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D01-37

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D01-37  [#permalink]

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New post 16 Sep 2014, 00:13
1
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A
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  15% (low)

Question Stats:

81% (01:18) correct 19% (01:17) wrong based on 124 sessions

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A box contains 11 hats, out of which 6 are red hats and 5 are green hats. If two hats are to be selected at random without replacement, what is the probability that at least one green hat will be selected?

A. \(\frac{10}{11}\)
B. \(\frac{8}{11}\)
C. \(\frac{7}{12}\)
D. \(\frac{5}{13}\)
E. \(\frac{2}{7}\)

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Re D01-37  [#permalink]

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New post 16 Sep 2014, 00:13
2
1
Official Solution:

A box contains 11 hats, out of which 6 are red hats and 5 are green hats. If two hats are to be selected at random without replacement, what is the probability that at least one green hat will be selected?

A. \(\frac{10}{11}\)
B. \(\frac{8}{11}\)
C. \(\frac{7}{12}\)
D. \(\frac{5}{13}\)
E. \(\frac{2}{7}\)

It's easier to find the probability of the opposite event and subtract it from 1. The opposite event would be if we select zero green hats, or which is the same if we select 2 red hats, so: \(P(G \ge 1)=1-P(RR)=1-\frac{6}{11}*\frac{5}{10}=\frac{8}{11}\).

Answer: B
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Collection of Questions:
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Re: D01-37  [#permalink]

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New post 12 Jan 2015, 22:43
Option B - (1-6C2/11C2)
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Re: D01-37  [#permalink]

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New post 15 Jan 2016, 17:33
Hi All!

Could someone please explain how I could solve this exercise without doing the opposite?

Should not the solution be 5/11 X 4/10?

Regards!

Bunuel wrote:
Official Solution:

A box contains 11 hats, out of which 6 are red hats and 5 are green hats. If two hats are to be selected at random without replacement, what is the probability that at least one green hat will be selected?

A. \(\frac{10}{11}\)
B. \(\frac{8}{11}\)
C. \(\frac{7}{12}\)
D. \(\frac{5}{13}\)
E. \(\frac{2}{7}\)

It's easier to find the probability of the opposite event and subtract it from 1. The opposite event would be if we select zero green hats, or which is the same if we select 2 red hats, so: \(P(G \ge 1)=1-P(RR)=1-\frac{6}{11}*\frac{5}{10}=\frac{8}{11}\).

Answer: B
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D01-37  [#permalink]

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New post 18 Jan 2016, 07:48
2
mestrec wrote:
Hi All!

Could someone please explain how I could solve this exercise without doing the opposite?

Should not the solution be 5/11 X 4/10?

Regards!

Bunuel wrote:
Official Solution:

A box contains 11 hats, out of which 6 are red hats and 5 are green hats. If two hats are to be selected at random without replacement, what is the probability that at least one green hat will be selected?

A. \(\frac{10}{11}\)
B. \(\frac{8}{11}\)
C. \(\frac{7}{12}\)
D. \(\frac{5}{13}\)
E. \(\frac{2}{7}\)

It's easier to find the probability of the opposite event and subtract it from 1. The opposite event would be if we select zero green hats, or which is the same if we select 2 red hats, so: \(P(G \ge 1)=1-P(RR)=1-\frac{6}{11}*\frac{5}{10}=\frac{8}{11}\).

Answer: B


Hi,
although you should solve by looking at the prob of opposite event, since it is less time consuming and less error prone..
the normal way would be..
there are 5 green and 6 red ...
so way you can pick up atleast one of red is..
both green=\(\frac{5}{11}*\frac{4}{10}\)..
one red and one green=\(\frac{6}{11}*\frac{5}{10}\) and \(\frac{5}{11}*\frac{6}{10}\) (2 ways one of each can be picked up)
total prob=\(\frac{5}{11}*\frac{4}{10} + \frac{6}{11} * \frac{5}{10} *2= \frac{(20+60)}{110}= \frac{8}{11}\)
hope it helps you
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Re: D01-37  [#permalink]

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New post 26 Jun 2017, 09:17
1
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Total Cases : 11C2 = 55
Prob. of Picking up One Green Hat = 5c1 * 6c1 = 30 (One from Green Hat Box and Other from Red Hat Box)
Prob. of Picking up Two Green Hat = 5c2 = 10 (Both from Green Hat Bucket)

P(x) = (30 + 10)/55
= 40/55 = 8/11
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Re: D01-37  [#permalink]

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New post 17 Nov 2017, 11:06
I am not sure what am i doing wrong in the problem? my approach is :
1 red and one green or both green
thus, (6/11*5/10) or (5/11*4/10)
=5/11

please help me know what i am missing in my approach
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Re: D01-37  [#permalink]

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New post 17 Nov 2017, 11:19
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jyotipes21@gmail.com wrote:
I am not sure what am i doing wrong in the problem? my approach is :
1 red and one green or both green
thus, (6/11*5/10) or (5/11*4/10)
=5/11

please help me know what i am missing in my approach



Hi..

Where you are going wrong is that you are taking two cases RG and GR as one case..
You can pick green then red OR pick red then green OR pick both green..
So 6/11*5/10+5/12*6/10+5/11*4/10

You will get your answer
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2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html


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Re: D01-37  [#permalink]

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New post 24 Dec 2017, 04:31
Can anyone explain ( Experts in particular)

Why has the order not matter here? Is it because of the word ( selection) used in Question stem which hints for a Combination question?

Atleast one green will be interpreted as ( Green, Green ) Or ( Green , Red) [ Now this can also be taken as ) --> (Red, Green)

But here we have only take these two ( Green, Green ) Or ( Green , Red) as possibilities and left out the third one. Kindly clarify.
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Re: D01-37  [#permalink]

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New post 24 Dec 2017, 04:38
mtk10 wrote:
Can anyone explain ( Experts in particular)

Why has the order not matter here? Is it because of the word ( selection) used in Question stem which hints for a Combination question?

Atleast one green will be interpreted as ( Green, Green ) Or ( Green , Red) [ Now this can also be taken as ) --> (Red, Green)

But here we have only take these two ( Green, Green ) Or ( Green , Red) as possibilities and left out the third one. Kindly clarify.



Hi..

Here the order matters and you have to take all 3 cases.
See solutions above
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1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html


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Re: D01-37 &nbs [#permalink] 24 Dec 2017, 04:38
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