ 2 = 18

now one ball is drawn according to the question

1. P(white or blue) = 5+4/ 18 => 9/18 => 1/2

2.P(red or black) = 7+4/18 => 11/ 18

3. P(not white) = 7+ 4+2/18 => 13/18

4. P(neither white nor black) = 7+ 2/18 => 9/18 => 1/2....