# Problem 2

If a ball is drawn at random from a bag containing 12 black and 15 white balls, what is the probability and odds that it is a) Black b) White?
Ans : a)
 4 9
, 4 : 5, 5 : 4; b)
 5 9
, 5 : 4, 4 : 5;

# Solution

Total number of balls in the bag

= 12 Black + 15 White

= 27

Experiment :

Drawing a ball at random from the bag containing 12 black and 15 white balls

Total Number of Possible Choices

= Number of ways in which one ball can be drawn from the total 27

⇒ n = 27C1
=
 27 1
= 27

Let

A : the event of the ball drawn being black

B : the event of the ball drawn being white

## For Event A

 Favorable (Black) Unfavorable (Others) Total Available 12 15 27 To Choose 1 0 1 Choices 12C1 15C0 27C1

Number of Favorable Choices

= Number of ways in which one black ball can be drawn from the total 12

⇒ mA = 12C1
=
 12 1
= 12

Probability of the ball drawn being black

⇒ 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
=
 12 27
=
 4 9

## Odds

Number of Unfavorable Choices

= Total Number of possible choices − Number of Favorable choices

 ⇒ mAc = n − mA = 27 − 12 = 15

## in favor

Odds in Favor of the ball drawn being black

⇒ Odds in Favor of Event A

= Number of Favorable Choices : Number of Unfavorable Choices

= mA : mAc

= 12 : 15

= 4 : 5

## against

Odds against the ball drawn being black

⇒ Odds against Event A

= Number of Unfavorable Choices : Number of Favorable Choices

= mAc : mA

= 15 : 12

= 5 : 4

## For Event B

 Favorable (White) Unfavorable (Others) Total Available 15 12 27 To Choose 1 0 1 Choices 15C1 12C0 27C1

Number of Favorable Choices

= Number of ways in which one white ball can be drawn from the total 15

⇒ mB = 15C1
=
 15 1
= 15

Probability of the ball drawn being 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
=
 15 27
=
 5 9

## Odds

Number of Unfavorable Choices

= Total Number of possible choices − Number of Favorable choices

 ⇒ mBc = n − mB = 27 − 15 = 12

## in favor

Odds in Favor of the ball drawn being white

⇒ Odds in Favor of Event B

= Number of Favorable Choices : Number of Unfavorable Choices

= mB : mBc

= 15 : 12

= 5 : 4

## against

Odds against the ball drawn being white

⇒ Odds against Event B

= Number of Unfavorable Choices : Number of Favorable Choices

= mBc : mB

= 12 : 15

= 4 : 5

## For Event B [Alternative I]

Since there are only two possible outcomes on drawing a ball i.e. the ball being either a black ball or a white ball

Probability of the ball drawn being white

= Probability of the ball drawn not being black.

⇒ P(B) = P (Ac)
= 1 − P (A)
= 1 −
 12 27
=
 27 − 12 27
=
 15 27

## For Event B [Alternative II]

There are only two possible outcomes on drawing a ball i.e. the ball being either a black ball or a white ball

The number of favourable choices for drawing a black ball would form the number of unfavorable choices for drawing a white ball and vice-versa

Number of Favorable Choices

= Number of Unfavorable Choices for Event A

 ⇒ mB = mAc = 1 − mA = 1 − 12 = 15

Probability of the ball drawn being 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
=
 15 27
=
 5 9
Author : The Edifier