BrentGMATPrepNow
Bunuel
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(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/
24number of integers that are divisible by both 2 and 3 OR are divisible by 7The 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/
24Answer: C
Cheers,
Brent
BrentGMATPrepNowThank you for this! I have a hypothetical question. What if the question asked for cards from 1 through 42. Because 42 is divisible by both 2, 3, and 7, would you count it twice within the odds? In other words, would the answer be 13/42? I am just confused with how to deal with the "or" if there is overlap with the specifications should there be a similar problem in this nature.
Divisible by 2 & 3: 6, 12, 18, 24, 30, 36, 42
Divisible by 7: 7, 14, 21, 28, 35, 42
7/42 + 6/42 = 13/42
Thanks so much!