12 Days of Christmas GMAT Competition - Day 10: Peter has two fair six

Question

12 Days of Christmas 🎅 GMAT Competition with Lots of Questions & Fun

Peter has two fair six-sided dice. One is labeled with the first six prime numbers, and the other with their reciprocals. If he rolls both dice once, what is the probability that the product of the resulting numbers is NOT a terminating decimal? (A terminating decimal is any decimal that has only a finite number of nonzero digits. For example, 24, 0.82, and 5.096 are three terminating decimals.)

A. 1/36
B. 1/6
C. 5/9
D. 2/3
E. 5/6

Bunuel · Accepted Answer

GMAT Club Official Explanation:

Peter has two fair six-sided dice. One is labeled with the first six prime numbers, and the other with their reciprocals. If he rolls both dice once, what is the probability that the product of the resulting numbers is NOT a terminating decimal? (A terminating decimal is any decimal that has only a finite number of nonzero digits. For example, 24, 0.82, and 5.096 are three terminating decimals.)

A. 1/36
B. 1/6
C. 5/9
D. 2/3
E. 5/6

A reduced fraction  \frac{a}{b}  (meaning that the fraction is already in its simplest form, so reduced to its lowest term) can be expressed as a terminating decimal if and only if the denominator  b  is of the form  2^n5^m , where  m  and  n  are non-negative integers. For example:  \frac{7}{250}  is a terminating decimal  0.028 , as the denominator  250  equals  2*5^3 . The fraction  \frac{3}{30}  is also a terminating decimal, as  \frac{3}{30}=\frac{1}{10}  and the denominator  10=2*5 .
 
 Note that if the denominator already consists of only 2s and/or 5s, then it doesn't matter whether the fraction is reduced or not. 
 
For example,  \frac{x}{2^n5^m} , (where  x ,  n , and  m  are integers) will always be a terminating decimal.
 
(We need to reduce the fraction in case the denominator has a prime other than 2 or 5, to see whether it can be reduced. For example, the fraction  \frac{6}{15}  has 3 as a prime in the denominator, and we need to know if it can be reduced.)
 
Thus, for the product of the numbers not to be a terminating decimal, its denominator, after reduction, must have primes other than 2 and 5.

We have two dice, one is labeled with the first six prime numbers: 2, 3, 5, 7, 11, and 13, and the other with their reciprocals: 1/2, 1/3, 1/5, 1/7, 1/11, and 1/13.

If the die with the reciprocals shows 1/2, or 1/5, then the product will always be a terminating decimal, regardless of the result on the second die.

If the die with the reciprocals shows 1/3, then the die with primes should show 2, 5, 7, 11, or 13, for the product not to be a terminating decimal. The probability of that is 1/6*5/6.
 
  If the die with the reciprocals shows 1/7, then the die with primes should show 2, 3, 5, 11, or 13, for the product not to be a terminating decimal. The probability of that is 1/6*5/6.
 
  If the die with the reciprocals shows 1/11, then the die with primes should show 2, 3, 5, 7, or 13, for the product not to be a terminating decimal. The probability of that is 1/6*5/6.
 
  If the die with the reciprocals shows 1/13, then the die with primes should show 2, 3, 5, 7, or 11, for the product not to be a terminating decimal. The probability of that is 1/6*5/6.

Therefore, the overall probability is 4*1/6*5/6 = 5/9.

