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

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

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22 Nov 2010, 23:01
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Question Stats:

71% (02:38) correct 29% (02:59) wrong based on 93 sessions

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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

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Re: There are two set each  [#permalink]

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23 Nov 2010, 03:22
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monirjewel wrote:
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

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

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Re: There are two set each  [#permalink]

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11 Aug 2012, 22:34
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).
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Re: There are two set each  [#permalink]

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12 Aug 2012, 00:46
LalaB wrote:
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).

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.
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Re: There are two set each with the number 1, 2, 3, 4, 5, 6. If  [#permalink]

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22 Jan 2018, 18:05
monirjewel wrote:
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

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,$$15/36$$.
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There are two set each with the number 1, 2, 3, 4, 5, 6. If  [#permalink]

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12 Feb 2019, 16:18
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
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Re: There are two set each with the number 1, 2, 3, 4, 5, 6. If  [#permalink]

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12 Feb 2019, 16:44
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

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Re: There are two set each with the number 1, 2, 3, 4, 5, 6. If   [#permalink] 12 Feb 2019, 16:44
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