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M17-17

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M17-17  [#permalink]

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New post 16 Sep 2014, 01:01
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If a bin contains twenty cards numbered 1, 2, ..., 20, what is the probability that the sum of the numbers on two cards extracted from the bin will be even?

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

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Re M17-17  [#permalink]

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New post 16 Sep 2014, 01:01
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Official Solution:

If a bin contains twenty cards numbered 1, 2, ..., 20, what is the probability that the sum of the numbers on two cards extracted from the bin will be even?

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

For the sum of the numbers to be even, either both cards have to bear even numbers or both cards have to bear odd numbers. The probability of the first scenario = \(\frac{10}{20}*\frac{9}{19} = \frac{9}{38}\). The probability of the second scenario = \(\frac{10}{20}*\frac{9}{19} = \frac{9}{38}\). The combined probability = \(\frac{9}{38} + \frac{9}{38} = \frac{18}{38} = \frac{9}{19}\).

Answer: D
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Re: M17-17  [#permalink]

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New post 10 Aug 2016, 12:22
1
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.
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Re: M17-17  [#permalink]

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New post 03 May 2017, 05:57
1
1 to 20

10 even and 10 odd

For a sum of even , we need either two even or two odd

so select 2 from 10 even = 10c2
same for odd = 10c2

So total = 10c2 * 2

for total cases = 20c2


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

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Re: M17-17  [#permalink]

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New post 03 May 2017, 07:12
Sum of two odd numbers or sum of two even numbers equals an even number. We have 10 odd and 10 even numbers from 1to 20. Hence , 2*(10c2/20c2) = 9/19 or Option D is the answer.
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Re: M17-17  [#permalink]

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New post 01 Jul 2017, 03:18
Nice question .

My solution.

Combination to find total= 20C2 = 190

We know, Odd+Odd=Even and Even + Even =Even

Total Odd number = 10 => 10C2=45
Total Even number = 10 => 10C2=45

Now, the numbers can be ODD "OR" EVEN

hence (45+45)/190=90/190

Hence the answer.
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Re: M17-17  [#permalink]

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New post 03 Jun 2018, 09:02
Are these questions named as "M17-17" series from GMAT Club test?

Bunuel wrote:
If a bin contains twenty cards numbered 1, 2, ..., 20, what is the probability that the sum of the numbers on two cards extracted from the bin will be even?

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

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Re: M17-17  [#permalink]

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New post 03 Jun 2018, 10:22
gmatbusters wrote:
Are these questions named as "M17-17" series from GMAT Club test?

Bunuel wrote:
If a bin contains twenty cards numbered 1, 2, ..., 20, what is the probability that the sum of the numbers on two cards extracted from the bin will be even?

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


Not sure what toy mean. This question is 17th question from set #17. You can check all sets here: https://gmatclub.com/forum/gmat-club-te ... 85003.html
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Re: M17-17  [#permalink]

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New post 27 Jun 2018, 17:33
Hi Bunuel

How would we know which case to consider, with replacement or without replacement? I think answer would vary in both cases.
Are we to assume without replacement scenario if the prompt does not mention it specifically?

Thanks.
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Re: M17-17  [#permalink]

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New post 27 Jun 2018, 20:11
pantera07 wrote:
Hi Bunuel

How would we know which case to consider, with replacement or without replacement? I think answer would vary in both cases.
Are we to assume without replacement scenario if the prompt does not mention it specifically?

Thanks.


I'd say yes. If nothing is mentioned than the case is without replacement.
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Re: M17-17  [#permalink]

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New post 27 Jun 2018, 20:11
pantera07 wrote:
Hi Bunuel

How would we know which case to consider, with replacement or without replacement? I think answer would vary in both cases.
Are we to assume without replacement scenario if the prompt does not mention it specifically?

Thanks.


If it is given you are picking 2 cards, it means you are picking them together so it will be without replacement..
Also if it is with replacement, it will be specifically mentioned or some word to that effect will be used.
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Re: M17-17  [#permalink]

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New post 27 Jun 2018, 20:21
Okay, I get it now.
Thank you Bunuel and chetan2u.
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Re: M17-17  [#permalink]

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New post 28 Aug 2018, 23:52
Bunuel chetan2u

I followed the method of taking the total number of o/c as 20C2
and favourable as 10C2 as there are 10 odd numbers.

But I haven't multiplied by 2.

What scenario am I missing out?

Can you pls explain why we are multiplying by 2.
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Re: M17-17  [#permalink]

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New post 28 Aug 2018, 23:56
psych77 wrote:
Bunuel chetan2u

I followed the method of taking the total number of o/c as 20C2
and favourable as 10C2 as there are 10 odd numbers.

But I haven't multiplied by 2.

What scenario am I missing out?

Can you pls explain why we are multiplying by 2.


The sum of two even numbers will also be even, no?
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: M17-17 &nbs [#permalink] 28 Aug 2018, 23:56
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