Drawing/Picking/Choosing Single/One black/white ball from a bag/urn/box

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 12C115C027C1

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 15C112C027C1

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