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There are two set each with the number 1, 2, 3, 4, 5, 6. If [#permalink]
22 Nov 2010, 22:01

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A

B

C

D

E

Difficulty:

75% (hard)

Question Stats:

48% (02:25) correct
52% (02:34) wrong based on 27 sessions

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?

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Re: There are two set each [#permalink]
23 Nov 2010, 02:22

1

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Expert's post

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

Re: There are two set each [#permalink]
11 Aug 2012, 21:34

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). _________________

Happy are those who dream dreams and are ready to pay the price to make them come true

Re: There are two set each [#permalink]
11 Aug 2012, 23:46

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
[#permalink]
11 Aug 2012, 23:46

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