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A box contains red and blue balls only. If there are 8 balls in total

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A box contains red and blue balls only. If there are 8 balls in total [#permalink]

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New post 31 Mar 2015, 05:26
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A box contains red and blue balls only. If there are 8 balls in total, how many red balls are in the box?

(1) If two balls are randomly selected without replacement, the probability that both balls are red is 5/14

(2) If two balls are randomly selected without replacement, the probability is 15/56 that the first ball is red and the second ball is blue.


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Re: A box contains red and blue balls only. If there are 8 balls in total [#permalink]

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New post 31 Mar 2015, 05:41
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Bunuel wrote:
A box contains red and blue balls only. If there are 8 balls in total, how many red balls are in the box?

(1) If two balls are randomly selected without replacement, the probability that both balls are red is 5/14

(2) If two balls are randomly selected without replacement, the probability is 15/56 that the first ball is red and the second ball is blue.


Kudos for a correct solution.


1. R(R-1)/(8*7)=5/14
R=5--->Sufficient

2. R(8-R)/(8*7)=15/56
R^2-8R+15=0
R=5 OR 3--->Insufficient

Answer: A
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A box contains red and blue balls only. If there are 8 balls in total [#permalink]

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New post 31 Mar 2015, 05:47
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Bunuel wrote:
A box contains red and blue balls only. If there are 8 balls in total, how many red balls are in the box?

(1) If two balls are randomly selected without replacement, the probability that both balls are red is 5/14

(2) If two balls are randomly selected without replacement, the probability is 15/56 that the first ball is red and the second ball is blue.


Kudos for a correct solution.



Answer A .

(1) If two balls are randomly selected without replacement, the probability that both balls are red is 5/14\
let N be the number of red balls.
\(\frac{^NC_2}{^8C_2} = \frac{5}{14}\)
N(N-1) = 20
N=5


(2) If two balls are randomly selected without replacement, the probability is 15/56 that the first ball is red and the second ball is blue.
let N be the number of red balls.
\(\frac{N}{8} * \frac{8-N}{7} = \frac{15}{56}\)
N=5,3
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Re: A box contains red and blue balls only. If there are 8 balls in total [#permalink]

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New post 31 Mar 2015, 22:02
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Suppose there are x red balls, and thus, 8-x blue balls.
1) P(both red w/o rep) = x/8 *(x-1)/7 = 5/14. since quad, we'll get 2 values of x, but the no of balls has to be positive. So we take the +ve one. SUFF
2) P(1st red, 2nd blue w/o rep) = x/8 * (8-x)/7 = 15/56 => x*(8-x) = 15. This gives 2 values of x : 5,3. Thus not SUFF, since red balls can be either 5 or 3.

Ans A.
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Re: A box contains red and blue balls only. If there are 8 balls in total [#permalink]

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New post 31 Mar 2015, 22:59
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1
Bunuel wrote:
A box contains red and blue balls only. If there are 8 balls in total, how many red balls are in the box?

(1) If two balls are randomly selected without replacement, the probability that both balls are red is 5/14

(2) If two balls are randomly selected without replacement, the probability is 15/56 that the first ball is red and the second ball is blue.


Kudos for a correct solution.


I went with A


1) Sufficient

x/8 * x/7 = 5/14 <--- solvable and gives ratio. I don't do the actual math, I just know that it's possible....

2) Insufficient
Since the 15/56 results from p(blue)*p(red) there is no way to determine one of them alone from the information given..
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A box contains red and blue balls only. If there are 8 balls in total [#permalink]

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New post 02 Apr 2015, 16:52
1
Bunuel wrote:
A box contains red and blue balls only. If there are 8 balls in total, how many red balls are in the box?

(1) If two balls are randomly selected without replacement, the probability that both balls are red is 5/14

(2) If two balls are randomly selected without replacement, the probability is 15/56 that the first ball is red and the second ball is blue.


Kudos for a correct solution.


number of red balls (R), number of Blue balls (B)
1) probability that two balls randomly selected without replacement, are red = \((R/8) * ({R-1}/7) = 5/14\)
we get R(R-1) = 20
R = 5,-4
R cannot be negative so number of red balls (R) = 5

Sufficient Alone.

