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M20-29

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M20-29 [#permalink]

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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 extracted 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. \(\frac{19}{20}\)
[Reveal] Spoiler: OA

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New post 16 Sep 2014, 01:09
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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 extracted 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. \(\frac{19}{20}\)

In only one outcome, when cards with 2 and 4 are selected, the sum will not be greater than 7. The probability of drawing 2 and 4 \(= \frac{1}{5}*\frac{1}{4} + \frac{1}{5}*\frac{1}{4} = \frac{1}{10}\) (2 and 4 can be drawn in a different order). The answer to the question is \(1 - \frac{1}{10} = \frac{9}{10}\).

Answer: D
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M20-29 [#permalink]

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New post 06 Apr 2015, 09:27
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

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Re: M20-29 [#permalink]

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New post 06 Apr 2015, 09:45
karant wrote:
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


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.
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Re: M20-29 [#permalink]

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New post 17 Dec 2016, 05:22
Bunuel wrote:
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 extracted 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. \(\frac{19}{20}\)

In only one outcome, when cards with 2 and 4 are selected, the sum will not be greater than 7. The probability of drawing 2 and 4 \(= \frac{1}{5}*\frac{1}{4} + \frac{1}{5}*\frac{1}{4} = \frac{1}{10}\) (2 and 4 can be drawn in a different order). The answer to the question is \(1 - \frac{1}{10} = \frac{9}{10}\).

Answer: D




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?
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Re: M20-29 [#permalink]

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New post 17 Dec 2016, 05:26
sidoknowia wrote:
Bunuel wrote:
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 extracted 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. \(\frac{19}{20}\)

In only one outcome, when cards with 2 and 4 are selected, the sum will not be greater than 7. The probability of drawing 2 and 4 \(= \frac{1}{5}*\frac{1}{4} + \frac{1}{5}*\frac{1}{4} = \frac{1}{10}\) (2 and 4 can be drawn in a different order). The answer to the question is \(1 - \frac{1}{10} = \frac{9}{10}\).

Answer: D




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?


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.
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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

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: M20-29 [#permalink]

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New post 27 May 2017, 06:45
Bunuel wrote:
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 extracted 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. \(\frac{19}{20}\)


Probability = no of favourable cases / total number of cases.

Let's find the total number of cases = 5 cards and we need to pick any two - then 5C2 = 10.

Let's find the favourable cases that are greater than 7..

consider with starting with 2..
(2,6)(2,8)(2,10)
With 4
(4,6)(4,8)(4,10)
with 6
(6,8) (6,10)
with 8
(8,10).... Total 9 cases..

Probability = 9 / 10...

Hope this simple and clear...:)

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Re: M20-29 [#permalink]

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New post 01 Jul 2017, 02:11
I think this way is also possible.

Two number picked
5C2=10

Now, we know that the only combination out of 10 that does not exceed 7 is (4,2).

Hence the answer. 9/10

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Re: M20-29 [#permalink]

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New post 08 Aug 2017, 08:54
Hi,

Can we do:

Total number of ways= 5!/(3!x2!) = choose 2 among 5 of the number unordered = 10
Only 1 combination unordered not favorable (2;4)

Not favorable probability = 1 /10

Favorable case = 1 - 1/10 = 9/10

?

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Re: M20-29 [#permalink]

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New post 13 Sep 2017, 18:20
Tpral wrote:
Hi,

Can we do:

Total number of ways= 5!/(3!x2!) = choose 2 among 5 of the number unordered = 10
Only 1 combination unordered not favorable (2;4)

Not favorable probability = 1 /10

Favorable case = 1 - 1/10 = 9/10

?


This is the case of Permutation, not only Combination. You use combination formula here, but fortunately you reach the same answer.

As Bunuel explanation, [2,4] and [4,2] counted as two different cases.
Total possible outcomes : 5P2, which is 20.
Probability then become =\(1 - \frac{2}{20} = \frac{18}{20} = \frac{9}{10}\)

Bunuel please correct my approach ya.
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Re: M20-29   [#permalink] 13 Sep 2017, 18:20
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