The first 13 prime numbers are 2,3,5,7,11,13,17,19,23,29,31,37,41

Probability Level 3

Each of the first thirteen prime numbers is written on an individual piece of paper and put in a box.

If 2 pieces of paper are drawn without replacement, what is the probability that the sum of the 2 numbers will be a prime number?

If the probability can be expressed as $\dfrac ab$ for coprime positive integers $a$ and $b$ , find the value of $a+b$ .

The answer is 14.

2 solutions

Brian Charlesworth
Sep 23, 2015

The sum of any two odd primes is even and greater than $2,$ and hence is not prime. So one of the numbers chosen must be $2$ in order for the sum to be prime. However, the odd prime chosen must be the lesser element of a twin prime pair in order for its sum with $2$ to be prime. There are $6$ such odd primes in the given list, namely $3,5,11,17,29$ and $41.$ As there are $\dbinom{13}{2}$ possible combinations of two numbers that can be chosen without restriction, the desired probability is

$\dfrac{6}{\dbinom{13}{2}} = \dfrac{6}{\dfrac{13*12}{2}} = \dfrac{1}{13}$ , and so $a + b = 1 + 13 = \boxed{14}.$

If you put the numbers on a chart, there are 169 (13*13) possible sums, and out of them are 12 primes out of the 169 possibilities. Shouldn't it just be 12/169? EDIT: Didn't read the question! My bad.

Raghav Arora - 5 years, 8 months ago

The numbers are chosen without replacement, so there are only $13*12 = 156$ permutations of two primes, $12$ of which sum to a prime. Thus the probability is $\frac{12}{156} = \frac{1}{13}.$

Brian Charlesworth - 5 years, 8 months ago

Same Method

Kushagra Sahni - 5 years, 8 months ago

No estoy de acuerdo. Solo 2 puede sumar un # primo, en este caso: 2+3, 2+5, 2+11, 2+17, 2+29, 2+41 total son seis posibles eventos. Como la pareja retirada no retorna a la bolsa y solo es posible retirar seis parejas solo en el evento que el #2 salga la probabilidad sería 1/6 y 1 +6 = 7. Puede ocurrir el evento cero, so el #2 no ocurre en las seis parejas. Mi respuesta es 7 ó 0

Jaime Maldonado - 5 years, 7 months ago

Rwit Panda
Sep 24, 2015

Only 2 is an even prime. Others are odd. As for a prime we ought to have odd if it's greater than 2, so 2 has to be there. We observe that we have 6 options with 2, i.e., with 3,5,11,17,29,41. So probability is 6/13c2 which is 1/13. So a=1 and b=13. So a+b=13+1=14. Ola!!!!!

The first 13 prime numbers are 2,3,5,7,11,13,17,19,23,29,31,37,41

The answer is 14.

2 solutions

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