Drawing a number which is a multiple of two numbers 3 or 4 having common multiples

Problem 5

An urn contains 25 balls numbered from 1 to 25. Find the probability that a ball selected at random is a ball with a number that is a multiple of 3 or 4.
Ans :
12
25

Solution

Total number of tickets in the urn

= 25

Experiment :

Drawing a ball from the urn containing balls numbered from 1 to 25

Total Number of Possible Choices

= Number of ways in which a ball can be drawn from the total 25

⇒ n = 25C1
=
25
1
= 25

A : the event of the number on the ball drawn being a multiple of 3 or 4.

For Event A

From 1 to 13

Multiples of 3 ⇒ 3, 6, 9, 12, 15, 18, 21, 24

Multiples of 4 ⇒ 4, 8, 12, 16, 20, 24

The numbers which are LCM of the two numbers and the multiples of LCM repeat. Strike off the repeated numbers from one to collect the required list.

Multiples of 3 or 4

= 12 {3, 4, 6, 8, 9, 12, 15, 16, 18, 20, 21, 24}

Favorable
(Multiples of 3 or 4)
Unfavorable
(Others)
Total
Available 12 13 25
To Choose 1 0 1
Choices 12C113C025C1

Number of Favorable Choices

= Number of ways in which a ball with a number which is a multiple of 3 or 4 can be drawn from the total 6 favorable balls

⇒ mA = 6C1
=
12
1
= 12

Probability of drawing a ball with a number which is a multiple of 3 or 4

⇒ 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
25

Odds

Number of Unfavorable Choices

= Total Number of possible choices − Number of Favorable choices

⇒ mAc = n − mA
= 25 − 12
= 13

in favor

Odds in Favor of drawing a bll with a number which is a multiple of 3 or 4

⇒ Odds in Favor of Event A

= Number of Favorable Choices : Number of Unfavorable Choices

= mA : mAc

= 12 : 13

against

Odds against drawing a ball with a number which is a multiple of 3 or 4

⇒ Odds against Event A

= Number of Unfavorable Choices : Number of Favorable Choices

= mAc : mA

= 13 : 12