If four fair six-sided dice are rolled, what is the probability that

Question

If four fair six-sided dice are rolled, what is the probability that, for each pair of dice, the product of the two numbers rolled on those dice is a multiple of 4?

A. 1/108

B. 1/32

C. 1/16

D. 31/432

E. 401/432

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chetan2u · Answer

Every dice has 6 sides with even number on three sides. Cases for the product to be multiple of 4 under given condition.

(1) Every dice shows an even number : Here each pair will be multiple of two even number, so a product of 4.
Ways = 3*3*3*3 = 81
(2) One dice shows odd while other three show a multiple of 4 : Here each pair will be a multiple of either two even number or one odd & other multiple of 4, so a product of 4. 
Ways = 3*4 ( The odd number could come on any of the 4 dice and the odd number could be any of the three digits.

Total ways = 81+12 = 93
Ways without restrictions = 6*6*6*6

Hence P= \frac{93}{6*6*6*6}=\frac{31}{432}

D

ashishgarg0027 · Answer

Hi  Bunuel ,  chetan2u ,

I approached this problem differently but didn't get the answer. Can you help me identify flaws in my approach?

4 dice are rolled. Number of pairs we can make out of 4 dices = 4C2 = 6.
Now, each pair can get a multiple of 4 in 15 different ways.  
Now, for each pair, to get a multiple of 4, probability = 15/36 = 5/12.
For 6 pairs, probability = (5/12)^6

This is the answer that i am getting. Let me know where i am making a mistake!

Thanks,
Ashish Garg

Kinshook · Answer

Asked: If four fair six-sided dice are rolled, what is the probability that, for each pair of dice, the product of the two numbers rolled on those dice is a multiple of 4?

Multiple of 4 can be achieved by multiplication of 2 even numbers or with one number as 4.
But for each pair of dice, the product of the two numbers rolled on those dice is a multiple of 4 then

Total ways = 6*6*6*6 = 1296

Case 1: number of each dice is even.
Number of favorable cases = 3*3*3*3 = 81

Case 2: 3 dice have 4 and number on other dice is odd.
Number of favorable cases = 4C3*1*1*1*3 = 12

Total number of favorable cases = 81 + 12 = 93

The probability that, for each pair of dice, the product of the two numbers rolled on those dice is a multiple of 4 = 93/1296 = 31/432

IMO D

chetan2u · Answer

Ashish,
You are having repetitions in your calculations. 
When we say any pair out of four means the events are dependent. But by taking each pair separately, you are taking them independent events.

Take an example with smaller set

Three coins have 1 and 2 written on opposite sides. Same question, ways pair will get a multiple of 4.

Only one way: when all show 2. =>  P=\frac{1}{2^3}

By taking each pair individually: (1,1),(1,2),(2,1),(2,2) so one way out of four=> Each pair probability =1/4, so three pairs will give  (\frac{1}{4})^3

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If four fair six-sided dice are rolled, what is the probability that

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