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If 2 different members are to be selected at random from a group of 8

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If 2 different members are to be selected at random from a group of 8  [#permalink]

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New post 08 Aug 2018, 06:42
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Question Stats:

63% (02:40) correct 38% (02:28) wrong based on 88 sessions

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If \(2\) different members are to be selected at random from a group of \(8\) people and if \(p\) is the probability that both members selected will be older than \(35\) years old,
is \(p > \frac{1}{3}\)?

(1) More than half of the group members are older than \(35\) years old.
(2) The probability that both members selected will be \(35\) years old or younger is greater than \(\frac{1}{10}\).
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Re: If 2 different members are to be selected at random from a group of 8  [#permalink]

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New post 08 Aug 2018, 07:29
If 22 different members are to be selected at random from a group of 88 people and if pp is the probability that both members selected will be older than 35 years old,
is p>13p>13?

(1) More than half of the group members are older than 35 years old.
(2) The probability that both members selected will be 35 years old or younger is greater than 1/10

1] No. of people >35 can be 5, 6, 7, 8

\(\frac{5c2}{8c2}\) =\(\frac{5*4}{8*7}\) = \(\frac{5}{14}\) > \(\frac{1}{3}\) = yes
\(\frac{6c2}{8c2}\) =\(\frac{6*5}{8*7}\) = \(\frac{15}{24}\) > \(\frac{1}{3}\) = yes
\(\frac{7c2}{8c2}\) =\(\frac{7*6}{8*7}\) = \(\frac{3}{4}\) > \(\frac{1}{3}\) = yes
\(\frac{8c2}{8c2}\) =\(\frac{8*7}{8*7}\) = \(\frac{1}{1}\) > \(\frac{1}{3}\) = yes

Hence A is sufficient

2] let's say n (> 35) = 4 & n1( <35 ) = 4
\(\frac{4c2}{8c2}\) + \(\frac{4c2}{8c2}\) = \(\frac{3}{7}\) > \(\frac{1}{10}\)
\(\frac{4c2}{8c2}\) = \(\frac{3}{14}\) > \(\frac{1}{3}\) = No

let's say n (> 35) = 6 & n1( <35 ) = 2

\(\frac{6c2}{8c2}\) + \(\frac{2c2}{8c2}\) = \(\frac{16}{28}\) > \(\frac{1}{10}\)
\(\frac{6c2}{8c2}\) = \(\frac{15}{28}\) > \(\frac{1}{3}\) = yes
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Re: If 2 different members are to be selected at random from a group of 8  [#permalink]

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New post 08 Aug 2018, 09:15
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If \(2\) different members are to be selected at random from a group of \(8\) people and if \(p\) is the probability that both members selected will be older than \(35\) years old,
is \(p > \frac{1}{3}\)?

(1) More than half of the group members are older than \(35\) years old.
So minimum number of members older than 35 is >8/2 or >4, hence 5
Thus minimum probability is \(\frac{5}{8}*\frac{4}{7}=\frac{5}{14}\), which is >5/15 or >1/3
Sufficient

(2) The probability that both members selected will be \(35\) years old or younger is greater than \(\frac{1}{10}\).
So minimum number of younger than 35 is 3 as 3/8*2/7=3/28>1/10
In this case >35 is 5/8*4/7=5/14...yes
But if <35 is 7 out of 8 so prob is 7/8*6/7=3/4 >1/10
Here >35 will be 1/8*0/7=0
Various combinations possible
Insufficient

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Re: If 2 different members are to be selected at random from a group of 8  [#permalink]

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New post 08 Aug 2018, 13:44
Is my method of evaluating statement B flawed?

(2) The probability that both members selected will be 35 years old or younger is greater than 1/10

Probability that both members are 35 yrs old or younger = (1 - probability that none selected are 35 yrs old or younger) >1/10
=>Probability that both members are 35 yrs old or younger = (1 - probability that both selected are older than 35) >1/10
=> probability that both selected are older than 35 < 9/10
being <9/10 doesn't mean its >1/3
hence insufficient
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Re: If 2 different members are to be selected at random from a group of 8  [#permalink]

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New post 09 Aug 2018, 07:59
1
Anshkarha wrote:
Is my method of evaluating statement B flawed?

