Each of the numbers from 1 to 30 is written on a card and placed in a bag.

Question

Each of the numbers from 1 to 30 is written on a card and placed in a bag. If one card is drawn at random, what is the probability that the number is a multiple of 3 or a multiple of 4? Round to the nearest percent.

A. 47%
B. 50%
C. 63%
D. 8%

Answer 1

Lily,

The number of cards is given as 30. 

The number of cards that are multiples of 3:  3,6,9,12,15,18,21,24,27,30. Therefore, P(M₃)=10/30

 

The number of cards that are multiples of 4: 4,8,12,16,20,24,28.

Therefore, P(M₄)=7/30

 

Since there are numbers that are both multiples of 3 and 4, such as, 12 and 24; the probability these cards are drawn is 

therefore, P(M₃ and M₄)=2/30

 

To calculate the probability, add the prob. of the cards being multiples of 3 and 4, then subtract when both occure at the same time-(to eliminate repetition)

 

P= (10/30)+(7/30)-(2/30)

P=15/30

In other words, 50%.

The correct choice is b.

 

Answer 2

Lily,

 

Here is your universal set: 1, 2, 3,...., 30 (30 elements)

Here is your subset: 3, 6, 9, 12, 15, 18, 21, 24, 27, 30 and 4, 8, 12, 20, 24, 28 (16 elements)

 

the probability that the number is a multiple of 3 or a multiple of 4 = 100*(16/30) = 53.3333 or 53%