Problem Solving

BTGmoderatorDC wrote: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

P = (good outcomes)/(all possible outcomes)

All possible outcomes:
Since there are 24 cards, there are 24 possible outcomes.

Good outcomes:
For a number to be divisible by 2 and 3, it must be a multiple of 6.
Multiples of 6 between 1 and 24, inclusive:
6, 12, 18, 24
For a number to be divisible by 7, it must be a multiple of 7.
Multiples of 7 between 1 and 24, inclusive:
7, 14, 21
Since a good outcome will be yielded by either the 4 blue options or the 3 red options, we get:
Good outcomes = blue options + red options = 4+3 = 7.

Resulting probability:
(good outcomes)/(all possible outcomes) = 7/24.

The correct answer is C.

Total outcomes possible= 24
Favourable outcome= Number divisible by 6(both by 2 and 3) or 7
So, possible multiples of 6 + multiples of 7 - Multiples of 6*7
= 4+3-0=7

So, probablity= 7/24