(2) The probability that both members selected will be 35 years old or younger is greater than 1/10

Probability that both members are 35 yrs old or younger = (1 - probability that none selected are 35 yrs old or younger) >1/10
=>Probability that both members are 35 yrs old or younger = (1 - probability that both selected are older than 35) >1/10
=> probability that both selected are older than 35 < 9/10
being <9/10 doesn't mean its >1/3
hence insufficient


Anshkarha
In my opinion,
Event A: Both members are more than 35 years old
Event B: one member is 35 or less than 35 years old, another member is more than 35 years old
Event C : Both members are 35yrs or less than 35 years old
\(P(A)+P(B)+P(C) =1\)
\(P(C)=1-P(A)-P(B)\)
i.e Statement 2: \(1-P(A)-P(B)> \frac{1}{10}\)
Let me know you agree to this?
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Re: If 2 different members are to be selected at random from a group of 8  [#permalink]

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New post 10 Aug 2018, 01:37
Bismarck wrote:
If \(2\) different members are to be selected at random from a group of \(8\) people and if \(p\) is the probability that both members selected will be older than \(35\) years old,
is \(p > \frac{1}{3}\)?

(1) More than half of the group members are older than \(35\) years old.
(2) The probability that both members selected will be \(35\) years old or younger is greater than \(\frac{1}{10}\).


This is a modified version of the following OG question: https://gmatclub.com/forum/if-2-differe ... 68280.html Hope it helps.
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Re: If 2 different members are to be selected at random from a group of 8  [#permalink]

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New post 08 Feb 2019, 12:09
Bismarck wrote:
If \(2\) different members are to be selected at random from a group of \(8\) people and if \(p\) is the probability that both members selected will be older than \(35\) years old,
is \(p > \frac{1}{3}\)?

(1) More than half of the group members are older than \(35\) years old.
(2) The probability that both members selected will be \(35\) years old or younger is greater than \(\frac{1}{10}\).


1) More than half the group means at least 5/8
p = 1st selection (5/8) * 2nd selection (4/7) = 20/56 = 10/28 = 5/14, can also be written as: 5C1*4C1 / 8C1*7C1
5/14 is larger than (1/3)*5/5 = 5/15 , since the denominator is smaller in the first fraction. Sufficient to answer p>1/3?

2) Gives us a range for the amount of people that could be <=35. If we take the same number, 5 for >35, then that means 3 are <=35. We get 3/8 * 2/7 = 6/56 = 3/28 which is bigger than 3/30, or 1/3. So in this case, it would be yes.

But if we take 4 for >35, then 4 are <=35. We get 1/2 * 3/7 = 3/14. So in this case find LCD, 3/14 or 9/42 is smaller than 1/3 or 14/42. So the answer would be no.
Not sufficient.


Very tough to solve quickly imo. I would say we can assume a good chance that 2) would be insufficient without doing the math.
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Re: If 2 different members are to be selected at random from a group of 8  [#permalink]

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New post 10 Feb 2019, 01:07
chetan2u wrote:
If \(2\) different members are to be selected at random from a group of \(8\) people and if \(p\) is the probability that both members selected will be older than \(35\) years old,
is \(p > \frac{1}{3}\)?

(1) More than half of the group members are older than \(35\) years old.
So minimum number of members older than 35 is >8/2 or >4, hence 5
Thus minimum probability is \(\frac{5}{8}*\frac{4}{7}=\frac{5}{14}\), which is >5/15 or >1/3
Sufficient

(2) The probability that both members selected will be \(35\) years old or younger is greater than \(\frac{1}{10}\).
So minimum number of younger than 35 is 3 as 3/8*2/7=3/28>1/10
In this case >35 is 5/8*4/7=5/14...yes
But if <35 is 7 out of 8 so prob is 7/8*6/7=3/4 >1/10
Here >35 will be 1/8*0/7=0
Various combinations possible
Insufficient

A


please explain statement B. I am getting lost at you initial statement that min. member are 3.
Is there any more convenient way to tackle such questions without following hit and trail techniques?
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Re: If 2 different members are to be selected at random from a group of 8   [#permalink] 10 Feb 2019, 01:07
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