Baltasar rolls a fair six-sided die three times. What is the

Question

Baltasar rolls a fair six-sided die three times. What is the probability that the product of the three results is a multiple of 4?

(A)  \frac{7}{16} 
(B)  \frac{1}{2} 
(C)  \frac{5}{9} 
(D)  \frac{9}{16} 
(E)  \frac{5}{8}

paragw · Answer

Here's my solution

HarshavardhanR · Answer

Total # of outcomes:

6 x 6 x 6

# of Favorable Outcomes:

Case 1

When all the 3 numbers picked are even numbers

# of ways -> 3 x 3 x 3 (for each of the three picks, any from (2,4,6) could be picked).

Case 2

When even numbers are picked for two of the rolls and the third number is an odd number

# of ways ->

- Odd, even, even -> 3 x 3 x 3
- Even, odd, even -> 3 x 3 x 3
- Even, even, odd -> 3 x 3 x 3

In total -> 3 x (3 x 3 x 3)

Case 3

When "4" is picked in one roll of the dice, and the other two rolls yield odd numbers

# of ways ->

- Odd, even, odd -> 3 x 1 x 3
- Odd, odd, even -> 3 x 3 x 1
- Even, odd, odd -> 1 x 3 x 3

In total -> 3 x (3 x 3 x 1)

Total # of favorable outcomes = Case 1 + Case 2 + Case 3 = (3 x 3 x 3) (1 + 3 + 1) = 5 x (3 x 3 x 3)

The required probability:
 
 \frac{5 * (3 * 3 * 3) }{ (6 * 6 * 6)}  =  \frac{5}{8} .

---
Harsha

kevincan · Answer

In this case, it is easier to find the probability that the product is not a multiple of 4 and subtract that probability from 1.

We can consider two disjoint cases:

Case 1: All three rolls result in odd numbers (1, 3, or 5).

\left(\frac{1}{2}\right)^3=\frac{1}{8}

Case 2: Two rolls result in odd numbers, and the other roll (either the first, second, or third) results in either 2 or 6.

3\left(\frac{2}{6}\right)\left(\frac{1}{2}\right)^2=\frac{1}{4}

Therefore, the probability that the product is not a multiple of 4 is

\frac{1}{8}+\frac{1}{4}=\frac{3}{8}

Hence, the probability that the product is a multiple of 4 is

1-\frac{3}{8}=\boxed{\frac{5}{8}}

yudhautomo · Answer

Asked: Probability of getting an arrangement that will produce a product that is divisible by 4.

Implied constraint: a positive integer is divisible by 4 if it has at least two 2s in its prime factor form.

Translation: 
# of arrangements that is divisible by 4 / # of total arrangements.

Process:
1) Find restricted cases/arrangements --> cases with zero 2s or just one 2 in its prime factorized form.
2) Subtract restricted cases from total possible cases to get allowed cases.
3) Allowed cases / total cases = answer

Process breakdown:

1) Find restricted cases:
a) Odd, Odd, Odd
--> 3*3*3 = 27 arrangements

b) Odd, Odd, 2
--> 3*3*1 = 9 arrangements 
--> 9 arrangements * (3!/2! to cover all possible positions) = 27 arrangements

c) Odd, Odd, 6
--> 3*3*1 = 9 arrangements 
--> 9 arrangements * (3!/2! to cover all possible positions) = 27 arrangements

From a, b, c, total restricted cases = 27*3 = 81 cases

2) Subtract restricted cases from total possible cases to get allowed cases.
Total # of outcome = 6^3 = 216 arrangements

Allowed cases = 216 - 81 = 135 arrangements

3) Allowed cases / total cases = answer
135/216 = 5/8

Baltasar rolls a fair six-sided die three times. What is the

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