If a number N is chosen at random from the set of two-digit integers

Question

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

pushpitkc · Accepted Answer

The 4 digits which are prime numbers are 2,3,5, or 7. N can have 16(4*4) possibilities.

For any number with 2 digits, the sum of the 2 digits must be divisible by 3.

There are 2 possibilities for the 2-digit number to be divisible by 3 
1. When digits differ - (2,7) and (5,7) - 4 such numbers | 2. Same digit - (3,3) - 1 such number

Therefore, the probability that N is divisible by 3 is  \frac{5}{16}  (Option D)

LeoGT · Answer

Prime number: 2,3,5,7
number of 2 digit integers whose digit are both prime: 4A2=12 number
number in set divisible by 3: 27,72, 57,75; 4 number
Probability that N is divisible by 7: 4/12=1/3
assuming 2 number is different=>wrong answer

soleimanian · Answer

N=_ _
there are 4 prime numbers less than 10: 2,3,5,7
therefore the total events equal 4*4=16
to be divisible by 3, sum of the digits of the number must be divisible by 3.
2+7=9
3+3=6
5+7=12
7+2=9
7+5=12

p=5/16
option D

CYp113 · Answer

All possibilities: 4*4 = 16
(23,25,27,32,35,37,52,53,57,72,73,75,22,33,55,77)

27,52,57,72, and 33 are div by 3.

GMATWhizTeam · Answer

Probability (Event) =  \frac{Number of outcomes favourable to the event }{ Total possible outcomes of the random experiment} .

A number N is chosen at random from a set of two-digit integers whose digits are both prime number.

Now, the only primes which are single-digit values are 2,3,5 and 7. Therefore, the set consists of all 2-digit numbers which can be made of these digits.

The tens’ digit of the required number can be filled in 4 ways. Since we want all the 2-digit numbers that can be formed from these digits, every digit can be repeatedly used; therefore, the units’ digit of the number can also be filled in 4 ways.

Number of 2-digit numbers which can be made of these digits = 4 * 4 = 16. The denominator has to be 16 or any of its factors. Neither 3 nor 25 are factors of 16; so there is no way of obtaining these in the denominator and hence answer options A and C can be eliminated.

75 is the example of a number which is formed using primes and is divisible by 3; this means that the favourable outcomes are not Zero and hence the probability is not zero, so answer option E can be eliminated.

A number is divisible by 3 if the sum of its digits is divisible by 3. Of the digits that we have, the sum of digits can be 6, 9 or 12.

The number that gives the sum of the digits as 6 is the number 33.
The numbers that give the sum of the digits as 9 are the numbers 27 and 72.
The numbers that give the sum of the digits as 12 are the numbers 57 and 75.

In total, of the 16 numbers that were formed, there are 5 numbers which are multiples of 3. 
Therefore, the required probability =  \frac{5 }{ 16} .

The correct answer option is D.

pikachu9 · Answer

if we apply combination formula then outcome will be 4C2=6 . why it is not correct

Bunuel · Answer

You’re mixing up two different situations here. 4C2 = 6 counts the number of ways to choose 2  different  digits out of 4 when repetition is not allowed and order doesn’t matter. But in the question, digits are allowed to repeat and order matters (for example, 2-2, 3-3, 5-5, 7-7 are all valid, and 2-3 is different from 3-2). That makes it 4 choices for the first digit * 4 choices for the second digit = 16 total outcomes, not 6.

If a number N is chosen at random from the set of two-digit integers

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If a number N is chosen at random from the set of two-digit integers

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