2) probability that two balls randomly selected without replacement, the first ball is red and the second ball is blue = \((R/8) * (B/7) = 15/56\)
we get, R*B = 15
we know that R+B = 8
So only possible values are 3 and 5 but which one is red is unclear.
Insufficient Alone.

Answer:A
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Re: A box contains red and blue balls only. If there are 8 balls in total [#permalink]

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New post 03 Apr 2015, 15:02
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R+B=8; R=?
1. 5/14=getting two R w/o replacement=>R(R-1)/8*7=> as R can be only positive so only one answer; Sufficient
2. 15/56=one R one B=>R(8-R)/8*7=> R has two positive values; Not sufficient
Hence answer is A

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A box contains red and blue balls only. If there are 8 balls in total [#permalink]

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New post 03 Apr 2015, 16:00
1
Bunuel wrote:
A box contains red and blue balls only. If there are 8 balls in total, how many red balls are in the box?

(1) If two balls are randomly selected without replacement, the probability that both balls are red is 5/14

(2) If two balls are randomly selected without replacement, the probability is 15/56 that the first ball is red and the second ball is blue.


Kudos for a correct solution.


For this type of question we should only know the principles of probability and don't need to make any calculations to find the correct answer:

We know that this box contain only red and blue balls (important: this approach will be work only for two types of items).
So if we have information about probability randomly selected balls of ONE color (any quantity) we can find how much balls of each color in box.

1) in this statement we have information about probability of two balls the SAME color - this is sufficient and we don't need any calculations, because this is DS task and we should save time

If we have probability for balls with DIFFERENT colors we can't find how much balls of each color in box
2) in this statement we have information about probability of two balls the same color - this is insufficient and we don't need any calculations, because this is DS task and we should save time

So answer is A
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Re: A box contains red and blue balls only. If there are 8 balls in total [#permalink]

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New post 06 Apr 2015, 05:39
1
Bunuel wrote:
A box contains red and blue balls only. If there are 8 balls in total, how many red balls are in the box?

(1) If two balls are randomly selected without replacement, the probability that both balls are red is 5/14

(2) If two balls are randomly selected without replacement, the probability is 15/56 that the first ball is red and the second ball is blue.


Kudos for a correct solution.


MAGOOSH OFFICIAL SOLUTION:
Attachment:
Count_and_Prod_redblue.png
Count_and_Prod_redblue.png [ 36.72 KiB | Viewed 2296 times ]

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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: A box contains red and blue balls only. If there are 8 balls in total [#permalink]

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New post 15 Jan 2018, 03:17
Bunuel wrote:
Bunuel wrote:
A box contains red and blue balls only. If there are 8 balls in total, how many red balls are in the box?

(1) If two balls are randomly selected without replacement, the probability that both balls are red is 5/14

(2) If two balls are randomly selected without replacement, the probability is 15/56 that the first ball is red and the second ball is blue.


Kudos for a correct solution.


MAGOOSH OFFICIAL SOLUTION:
Attachment:
Count_and_Prod_redblue.png




Hi All,
WRT official soln

is it not (RB / 56 ) = 1/2 * ( 15/56) because there are two ways ...

1. Non red & then red
2. Red and then non red.


Kindly clarify
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Re: A box contains red and blue balls only. If there are 8 balls in total [#permalink]

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New post 19 Jan 2018, 07:29
coolnaren wrote:
Bunuel wrote:
Bunuel wrote:
A box contains red and blue balls only. If there are 8 balls in total, how many red balls are in the box?

(1) If two balls are randomly selected without replacement, the probability that both balls are red is 5/14

(2) If two balls are randomly selected without replacement, the probability is 15/56 that the first ball is red and the second ball is blue.


Kudos for a correct solution.


MAGOOSH OFFICIAL SOLUTION:
Attachment:
Count_and_Prod_redblue.png




Hi All,
WRT official soln

is it not (RB / 56 ) = 1/2 * ( 15/56) because there are two ways ...

1. Non red & then red
2. Red and then non red.


Kindly clarify



Hi

Second statement clearly specifies that the probability of selecting first ball as red and second ball as blue is = 15/56. So its only the second case:- out of the two cases specified by you.
Re: A box contains red and blue balls only. If there are 8 balls in total   [#permalink] 19 Jan 2018, 07:29
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