GRE Prep Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
What is the probability that the sum of two different
[#permalink]
27 Feb 2019, 21:38
27 Feb 2019, 21:38
Expert Reply
1
Bookmarks
00:00
Question Stats:
46% (01:25) correct
53% (01:28) wrong based on 96 sessions
Hide Show timer Statistics
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
Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Given Kudos: 136
Re: What is the probability that the sum of two different
[#permalink]
28 Feb 2019, 10:11
28 Feb 2019, 10:11
1
1
Bookmarks
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
