$\begingroup$
I'm stuck with this problem:
$\endgroup$ single digit primes = 2, 3, 5, 7 2, 3 ⇒ 31 not possible 2 5 ⇒ 31 not possible 2, 7 ⇒ 31 not possible 3, 5 ⇒ (3, 5, 6 ,8 ,9) 3 ,7 ⇒ (3 , 4 , 7 , 8 , 9) 5 , 7 ⇒ (4, 5 , 6, 7 , 9 ) P = (3/6) = (1/2)
 Text Solution Solution : We know single digit natural number are from `1` to `9`.<br> Now, the maximum sum that we can get is `9+8=17` as we have to use different numbers.<br> Now, possible two digits having a sum equal to prime number are,<br> `(1+2),(1+4),(2+3),(1+6),(2+5),(3+4),(2+9),(3+8),(4+7),(5+6),`<br>`(4+9),(5+8),(6+7),(8+9)`.<br> So, there are `14` possible values.<br> Now, number of ways selecting these `2` digits from `9` digits is `C(9,2)`.<br> `:.` the required probability ` = 14/(C(9,2)) = 14/36 = 7/18`<br>
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What is the probability that the sum of two different [#permalink]
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Re: What is the probability that the sum of two different [#permalink]
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 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: CCheers,Brent _________________
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Re: What is the probability that the sum of two different [#permalink]
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 -------CONCEPT------------Single digit prime numbers are2,3,5,7Now, 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 = 52+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 = 6So, probability for sum NOT to be prime number = 4/6 = 2/3
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What is the probability that the sum of two different single digit prime numbers will not be prime
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