A math teacher assigns a distinct prime number, starting with the lowest, for each student in her class. If she chooses two students at random, the probability the sum of their numbers is not even is less than 1/10. What is the fewest possible number of students in her class?

(A) 19 (B) 20 (C) 21 (D) 71 (E) 73

Explanation:

A good way to approach this question is by back solving. That is, take one of the answer choices and “put it back” into the question. For instance, let’s assume the answer is (B) 20, which is a nice easy number to work with. If there are 20 students in the classroom, one of them has to be assigned the number 2. This is important because ‘2’, the only even prime, is the only way to obtain an odd number when summing two primes (Odd + Even = Odd).

The chances of choosing that student given 20 total students, 19 of whom have odd prime numbers, can be described as the following:

# of desired outcomes # of total outcomes

The one student with an even number can be paired with 19 other students, giving us 19 in the numerator.

The total number of pairings can be found by using combinations, not permutations, since we don’t want to count the same student twice (Student 1 and Student 2 in a group is the same as Student 2 and Student 1 in a group).

20!/18!2! = 190 = # of total outcomes

19/190 = 1/10

We know that the odds have to be less than 1/10. By adding one more student the odds the probability would have to dip ever so slightly below 1/10, since the more students you have in the class the less likely you’ll select the one and only student who has the even prime number, 2.

Therefore, by adding that one extra student we will have a probability that is slightly less than 1/10, meaning that 21 is the fewest number of students we can have in the class. Answer: (C).

I should mention one more thing: the question doesn’t specify whether the primes after the lowest prime, the number 2, are in specific order. But since we are only dealing with the odd and even properties of prime numbers, and every prime number after ‘2’ is odd, it doesn’t matter which primes those are.

I adopted the direct approach to tackling this problem Say there are n students in the class students : 1 2 3 4 5…. prime no. assigned : 2 3 5 7 11…. no, of ways to select 2 students from n students is nC2 Since the sum of the selected prime nos. has to be odd. 1 no. has to be even and the other odd. The only even prime no. is 2. Since we have 1 prime number and hence student, fixed. we need to select 1 additional student from n-1 remaining students. That yields (n-1)C1 Thus (n-1)C1/nC2 < 1/10 on further solving 2/n 20 so the minimum no. greater than 20 is 21.

I am confused by the explanation as well. Can you break it down a bit more? Where did 20!/18!2! = 190 = # of total outcomes come from? Specifically, the 18!2!. I don’t find these twisters particularly helpful if they aren’t broken down and showing all work. Thanks.

Magoosh Test Prep ExpertFebruary 8, 2016 at 9:43 am#

Hi Whitney 🙂

I’m happy to explain how we get 20!/(18!2!) in this problem. As Chris mentions, this is a combinations question, as the order in which we choose the two students doesn’t matter. So, we can use the combinations formula to determine the total number of possible outcomes:

n!/(n-r)!(r!)

where n is the total number of items and r is the number of items to be selected.

In this problem, there are 20 students (n=20) and we are going to select 2 of these students (r=2):

So i got little bit confused about this . I have used another approach and its not matching with yours , I am not able to understand where i am wrong . Lets say there are 20 Numbers then Probability of choosing first even number is 1/20 and the probability of choosing second odd number is 19/19. So the probability will be 1/20*19/19 = 1/20 and not 1/10.

I think the only difference is that i am counting twice by looking at your solution , but how ?

Magoosh Test Prep ExpertFebruary 8, 2016 at 9:34 am#

Good question! For this problem, it’s important to recognize that the order in which we choose the two students doesn’t matter. In other words, this is a combinations question in which we need to choose two students: (1) the student assigned 2 and (2) another student in the class. There are two ways we can do this:

1. First we select the student assigned 2 and then another student

or

2. First we select a student not assigned 2 and then the student assigned 2.

The probability of the first case occurring, as Gaurav mentioned, is 1/20*19/19 = 1/20. And the probability of the second case is 19/20*1/19 = 1/20. To find the total probability of selecting the desired pair of students, we add the probabilities of the two cases together:

1/20 + 1/20 = 2/20 = 1/10

And that’s the same probability that Chris got when solving this question 🙂 Hope this clears up your doubts!

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I adopted the direct approach to tackling this problem

Say there are n students in the class

students : 1 2 3 4 5….

prime no. assigned : 2 3 5 7 11….

no, of ways to select 2 students from n students is nC2

Since the sum of the selected prime nos. has to be odd. 1 no. has to be even and the other odd. The only even prime no. is 2. Since we have 1 prime number and hence student, fixed. we need to select 1 additional student from n-1 remaining students. That yields

(n-1)C1

Thus (n-1)C1/nC2 < 1/10

on further solving 2/n 20

so the minimum no. greater than 20 is 21.

Love u magoosh !!! u make GRE very easy

I am confused by the explanation as well. Can you break it down a bit more? Where did 20!/18!2! = 190 = # of total outcomes come from? Specifically, the 18!2!. I don’t find these twisters particularly helpful if they aren’t broken down and showing all work. Thanks.

Hi Whitney 🙂

I’m happy to explain how we get 20!/(18!2!) in this problem. As Chris mentions, this is a combinations question, as the order in which we choose the two students doesn’t matter. So, we can use the combinations formula to determine the total number of possible outcomes:

n!/(n-r)!(r!)

where n is the total number of items and r is the number of items to be selected.

In this problem, there are 20 students (n=20) and we are going to select 2 of these students (r=2):

n!/(n-r)!(r!) –> 20!/(20-2)!(2!) = 20!/(18!)(2!)

For a more in-depth review on combinations, definitely check out this post from our GRE blog on the topic: GRE Math: Combinations and Permutations

Hope this helps 🙂

Hey Chris,

So i got little bit confused about this . I have used another approach and its not matching with yours , I am not able to understand where i am wrong . Lets say there are 20 Numbers then Probability of choosing first even number is 1/20 and the probability of choosing second odd number is 19/19. So the probability will be 1/20*19/19 = 1/20 and not 1/10.

I think the only difference is that i am counting twice by looking at your solution , but how ?

I have the exact same doubt! Help 🙁

Good question! For this problem, it’s important to recognize that the order in which we choose the two students doesn’t matter. In other words, this is a combinations question in which we need to choose two students: (1) the student assigned 2 and (2) another student in the class. There are two ways we can do this:

1. First we select the student assigned 2 and then another student

or

2. First we select a student not assigned 2 and then the student assigned 2.

The probability of the first case occurring, as Gaurav mentioned, is 1/20*19/19 = 1/20. And the probability of the second case is 19/20*1/19 = 1/20. To find the total probability of selecting the desired pair of students, we add the probabilities of the two cases together:

1/20 + 1/20 = 2/20 = 1/10

And that’s the same probability that Chris got when solving this question 🙂 Hope this clears up your doubts!

nice and clear explanation….thanks