There are two set each with the number 1, 2, 3, 4, 5, 6. If

Question

There are two set each with the number 1, 2, 3, 4, 5, 6. If randomly choose one number from each set, what is the probability that the product of the 2 numbers is divisible by 4?

(A) 4/5
(B) 3/7
(C) 8/9
(D) 15/35
(E) 15/36

Bunuel · Accepted Answer

Let's count the probability of the opposite event and subtract it from 1.

The product of 2 integers won't be be a multiple of 4 if both are odd,  P(odd, \ odd)=\frac{1}{2}*\frac{1}{2}=\frac{1}{4} ,  or  one is odd and another is even but not multiple of 4 (note we'll have two cases odd from 1st set and even but not 4 from 2nd and vise-versa),  P(odd, \ even \ but\ not \ 4)=\frac{1}{2}*\frac{2}{6}+\frac{2}{6}*\frac{1}{2}=\frac{1}{3} .

So,  P=1-(\frac{1}{4}+\frac{1}{3})=\frac{5}{12}=\frac{15}{36} .

Answer: E.

LalaB · Answer

could you please explain me the following -

I found this solution in the internet-
Picking 2 numbers from each set : 6c1*6c1=36
Favorable outcomes = 15
(1,4)(2,2),(2,4),(2,6)(3,4)(4,1),(4,2),(4,3),(4,4),(4,5),(4,6)(5,4)(6,2),(6,4),(6,6)

Therefore Required probability = 15/36

my question is why do we take into account ,say, both (2;4) and (4;2)? I understand that they are different,since they are taken from dif.sets, but their products are the same. (always have a problem with such kind of things).

EvaJager · Answer

The same as you count 6C1*6C1: order matters, you count (2,4) and (4,2) as different pairs. Then you should also distinguish between them when considered for their product. You count favorable pairs and not only product values. They give the same product, but they come from different sources.

rahul16singh28 · Answer

Total Favorable outcomes =  15 
 (1,4)(2,2),(2,4),(2,6)(3,4)(4,1),(4,2),(4,3),(4,4),(4,5),(4,6)(5,4)(6,2),(6,4),(6,6) .. Total sample space is  36 . hence, 15/36 .

energetics · Answer

Choose x from {1,2,3,4,5,6} 
 and y from {1,2,3,4,5,6}.
When is xy/4?

Denominator total cases = 6*6 = 36

Numerator, we can check each value, Using 1-x to find the NO cases (1 has 5 possibilities, 2 has 3, 3 has 5, 4 has 0, 5 has 5 and 6 has 3). Can see a pattern here, the ODD# can take any number besides 4, the EVEN# besides 4 can only take other odd numbers, and 4 takes 0 numbers.
So, 5+3+5+0+5+3 = 21 where it's not a multiple of 4.
1 - 21/36 = 15/36

MartyTargetTestPrep · Answer

Question: There are two set each with the number 1, 2, 3, 4, 5, 6. If randomly choose one number from each set, what is the probability that the product of the 2 numbers is divisible by 4?

Since we are randomly choosing 1 number from each of 2 sets of 6 numbers, the total number of outcomes is 6 x 6 = 36.

In order for the product of the pair of numbers that we choose to be divisible by 4, the pair has to have either of the following sets of characteristics.

- It includes 2 even numbers. (The prime factors of a product of 2 even numbers will always include (2, 2). Thus, the product of any two even numbers is a multiple of 4.)

- It includes an odd number and 4.

First calculate the total number of possible pairs that include 2 even numbers.

Number of even numbers: 3 per set

Total number of ways to a pair of even numbers: 3 x 3 = 9

Now calculate the total number of pairs that include 4 and an odd number.

The number 4 from the first set can go with 3 odd numbers from the second set: 3

The number 4 from the second set can go with 3 odd numbers from the first set: 3

Total number of pairs that include 4 and an odd number: 3 + 3 = 6

Total number of pairs the product of which is a multiple of 4: 9 + 6 = 15

Favorable Outcomes/Total Outcomes = 15/36

The correct answer is (E).

Basshead · Answer

The question asks for the probability that 2 integers are divisible by 4.

We know: 
- The product of two even integers will be divisible by four.
- Any multiple of four will be divisible by four (1 * 4 = 4, 2 * 4 = 8, etc)

Thus, the only way for the product of 2 numbers from 1-6 to NOT be divisible by 4 is if they don't fall into the above categories. If the two integers are odd or if one integer is odd and the other is even but not a 4, then the product will not be divisible by 4.

P(odd) * P(odd) = 3/6 * 3/6 = 1/4
P(odd) * P(even but not 4) = 3/6 * 1/3 = 1/6
Since the order does not matter here we multiply 1/6 by 2C1 = 1/3

1/4 + 1/3 = 7/12

1 - 7/12 = 5/12 =  15/36

navderm · Answer

Let's first calculate the total number of possible combinations.  We know that it is 6C1 * 6C1 = 36.

Now,  instead of calculating the numerator (total number of combinations of two numbers divisible by 4), look at the answer choices.  We can see that possible answers are:
C) 32/36 (multiplying by 4 in both numerator and denominator)
E) 15/36. 
 No other answer choice has a factor of 36 in the denominator.

And we know the numerator cannot be 32 because that would be way too many possible combinations divisible by 4.

Hence E.

MathRevolution · Answer

Each set has '6' numbers. Picking '1' number each from sets:  ^6{C_1} * ^6{C_1}  = 6 * 6 = 36

Product of '2' numbers selected divisible by 4

=> (1,4) = 2 arrangements (1,4) or (4,1)
=> (2,2) = 1 arrangement
=> (2,4) = 2 arrangements (2,4) or (4,2)
=> (2,6) = 2 arrangements (2,6) or (6,2)
=> (3,4) = 2 arrangements (3,4) or (4,3)
=> (4,4) = 1 arrangement
=> (4,5) = 2 arrangements (4,5) or (5,4)
=> (4,6) = 2 arrangements (4,6) or (6,4)
=> (6,6) = 1 arrangement

Total: 15

Probability:  \frac{15}{36}

Answer E

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