Set of numbers

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alex.gellatly: Master | Next Rank: 500 Posts; Posts: 435; Joined: Wed Nov 16, 2011 7:27 amFollowed by: 16 members; Thanked: 48 times

by alex.gellatly » Mon Jul 30, 2012 11:55 pm

This is from 300+ Math questions and best solutions:

6. 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?

[spoiler]Solution:
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.[/spoiler]

I was wondering if there was a different way than simply picking favorable outcomes. Also.. I am a little confused why we count both (2,4) and (4,2), but then only (2,2) once. Shouldn't we count (2,2) twice from the same logic we count (2,4) twice?

Thanks

A useful website I found that has every quant OG video explanation:

https://www.beatthegmat.com/useful-websi ... tml#475231

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Source: Beat The GMAT — Problem Solving | Mon Jul 30, 2012 11:55 pm

armand_h: Junior | Next Rank: 30 Posts; Posts: 15; Joined: Wed Jul 25, 2012 9:02 am; Thanked: 1 times

by armand_h » Tue Jul 31, 2012 12:45 am

P(x4) = 1- P(not x4)
P (not x4) is equal to:
probability that the 1st number is odd and the second is not 4
or
the 1st number is even (but not 4) and the second is odd
=3/6*5/6 + 2/6*3/6
=21/36

P(x4)=1-21/36=15/36

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eagleeye: Legendary Member; Posts: 520; Joined: Sat Apr 28, 2012 9:12 pmFollowed by: 49 members; Thanked: 339 times; GMAT Score:770

by eagleeye » Wed Aug 01, 2012 10:18 am

alex.gellatly wrote:This is from 300+ Math questions and best solutions:

6. 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?

[spoiler]Solution:
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.[/spoiler]

I was wondering if there was a different way than simply picking favorable outcomes. Also.. I am a little confused why we count both (2,4) and (4,2), but then only (2,2) once. Shouldn't we count (2,2) twice from the same logic we count (2,4) twice?

Thanks

Yes, we can do it without counting the individual cases:

The product will be a multiple of 4 in two cases:

1. Both numbers are even. No. of such cases = 3(choices of even for 1st)*3(choices of even for second) = 3*3 = 9
2. One of the numbers is odd, and other is a 4. Number of such cases = 2(selecting which set is the 4 chosen from)*1 (the choice for 4)*3(the choice for odd numbers) = 2*1*3 = 6

Hence probability = (9+6)/(6*6) = 15/36 = 5/12.

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