probability

[sud21]

If one number x between 2,3,4,5,6 and another number y between 2,3,4,5,6 are selected at random, what is the probability that xy is divisible by 4?

[LalaB]

X 2,3,4,5,6
Y 2,3,4,5,6

X*Y-
2*2
2*4
2*6
3*4
4*2
4*3
4*4
4*5
4*6
5*4
6*2
6*4
6*6
total of XY=8

answ-8/25

[ronnie1985]

The question asks how many ways one can select a 2^2 from the 2 given sets of no.
1. 2 or 6 from each set
2. 4 from any one set

P = P(1)+P(2) = 2C1*2C1/(5C1*5C1)+1C1*5C1/(5C1*5C1)*2 = 4/25+10/25 = 14/25

[Whitney Garner]

The question asks how many ways one can select a 2^2 from the 2 given sets of no.
1. 2 or 6 from each set
2. 4 from any one set

P = P(1)+P(2) = 2C1*2C1/(5C1*5C1)+1C1*5C1/(5C1*5C1)*2 = 4/25+10/25 = 14/25

Very close, but you need to subtract for double counting a 4 in BOTH. If we break the possibilities into groups, it might be easier to see:

I'll list possibilities in the following notation: {set 1 choice}[set 2 choice(s)]

{2}[2,6] = 2 outcomes
{6}[2,6] = 2 outcomes
{4}[2,3,5,6] = 4 outcomes
{2,3,5,6}[4] = 4 outcomes
{4}[4] = 1 outcome

13 outcomes we want out of a total (5*5)=25 choices.

The probability of divisibility by 4 is [spoiler]13/25[/spoiler].

Hope this helps!

Whit

[LalaB]

Whitney Garner, I have counted out these 5 multiplications (please see above), since I thought they were double counted.dont u think that 13/25 is x*y and y*x ?

[Whitney Garner]

Whitney Garner, I have counted out these 5 multiplications (please see above), since I thought they were double counted.dont u think that 13/25 is x*y and y*x ?

Hi LalaB!

That is a GREAT question and one that people get stuck on all the time! Okay, so we have 2 distinct sets, X and Y...but the confusion is that they each contain the same numbers. I think that is what makes you feel like the X*Y for 2*4 and 4*2 might be the same thing. But they are actually DIFFERENT. Let me ask you this. If I were sitting in front of you with 2 bowls of number tiles - an X bowl and a Y bowl. If I reached my hand into the X bowl and pulled out a 2, would that be a different outcome than reaching into the same bowl and pulling out a 4? It actually doesn't really matter what I do with the Y bowl, if I pull out different numbers from the X bowl, it is, by definition a NEW outcome. So (X=2,Y=4) is DIFFERENT than (X=4,Y=2).

It also might be easier to think of this issue using something that has real value. Let's pretend that you are X and I am Y, and we are making desserts for an event (jello, pie, cake, muffins, pudding). Now let's consider 2 cases:

You make the pie, and I make muffins.
vs.
You make muffins and I make the pie.

Sure, muffins and pie are both going to end up at the event, but these are different events for you and I.


But let's say we both have to make muffins. There is no way to "reverse" the choice to get a NEW outcome.

You make muffins and I make muffins
vs.
You make muffins and I make muffins....

These are the same, therefore we cannot count them twice!!

So back to our earlier example, in the 2*4 and 4*2 case, sure, these do have the same product, but we got there via different routes. In the first, we selected the 2 from X and the 4 from Y. This is a different choice than selecting the 4 from X and the 2 from Y (albeit a seemingly silly choice, but a choice nonetheless!)

So your complete list (NOT throwing out the entries in bold) would be the correct summary of all outcomes that would make us happy. Therefore, 13 different outcomes are great, out of 25 total, so we have a probability of success of 13/25.

I hope this helps clear it up!

Whit

[LalaB]

oh, yes, u r right! I have messed everything up. this question is like "two dice are thrown. what is the probability that their sum is 5 or 7" (https://www.beatthegmat.com/probablity-q ... 77197.html)
the same logic is used there.

thnx a lot:)

[pemdas]

another GMAT question testing more than one concept: probability+number properties

sets x and y are the same and contain primes factorized as following {2,3,2*2,5,2*3} we need divisibility by 4, hence two 2s must sit there at least. There are 5*5 total counts here or 25 outcomes. The number of favorable outcomes will be 3*3 Plus the number 4 is divisible in itself (when multiplied by 3 and 5) and will add up, making the number of favorable outcomes (3*3 + 4) The required probability is 13/25

If one number x between 2,3,4,5,6 and another number y between 2,3,4,5,6 are selected at random, what is the probability that xy is divisible by 4?

[gmatdriller]

Whitney Garner, I have counted out these 5 multiplications (please see above), since I thought they were double counted.dont u think that 13/25 is x*y and y*x ?

A: (2, 3, 4, 5, 6)

B: (2, 3, 4, 5, 6)



picking a 2(set A) with 4(set B) is different from picking a 4(set A) with 2(set B)
though outcome is the same.


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