How to guess the least number of spheres in a problem involving taking out objects from a jar?

Question

The problem is as follows:

A porcelain jar has inside $20$ spheres labeled with integer numbers from $1$ to $20$. How many spheres at least do it has to be taken out of the jar randomly to be certain that the sum of two of them is a prime number less than $20$?

This situation I also found lost during exams as I don't know where to begin. I sketched the situation as this:

The spheres are labeled from $1$ to $20$

$1,2,3,4....20$

and they are asking the sum is prime so, to have the least number or spheres I can consider the worst case scenario which would mean that each time I take out a sphere I will get a non prime number so their sum is also a non prime.

The non prime numbers in that range are

$4,6,9,10,12,14,15,16,18,20$

So the least number is $10$?

I feel that the answer is incomplete or not correct. Can somebody help me to solve this situation in a more logical way without guessing?

I'd like to learn what is it behind this probability question. Am I on the right track.

What can I conclude to get a prime number, the sum of two primes is also a prime (seems logical, but is it true?). The sum of two non primes is a non prime also logical, but can I conclude that?.

I need help to clear these ideas.

Answer 1

You are asked for the number you need to draw to make sure the sum of two of them is a prime less than $20$, but answered the number you need to draw to make sure you get a single sphere that is a prime less than $20$. For the question you answered, you missed that $1$ is not prime and you need to draw one more to be the prime, so you need to draw $12$.

For drawing numbers such that there is not a pair that sums to a prime less than $20$, you want all the big numbers you can get. You can draw $20$ and $19$ without fear that you will get a prime less than $20$. In fact you can draw $10$ through $20$ without getting a sum less than $20$, let alone a prime. Having done that, any number you draw will let you get a sum of $19$, so you need to draw at least $12$ balls. This is not a proof that $12$ is the answer, but it is a lower bound.

To prove that $12$ is an upper bound, pair up $n$ and $19-n$ for $n \in [1,9]$. There are nine pairs here, each summing to $19$, so you can only pull one of each pair and avoid two numbers summing to a prime less than $20$. You can add $19$ and $20$ to make $11$ numbers as the maximum you can have without having a sum of $19$, so drawing $12$ numbers guarantees there is a pair summing to a prime less than $20$.

Answer 2

It appears the solution we want is the least number of draws to be absolutely certain that we have any two spheres that add up to a prime number. And also, note that we can have two non primes add up to a prime number, for example $9+8=17$

Also see that two primes can be added together to make a non prime, like $13+3=16$

To figure out this problem, we need to look at the different possible combinations of spheres and see if every possible combination of a certain size $k$ contains two numbers that add up to a prime less than $20$. And any time we can find a combination of size $k$ that doesn't fit the criteria, then we go up to the next size $k+1$.

For example, see that we definitely need to pick more than $3$ spheres if we want to guarantee we follow the criteria because if we picked the combination $\{3,5,7\}$ we can't take any two of those numbers and add them up to a prime number less than $20$.

A computer program of some sort may be needed to successfully solve this problem as writing out all the possible combinations by hand would be tedious. We can confirm that the number of spheres picked definitely needs to be more than $3$ (as shown above) and of course it can't be over $20$.

Ross Millikan gives a great answer as well (+1 to that) but I hope both of our answers together helps you understand the problem better