# Problem 4

A ball is drawn at random from a box containing 16 red balls, 24 white balls and 15 blue balls. Determine the probability that it is (i) Not red, (ii) Red or white, (iii) Neither Red Nor Blue
Ans : i)
 39 55
; ii)
 8 11
; iii)
 24 55
;

# Solution

Total number of balls in the bag

= 16 Red + 24 White + 15 Blue

= 55

Experiment :

Drawing a ball at random from the bag containing 16 red, 24 white and 15 blue balls

Total Number of Possible Choices

= Number of ways in which one ball can be drawn from the total 55

⇒ n = 55C1
=
 55 1
= 55

Let

A : the event of the ball drawn being not red

B : the event of the ball drawn being red or white

C : the event of the ball drawn being neither red nor blue

## For Event A

 Favorable (White + Blue) Unfavorable (Red) Total Available 39 16 55 To Choose 1 0 1 Choices 39C1 16C0 55C1

Number of Favorable Choices

= Number of ways in which a ball which is not red can be drawn from the total 39

⇒ mA = 39C1
=
 39 1
= 39

Probability of the ball drawn being not red

⇒ Probability of occurrence of Event A

=
 Number of Favorable Choices for the Event Total Number of Possible Choices for the Experiment
⇒ P(A) =
 mA n
=
 39 55

## Odds

Number of Unfavorable Choices

= Total Number of possible choices − Number of Favorable choices

 ⇒ mAc = n − mA = 55 − 39 = 16

## in favor

Odds in Favor of the ball drawn being not red

⇒ Odds in Favor of Event A

= Number of Favorable Choices : Number of Unfavorable Choices

= mA : mAc

= 39 : 16

## against

Odds against the ball drawn being not red

⇒ Odds against Event A

= Number of Unfavorable Choices : Number of Favorable Choices

= mAc : mA

= 16 : 39

## For Event B

 Favorable (Red+ White) Unfavorable (Blue) Total Available 40 15 55 To Choose 1 0 1 Choices 40C1 15C0 55C1

Number of Favorable Choices

= Number of ways in which a ball which is red or white can be drawn from the total 40

⇒ mB = 40C1
=
 40 1
= 40

Probability of the ball drawn being red or white

⇒ Probability of occurrence of Event B

=
 Number of Favorable Choices for the Event Total Number of Possible Choices for the Experiment
⇒ P(B) =
 mB n
=
 40 55
=
 8 11

## For Event C

The ball drawn being neither red nor blue

⇒ the ball drawn is white

 Favorable (White) Unfavorable (Red + Blue) Total Available 24 31 55 To Choose 1 0 1 Choices 24C1 31C0 55C1

Number of Favorable Choices

= Number of ways in which a ball which is neither red nor blue can be drawn from the total 24

⇒ mC = 24C1
=
 24 1
= 24

Probability of the ball drawn being neither red nor blue

⇒ Probability of occurrence of Event C

=
 Number of Favorable Choices for the Event Total Number of Possible Choices for the Experiment
⇒ P(C) =
 mC n
=
 24 55
Author : The Edifier