M20-29

Question

If a bin contains five cards numbered 2, 4, 6, 8, and 10, what is the probability that the sum of the numbers on two cards, randomly drawn together from the bin, will be greater than 7?

A.  \frac{1}{2}  
B.  \frac{3}{4}  
C.  \frac{4}{5}  
D.  \frac{9}{10}  
E.  \frac{19}{20}

Bunuel · Answer

Official Solution:

If a bin contains five cards numbered 2, 4, 6, 8, and 10, what is the probability that the sum of the numbers on two cards, randomly drawn together from the bin, will be greater than 7?

A.  \frac{1}{2}  
B.  \frac{3}{4}  
C.  \frac{4}{5}  
D.  \frac{9}{10}  
E.  \frac{19}{20}

There is only one outcome, selecting the cards with 2 and 4, where the sum is not greater than 7. The probability of drawing 2 and 4 is  \frac{1}{5} *\frac{1}{4} + \frac{1}{5}* \frac{1}{4} = \frac{1}{10} . This accounts for both possible orders: first drawing 2 then 4, and first drawing 4 then 2. Thus, the probability that the sum of the numbers on two randomly selected cards exceeds 7 is  1 - \frac{1}{10} = \frac{9}{10} .

Answer: D

karant · Answer

Hi Bunuel,

I tried to solve the question with another approach but it seems to be wrong.

I tried to count the no of cases
The no of cases whose sum is greater than 7 when 2 is included : (2,6) , (2,8) , (2,10) = 3
The no of cases whose sum is greater than 7 when 4 is included : (4,6) , (4,8) , (4,10) = 3
The no of cases whose sum is greater than 7 when 6 is included : (6,4), (6,8) , (6,10) = 3
The no of cases whose sum is greater than 7 when 8 is included : (8,2), (8,4), (8,6), (8,10) = 4
The no of cases whose sum is greater than 7 when 10 is included

10,2),(10,4),(10,6),(10,8)= 4

So total cases = 3+3+3+4+4 = 17.

And total number of outcomes can be calculated as 5C2.

Please help where I am going wrong ? Is 2,8 and 8,2 one case ?

Thanks

10,2),(10,4),(10,6),(10,8)= 4 So total cases = 3+3+3+4+4 = 17. And total number of outcomes can be calculated as 5C2. Please help where I am going wrong ? Is 2,8 and 8,2 one case ? Thanks

Bunuel · Answer

1. You are missing (6, 2) case there. This will make the number of favorable outcomes equal to 18.
2. The total number of outcomes is 5*4 = 20, not 10. 5 options for first number and 4 options for the second number.

Thus, P = 18/20 = 9/10.

Hope it's clear.

sidoknowia · Answer

Hi,

What I tried was
there are 10 ways to choose 2 cards from 5. Ot these choices only 2 gives sum less than 7. 
2 & 4 or 4&2

so probability of those choices is 2/10 = 1/5
hence probability of getting sum > 7 is 1-1/5 = 4/5

Can you please tell me, where I'm going wrong?

Bunuel · Answer

In the first case you count groups of two so in the second case you also should count in groups of two. So, 2 and 4 is only one group. 1 - 1/10 = 9/10.

axezcole · Answer

Is the word numbered a hint to use permutation?

Can any expert tell me when to decide whether order is imp or not?

Explaining a few cases would help me a lot.

Bunuel ,  chetan2u ,  GMATPrepNow ,  MartyTargetTestPrep ,  ScottTargetTestPrep ,  VeritasKarishma

ScottTargetTestPrep · Answer

Since there's no replacement in this problem, it does not matter which method you use for this question; it can be considered as either a combination problem or a permutation problem (hence the same answer for both methods).

If you interpret the experiment as drawing the first card from a total of five and then drawing another card from the remaining four, the problem becomes a permuation problem. In this interpretation, drawing a 2 first and an 8 next is different from the outcome of drawing an 8 first and a 2 next.

If, on the other hand, you interpret the experiment as drawing two cards at the same time from the total of five, then the question becomes a combination problem. When the two cards you draw are a 2 and an 8, that's a single outcome as you draw the two cards at the same time.

In the first interpretation, the total number of outcomes and the number of favorable outcomes are multiplied by two; since each outcome can happen in exactly two ways (such as (2, 8) or (8, 2)). That's why the probability we need to calculate will be the same regardless of whether we use the first interpretation or the second interpretation.

Had there been replacement, it would have not worked out like above. With replacement, you also need to take into account outcomes of (2, 2), (4, 4) etc.; thus you could no longer use the combination method.

It is also interesting to think about how would each method work if you picked not two but three cards.

Bunuel · Answer

I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.

shubhamagarwal44 · Answer

Can someone please help how we can solve this using combination? I tried doing this -> 1 minus probability of picking 2 & 4.

So 5C1 * 4C1 / 5C2 - but clearly this is wrong. Please help.

Bunuel · Answer

You're right to subtract the one pair that sums to 6, but the correct way is:

There are 5C2 = 10 total pairs. Only one of them (2 and 4) has a sum not greater than 7.

So the probability = 1 - 1/10 = 9/10.

Shaurs · Answer

I like the solution - it’s helpful.

Confusion · Answer

I found denominator as 5C2 and numerator as ways or combo less than sum 7 which is 1 - (2,4).
So Numerator = denominator - less than 7 sum observations = 9

shaliny · Answer

Here why order matters because two cards drawn together?

Bunuel · Answer

Mathematically, it does not matter whether we treat the two cards as drawn together or as drawn one after the other. The solution uses the probability approach, so it treats the selection as sequential and counts both possible orders: 2 then 4 and 4 then 2.

If we use the combination method instead, we get the same result: there are 5C2 = 10 total pairs. Only one of them, 2 and 4, has a sum not greater than 7.  So the probability = 1 - 1/10 = 9/10.

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