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.
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