M17-17

Question

In a bin, there are 20 cards numbered from 1 to 20. If two cards are drawn without replacement, what is the probability that the sum of their numbers is even?

A.  \frac{1}{3}  
B.  \frac{7}{18}  
C.  \frac{4}{9}  
D.  \frac{9}{19}  
E.  \frac{1}{2}

Bunuel · Accepted Answer

Official Solution:

In a bin, there are 20 cards numbered from 1 to 20. If two cards are drawn without replacement, what is the probability that the sum of their numbers is even?

A.  \frac{1}{3}  
B.  \frac{7}{18}  
C.  \frac{4}{9}  
D.  \frac{9}{19}  
E.  \frac{1}{2}

To achieve an even sum, both cards must either have even numbers or odd numbers. The probability of drawing two even numbers is  \frac{10}{20}* \frac{9}{19} = \frac{9}{38} . Similarly, the probability of drawing two odd numbers is  \frac{10}{20} * \frac{9}{19} = \frac{9}{38} . Therefore, the combined probability is  \frac{9}{38} + \frac{9}{38} = \frac{18}{38} = \frac{9}{19} .

Answer: D

gmatravi · Answer

2 * 10c2 /20c2 =>9/19

2*10c2 => there are 10 even and 10 odd. So you can have either 2 even number cards or 2 odd number cards. 
20c2 =>total combinations.

prasunsaha3434 · Answer

Hi Bunuel,

I have a small doubt in the official solution as given :

The probability of the first scenario = 10/20∗9/19=9/38

when you considered Odd numbers then total number of odds will be 10 out of 19 remaining numbers ( first number is even number ,but we are interested to choose "odd+odd " case , so 2nd number will be an odd number)

But you chose 9 odd numbers . May I request you to explain me how you multiplied 9/19 .

Thank you.

Bunuel · Answer

The first case is the probability of both cards being even. P(even, even) = 10/20∗9/19. Initially there are 10 even numbers, after one is picked you are left with 9 even numbers out of 19.

The second case is the probability of both cards being odd. P(odd, odd) = 10/20∗9/19. Initially there are 10 odd numbers, after one is picked you are left with 9 odd numbers out of 19.

Bunuel · Answer

I've revised the question and solution, incorporating additional details for improved clarity. I trust this makes it more comprehensible.

Nipunh · Answer

Hi Bunuel,

I initially thought of this question using filling spaces methods and I got 180 favourable outcome ((10x9)x2)
but I thought of the denominator as 20C2 (total outomes for choisng 2 no.s out of a set of 20)... I am unable to understand why and when should we not use this approach.. also where exactly did I falter?

please help

Bunuel · Answer

The point is that 10 * 9 + 10 * 9 gives ordered pairs of even and odd numbers, meaning it counts both (2, 4) and (4, 2) as distinct pairs. However, the denominator 20C2 gives unordered pairs. To maintain consistency, you either need to divide the numerator by 2 to make it unordered or multiply the denominator by 2 to account for the order.

phmahi1997 · Answer

I like the solution - it’s helpful.

sb995 · Answer

I did not quite understand the solution. Could you please explain why we don't need to multiply each part of the sum by 2 to take into account picking the second number first? Trying to clarify my understanding.

chetan2u · Answer

10/20 means any of the 10 cam be picked and 9/19 means any of the remaining 9 can be picked. This scenario includes all possibilities as the order matters here.. it is basically 10P2/20P2 or 10*9/(20*19)

The scenario which you are saying would be 10C2 that is pick any two out of 10. To arrange them we would multiply by 2, so 10C2 * 2

sb995 · Answer

I did not quite understand the solution. Hi, in what case would the probability be 1/2? Said differently, what would have to change in the problem for the probability of the sum of the numbers to be even? Thanks.

Bunuel · Answer

If the problem were changed to  with replacement  (you put the first card back before drawing the second), then every draw would always be 10 even and 10 odd, so the events would be independent. In that case:

P(both even) = 10/20*10/20 = 1/4 
 P(both odd) = 10/20*10/20 = 1/4 
 Total = 1/4 + 1/4 = 1/2

So the probability would be 1/2 if the cards were drawn  with replacement .

horrorslive · Answer

Bunuel  is this method correct?
 
10c2/20c2 + 10c2/20c2 = 90/190=9/19?

M17-17

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