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What is the probability that the sum of two different
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27 Feb 2019, 21:38
27 Feb 2019, 21:38
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What is the probability that the sum of two different single-digit prime numbers will NOT be prime?
(A) 0
(B) \(\frac{1}{2}\)
(C) \(\frac{2}{3}\)
(D) \(\frac{5}{6}\)
(E) 1
Diagnostic # 11
(A) 0
(B) \(\frac{1}{2}\)
(C) \(\frac{2}{3}\)
(D) \(\frac{5}{6}\)
(E) 1
Diagnostic # 11
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Re: What is the probability that the sum of two different
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28 Feb 2019, 10:11
28 Feb 2019, 10:11
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chetan2u wrote:
What is the probability that the sum of two different single-digit prime numbers will NOT be prime?
(A) 0
(B) \(\frac{1}{2}\)
(C) \(\frac{2}{3}\)
(D) \(\frac{5}{6}\)
(E) 1
Diagnostic # 11
(A) 0
(B) \(\frac{1}{2}\)
(C) \(\frac{2}{3}\)
(D) \(\frac{5}{6}\)
(E) 1
Diagnostic # 11
Single-digit prime numbers: 2, 3, 5, 7
Since there are so few numbers, we can list all 6 possible combinations.
#1) 2 & 3 (sum = 5)
#2) 2 & 5 (sum = 7)
#3) 2 & 7 (sum = 9)
#4) 3 & 5 (sum = 8)
#5) 3 & 7 (sum = 10)
#6) 5 & 7 (sum = 12)
Of the 6 possible outcomes, 4 outcomes yield a NON-PRIME sum
P(sum is NOT prime) = 4/6 = 2/3
Answer: C
Cheers,
Brent
Re: What is the probability that the sum of two different
[#permalink]
12 Jun 2019, 02:16
12 Jun 2019, 02:16
1
chetan2u wrote:
What is the probability that the sum of two different single-digit prime numbers will NOT be prime?
(A) 0
(B) \(\frac{1}{2}\)
(C) \(\frac{2}{3}\)
(D) \(\frac{5}{6}\)
(E) 1
Diagnostic # 11
(A) 0
(B) \(\frac{1}{2}\)
(C) \(\frac{2}{3}\)
(D) \(\frac{5}{6}\)
(E) 1
Diagnostic # 11
-------CONCEPT------------
Single digit prime numbers are
2,3,5,7
Now, except 2 all are odd so we can only get an odd sum if one of our prime number is even while other is odd (Only an odd number can be prime).
2+3 = 5
2+5 = 7
are only 2 cases to satisfy the condition.
while 2 + 7 = 9, it is not prime
& Now the total cases to get a sum by selecting any 2 prime numbers out of these 4 are 4C2 = 6
So, probability for sum NOT to be prime number = 4/6 = 2/3