Answer: C.

limuengaoey · Answer

the first dice has face labels as 2,3,5,7,11,13
the second dice hac face labels as 1/2,1/3,1/5,1/7,1/11,1/13

we would like to get not terminating decimal
if the first dice we get 2 prob is 1/6
the second dice must be 1/3,1/7,1/11,1/13 prob 4/6
hence 1/6*2/3 = 2/18 =1/9

if the first dice is 3 prob =1/6
2nd dice must be 1/7,1/11,1/13 = 3/6
hence 1/6*1/2 = 1/12

if the first dice is 5 prob = 1/6
2nd dice must be 1/3,1/7,1/11,1/13 = 4/6 = 1/9

if the first dice is 7 prob =1/6
2nd dice must be 1/3,1/11,1/13 = 3/6
hence 1/6*1/2 = 1/12

if the first dice is 11 prob =1/6
2nd dice must be 1/3,1/7,1/13 = 3/6
hence 1/6*1/2 = 1/12

if the first dice is 13 prob =1/6
2nd dice must be 1/3,1/7,1/11 = 3/6
hence 1/6*1/2 = 1/12

hence all is 5/9 --> c

sj296 · Answer

If we get 
2 on one then we should get 1/3 or 1/7 or 1/11 or 1/13 with probability= 1/6*4/6

Similarly, for 3 the prob = 1/6 * 3/6

For 5, prob = 1/6 * 4/6

For 7, prob = 1/6 * 3/6

For 11, prob = 1/6 * 3/6

For 13, prob = 1/6 * 3/6

Total prob = 20/36 = 5/9

Option C

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andreagonzalez2k · Answer

First six prime numbers: 2, 3, 5, 7, 11, 13
Reciprocals of the first six prime numbers: 1/2, 1/3, 1/5, 1/7, 1/11, 1/13

1/3, 1/7, 1/11, 1/13 are NOT a terminating decimal.

If we multiply 1/3 by an integer which is not a multiple of 3 we obtain another NOT a terminating decimal.
Similarly with 1/7, 1/11 and 1/13.

So we have:
- 1/3 with 2, 5, 7, 11, 13: 5 cases
- 1/7 with 2, 3, 5, 11, 13: 5 cases
- 1/11 with 2, 3, 5, 7, 13: 5 cases
- 1/13 with 2, 3, 5, 7, 11: 5 cases

Total favorable cases = 5*4 = 20

Total cases = 6x6 = 36

Probablity = 20/36 = 5/9

IMO C

$!vakumar.m · Answer

Peter has two fair six-sided dice. One is labeled with the first six prime numbers, and the other with their reciprocals. If he rolls both dice once, what is the probability that the product of the resulting numbers is NOT a terminating decimal?

the first six prime numbers are - 2, 3, 5, 7, 11, 13
their reciprocals are - 1/2, 1/3, 1/5, 1/7, 1/11, 1/13
If both dice are rolled once -  we have 36 ways to get two numbers and their products. 
For products with non terminating decimals - the denominator should not consist of 2 or 5 or both.

In this case, we have  20 ways of obtaining a product with non terminating decimals  -
if one dice has 2 or 5 - the other dice with fractions 1/3, 1/7, 1/11, 1/13 results in 4 non terminating decimals - total 8 ways.
if one dice has 3 (or 7, 11 , 13) - the other dice with fractions 1/7, 1/11, 1/13 results in 3 non terminating decimals - hence total of 12 ways.
desired probability = 20/36 = 5/9
Answer C

Adarsh_24 · Answer

Probability = 36-(terminating)/36
Terminating if perfectly divisible or denominator has 2 or 5 while numerator doesn’t have multiples of 2 and 5.
Terminating= 6(eg:2/2,3/3,5/5,..)+5(eg:3/2,5/2,..)+5(eg:2/5,3/5,7/5,...)
=36-16 / 36 =5/9

zerowing · Answer

With 2 dice, we have 6*6 = 36 total possible pair

For terminating decimal we have 2 scenarios:
1) If the results include 1/2 and 1/5: 2*6 = 12 
2) If the results matching with their reciprocal: 3*1/3, 7*1/7, 11*1/11, 13* 1/1 = 4 total

So there is 16 way for the results to be terminating => non-terminating 36-16 = 20

%Prob = 20/36 = 5/9

12 Days of Christmas GMAT Competition - Day 10: Peter has two fair six

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12 Days of Christmas GMAT Competition - Day 10: Peter has two fair six

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