Of a set of 25 consecutive integers beginning with 4, what is the probability that a number selected at random will be divisible by 3?
A. 13/25
B. 9/25
C. 8/25
D. 6/25
E. 3/8
The OA is C .
I don't like the probabilities questions. Experts, may you help me please?
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Of a set of 25 consecutive integers beginning with 4
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Hi M7MBA,
We're told to consider a set of 25 consecutive integers beginning with 4 (meaning the integers from 4 to 28, inclusive). We're asked for the probability that a number selected at random from this group will be divisible by 3.
The numbers divisible by 3 would be 6, 9, 12, 15, 18, 21, 24 and 27, meaning 8 out of 25 would be divisible by 3.
Final Answer: C
GMAT assassins aren't born, they're made,
Rich
We're told to consider a set of 25 consecutive integers beginning with 4 (meaning the integers from 4 to 28, inclusive). We're asked for the probability that a number selected at random from this group will be divisible by 3.
The numbers divisible by 3 would be 6, 9, 12, 15, 18, 21, 24 and 27, meaning 8 out of 25 would be divisible by 3.
Final Answer: C
GMAT assassins aren't born, they're made,
Rich
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Jeff@TargetTestPrep
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We are given that there is a set of 25 consecutive integers, starting with the number 4. Thus, the set of integers is from 4 to 28, inclusive. We need to determine how many multiples of 3 fall within that range. To do so, we can use the following formula:M7MBA wrote:Of a set of 25 consecutive integers beginning with 4, what is the probability that a number selected at random will be divisible by 3?
A. 13/25
B. 9/25
C. 8/25
D. 6/25
E. 3/8
# of multiples of 3 = [(largest multiple of 3 in the set - smallest multiple of 3 in the set)/3] + 1
# of multiples of 3 = [(27 - 6)/3] +1 = 21/3 + 1 = 8
Thus, the probability of selecting a multiple of 3 is 8/25.
Answer: C
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