In a set of 24 cards, each card is numbered with a different positive integer from 1 to 24.

[BTGModeratorVI]

In a set of 24 cards, each card is numbered with a different positive integer from 1 to 24. One card will be drawn at random from the set. What is the probability that the card drawn will have either a number that is divisible by both 2 and 3 or a number that is divisible by 7 ?

A. 3/24

B. 4/24

C. 7/24

D. 8/24

E. 17/24

Answer: C
Source: Official guide

[Scott@TargetTestPrep]

In a set of 24 cards, each card is numbered with a different positive integer from 1 to 24. One card will be drawn at random from the set. What is the probability that the card drawn will have either a number that is divisible by both 2 and 3 or a number that is divisible by 7 ?

A. 3/24

B. 4/24

C. 7/24

D. 8/24

E. 17/24

Answer: C
Source: Official guide

Solution:

The numbers that are divisible by both 2 and 3 are 6, 12, 18, and 24.

The numbers that are divisible by 7 are 7, 14, and 21.

So the probability is 7/24.

Answer: C

[Brent@GMATPrepNow]

In a set of 24 cards, each card is numbered with a different positive integer from 1 to 24. One card will be drawn at random from the set. What is the probability that the card drawn will have either a number that is divisible by both 2 and 3 or a number that is divisible by 7 ?

A. 3/24

B. 4/24

C. 7/24

D. 8/24

E. 17/24

Answer: C
Source: Official guide

P(event) = (number of outcomes that meet the given condition)/(total number of possible outcomes)

So, P(selected number is divisible by both 2 and 3 OR is divisible by 7) = number of integers that are divisible by both 2 and 3 OR are divisible by 7/24

number of integers that are divisible by both 2 and 3 OR are divisible by 7
The integers that meet this condition are: 6, 7, 12, 14, 18, 21, 24
There are 7 such numbers

So, P(selected number is divisible by both 2 and 3 OR is divisible by 7) = 7/24

Answer: C

Cheers,
Brent