A bag contains 5 white balls, 7 red balls, 4 black balls, and 2 blue balls.

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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. What is the probability that the ball drawn is :
(i) white or blue
(ii) red or black
(iii) not white
(iv) neither white nor black?

Solution:

Number of total balls = 5 + 7 + 4 + 2 = 18
Number of white balls = 5
number of red balls = 7
number of black balls = 4
and number of blue balls = 2.
(i) Number of white and blue balls = 5 + 2 = 7
Probability of white or blue balls will be
P(E) = \frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } = \\ \frac { 7 }{ 18 }
(ii) Number of red and black balls = 7 + 4 = 11
Probability of red or black balls will be
P(E) = \frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } = \\ \frac { 11 }{ 18 }
(iii) Number of ball which are not white = 7 + 4 + 2 = 13
Probability of not white balls will be
P(E) = \frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } = \\ \frac { 13 }{ 18 }
(iv) Number of balls which are neither white nor black = 18 – (5 + 4) = 18 – 9 = 9
Probability of ball which is neither white nor black will be
P(E) = \frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } = \\ \frac { 9 }{ 18 } = \\ \frac { 1 }{ 2 }

A box contains 20 balls bearing numbers 1, 2, 3, 4,……, 20. A ball is drawn at random from the box. What is the probability that the number on the ball is
(i) an odd number
(ii) divisible by 2 or 3
(iii) prime number
(iv) not divisible by 10?

Solution:

In a box, there are 20 balls containing 1 to 20 number
Number of possible outcomes = 20
(i) Numbers which are odd will be,
1, 3, 5, 7, 9, 11, 13, 15, 17, 19 = 10 balls.
Probability of odd ball will be
P(E) = \frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } = \\ \frac { 10 }{ 20 } = \\ \frac { 1 }{ 2 }
(ii) Numbers which are divisible by 2 or 3 will be
2, 3, 4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20 = 13 balls
Probability of ball which is divisible by 2 or 3 will be
P(E) = \frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } = \\ \frac { 13 }{ 20 }
(iii) Prime numbers will be 2, 3, 5, 7, 11, 13, 17, 19 = 8
Probability of prime number will be
P(E) = \frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } = \\ \frac { 8 }{ 20 } = \\ \frac { 2 }{ 5 }
(iv) Numbers not divisible by 10 will be
1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 12, 13, 14, 15, 16, 17, 18, 19 = 18
Probability of prime number will be
P(E) = \frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } = \\ \frac { 18 }{ 20 } = \\ \frac { 9 }{ 10 }

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