M23-22

Official Solution:

If two different numbers are randomly selected from set \(\{1, 2, 3, 4, 5\}\), what is the probability that the sum of the two numbers is greater than 4?

A. \(\frac{3}{5}\)
B. \(\frac{7}{10}\)
C. \(\frac{4}{5}\)
D. \(\frac{9}{10}\)
E. \(\frac{19}{20}\)


Let's find the probability of the opposite event and subtract that value from 1. The opposite event would be if we choose 2 different number so that their sum will be less or equal to 4.

The number of total outcomes is \(C^2_5=10\) (choosing 2 different numbers from the set of 5 different numbers);

The number of outcomes when \(sum\leq{4}\) is 2: only (1,2) and (1,3);

\(P=1-\frac{2}{10}=\frac{4}{5}\).


Answer: C
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Thank you for this great questions, gmat club is really good!
This said,
why do I have only 2 possible outcomes? (1,2) and (1,3).. why doesn't (2,1) and (3,1) work? I can randomly select them in this order as well?

By definition a subset contains only different elements of the subset. So {1,2} and {2,1} is the same subset.

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The answer is wrong
\(P=1-\frac{2}{10}=\frac{8}{5}\). and not\(\frac{4}{5}\)

Really? Check again, please.
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You are right its \(\frac{4}{5}\)

How is it not B =7/10?

When you consider the case , how many pairs of numbers whose sum is greater than 4, you get

(1,4),(1,5).(2,4),(2,5),(2,3),(3,4),(3,5),(4,5) which is 7 outcomes out of total possible 10 outcomes.

Please explain.

It's not that hard to do the following:
1 - (1,4)
2 - (1,5)
3 - (2,4)
4 - (2,5)
5 - (2,3)
6 - (3,4)
7 - (3,5)
8 - (4,5)
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Only (1,2) and (1,3) are less than or equal to 4. The required probability is 1-(2/5C2). The answer is 4/5.
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Bunuel can you please explain why the order does not matter?

A quick way to test whether or not order matters is to try making the same selection in two different orders to see if you get a new result.

For example, if I am making a 3-person group from A, B, C, D, and E and I select A, then B, then C, I get the same group as when I select C, then B, then A. Same result = order doesn't matter = combination.

On the other hand, if I am selecting 1st, 2nd, and 3rd place from A, B, C, D, and E, and I select A, then B, then C, I get a different order than when I select C, then B, then A. Different result = order does matter = permutation.

Trying that here, if I am picking 2 numbers from {1,2,3,4,5}, and I pick 1, then 2, I get the same sum (3) as when I pick 2, then 1. Same result = order doesn't matter = combination.

While there are a few key words that indicate whether or not order matters, we don't always get them, so the best way to figure it out is to test it yourself.

the answer calculated is based on assumption greater than or equal to 4 not only greater than 4.

If only greater than 4 the answer is 7/10 for sure.

Thanks

No, the answer is correct as it is. We can get the sum less than or equal to 4 in only two cases (1,2) and (1,3), the probability of that is 2/10. All other cases give the sum strictly MORE than 4, so the probability of more than 4 is 1 - 2/10 = 4/5.
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total pairs ; 5c2 ; 10
and <=4 ; (1,2) & ( 1,3)
so pairs; 2/10 ; 1/5 and pairs >4 1-1/5; 4/5
IMO C

I think that it is possible to tackle this with what I call the probability method as well (is there a name for this approach?).

First we must notice that this is only possible with 1-2 and 1-3.

Let's take each scenario in kind and subtract the resultant probability from 1 to obtain the desired probability.

1-2

Probability of selecting EITHER one or two*Probability of selecting the other one

2/5*1/4=2/20=1/10

1-3

The problem is identical to the last one, so the probability of this will be 1/10

2*(1/10)=2/10 and the desired probability is 1-(2/10) i.e. 4/5

Please tell me if I am missing something or if I am "accidentally" getting it right by using this method.

Bunuel Can this be solved through probability approach rather than combinations? I understand that combination approach is easier but just for my knowledge.

I tried to solve it the following way
If the first number we pick is 1 then we have two other options (2,3) to get answer <=4 . So 1/5 * 2/4 = 2/20
If the first number we pick is 2 then we have only one option (1) so 1/5 * 1/4 = 1/20

So adding both 2/20 + 1/20 = 3/20
And our answer we want is = 1 - 3/20 = 17/20

Where am I making a mistake?
Thanks in advance

Cases are:
1, 2
1, 3
2, 1
3, 1

So, you are missing one case: if the first number is 3 and the second number is 1: 1/5*1/4 = 1/20

2/20 + 1/20 + 1/20 = 1/5.

1 - 1/5 = 4/5.

Hope it's clear.
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