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
In a set of 24 cards, each card is numbered with a different positive integer from 1 to 24.
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Solution:BTGModeratorVI wrote: ↑Thu Jul 23, 2020 6:30 amIn 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
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
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P(event) = (number of outcomes that meet the given condition)/(total number of possible outcomes)BTGModeratorVI wrote: ↑Thu Jul 23, 2020 6:30 amIn 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
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