Probability

[BTGmoderatorRO]

If a number N is chosen at random from the set of two-digit integers whose digits are both prime numbers, what is the probability that N is divisible by 3?

A. 1/3
B. 1/4
C. 9/25
D. 5/16
E. 0

OA is D
What is the best possible way to solve this? An Expert contribution please.

[ErikaPrepScholar]

We want both digits of N to be prime. There are 4 single-digit prime numbers (2, 3, 5, 7). This means that there are 4*4=16 total options for N. We want to find how many of those 16 options are divisible by 3. So our probability will either be something/16 or - if the fraction can be reduced - something/a factor of 16. So we can eliminate A and C right off the bat because a fraction with 16 in the denominator can't be equivalent to a fraction with 3 or 25 in the denominator.

To find which of our are divisible by 3, we need to think about each number individually. We only have 16 options, and we can eliminate answers as we go, so this shouldn't take too long. Let's start with numbers that begin with 2:

22 - no
23 - no
25 - no
27 - yes, 3*9=27

Since we've found at least one option that is divisible by 3, we know that our probability is at least 1/16. This means we can eliminate answer choice E.

Now we're between B and D - a 1/4 probability means that 4 out of 16 options are divisible by 3, while a 5/16 probability means that 5 out of 16 options are. Let's go through our remaining options:

32 - no
33 - yes, 3*11=33
35 - no
37 - no

52 - no
53 - no
55 - no
57 - yes, 3*19=57

72 - yes, 3 *24=72
73 - no
75 - yes, 3*25=75
77 - no

There were 5 options in total that were divisible by 3, so our probability is 5/16, or answer choice D.

[[email protected]]

Hi Roland2rule,

We're told that a number N is chosen at random from the set of two-digit integers whose digits are both PRIME numbers. We're asked for the probability that N is divisible by 3.

To start, we know that each of the two digits in each number must be 2, 3, 5 or 7. The number of two-digit numbers that fit this restriction is (4)(4) = 16, so the correct answer must be some fraction out of 16. If you had to guess at this point, then you could eliminate Answers A and C.

There's a quick way to determine whether a number is divisible by 3: "the rule of 3" - if the DIGITS of a number SUM to a total that is divisible by 3, then that number is divisible by 3... For example:

12 is divisible by 3 because 1+2 = 3 is divisible by 3
25 is NOT divisible by 3 because 2+5 = 7 and 7 is NOT divisible by 3

Given the possible digits of the two-digit number, the following 'pairs' of digits would give us a number that's divisible by 3:

2 & 7... giving us 27 and 72
3 & 3... giving us 33
5 & 7... giving us 57 and 75

5 out of the 16 numbers 'fit' what we're looking for, so the probability is 5/16.

Final Answer: D

GMAT assassins aren't born, they're made,
Rich

[Scott@TargetTestPrep]

If a number N is chosen at random from the set of two-digit integers whose digits are both prime numbers, what is the probability that N is divisible by 3?

A. 1/3
B. 1/4
C. 9/25
D. 5/16
E. 0

OA is D
What is the best possible way to solve this? An Expert contribution please.

The single-digit prime numbers are 2, 3, 5, and 7. So there are 4 x 4 = 16 two-digit integers that can be formed using these 4 digits. Of these 16 integers, only 33, 27, 72, 57, and 75 are divisible by 3. So the probability is 5/16.

Answer: D