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bipolarbear
Joined: 11 Dec 2008
Last visit: 16 Sep 2013
Posts: 353
Given Kudos: 12
Location: United States
Schools: HBS '15 (D) Stanford '15 (WL) Wharton '15 (WL)
GMAT 1: 760 Q49 V44
GPA: 3.9
06 Aug 2009, 10:38
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
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Question Stats:
50% (01:53) correct
50%
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wrong
based on 246
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If two integers are selected, one after the other with replacement, from a set of integers from 20 to 29, inclusive, what is the probability that one of the selected integers is a prime number, and the other is a multiple of 3?
A. 0.06
B. 0.12
C. 0.15
D. 0.18
E. 0.20
A. 0.06
B. 0.12
C. 0.15
D. 0.18
E. 0.20
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20 Sep 2023, 08:45
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If two integers are selected, one after the other with replacement, from a set of integers from 20 to 29, inclusive, what is the probability that one of the selected integers is a prime number, and the other is a multiple of 3?
A. 0.06
B. 0.12
C. 0.15
D. 0.18
E. 0.20
Start by identifying the prime integers and multiples of 3 in the range 20 to 29. The prime integers are 23 and 29, while the multiples of 3 are 21, 24, and 27.
We can then consider two cases:
Case 1: The first number selected is prime, and the second is a multiple of 3. The probability of this is (2/10) * (3/10) = 0.06. Notice, that since we are selecting two integers with replacement, the denominator for selection case is 10 (the total number of integers in the given range).
Case 2: The first number selected is a multiple of 3, and the second is prime. The probability of this is (3/10) * (2/10) = 0.06.
Therefore, the total probability is the sum of the probabilities of the above two cases: 0.06 + 0.06 = 0.12.
Answer: B
A. 0.06
B. 0.12
C. 0.15
D. 0.18
E. 0.20
Start by identifying the prime integers and multiples of 3 in the range 20 to 29. The prime integers are 23 and 29, while the multiples of 3 are 21, 24, and 27.
We can then consider two cases:
Case 1: The first number selected is prime, and the second is a multiple of 3. The probability of this is (2/10) * (3/10) = 0.06. Notice, that since we are selecting two integers with replacement, the denominator for selection case is 10 (the total number of integers in the given range).
Case 2: The first number selected is a multiple of 3, and the second is prime. The probability of this is (3/10) * (2/10) = 0.06.
Therefore, the total probability is the sum of the probabilities of the above two cases: 0.06 + 0.06 = 0.12.
Answer: B
General Discussion
14 Jan 2016, 05:20
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shreyashid
This is how I approached it. Please tell me if this is not right and why?
Total integers between 20-29=10
Number of prime =2 (23 & 29)
Number of integers divisible by 3 = 3 (21,24 & 27)
Total ways 2 integers can be picked from 10 integers : 10C2
Thus probability : 2*3/ 10C2 = 2/15
Total integers between 20-29=10
Number of prime =2 (23 & 29)
Number of integers divisible by 3 = 3 (21,24 & 27)
Total ways 2 integers can be picked from 10 integers : 10C2
Thus probability : 2*3/ 10C2 = 2/15
No. Note that the numbers are picked with replacement. 10C2 gives you the number of ways of picking the two numbers one after the other without replacement.
You can pick the prime number in 2 ways and then the multiple of 3 in 3 ways. This gives you 2*3 = 6 ways.
But you can also pick the multiple of 3 first and then the prime number. This gives you another 3*2 = 6 ways.
No of ways of picking the first number is 10 and the second number is 10 again. So total ways = 100 ways
Probability = 12/100
or use probability
20-29 are 10 numbers of which 2 are prime (23 and 29) and 3 are multiples of 3 (21, 24 and 27)
First, you can select a prime with probability 2/10 and then a multiple of 3 with probability 3/10.
Or you can select a multiple of 3 with probability 3/10 and then a prime with probability 2/10.
Total probability = (2/10)*(3/10) + (3/10)*(2/10) = 3/25
