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

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Senior Manager
Joined: 18 Jun 2018
Posts: 267
If 2 different members are to be selected at random from a group of 8  [#permalink]

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08 Aug 2018, 06:42
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Difficulty:

65% (hard)

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 $$\frac{1}{10}$$.
Manager
Joined: 23 May 2017
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Concentration: Finance, Accounting
WE: Programming (Energy and Utilities)
Re: If 2 different members are to be selected at random from a group of 8  [#permalink]

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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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08 Aug 2018, 09:15
1
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
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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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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
Senior Manager
Joined: 18 Jun 2018
Posts: 267
Re: If 2 different members are to be selected at random from a group of 8  [#permalink]

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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?
Math Expert
Joined: 02 Sep 2009
Posts: 55623
Re: If 2 different members are to be selected at random from a group of 8  [#permalink]

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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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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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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?
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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