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Math Expert V
Joined: 02 Sep 2009
Posts: 59622
A box contains red and blue balls only. If there are 8 balls in total  [#permalink]

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9 00:00

Difficulty:   65% (hard)

Question Stats: 58% (02:03) correct 42% (01:58) wrong based on 215 sessions

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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.

Kudos for a correct solution.

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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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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

Senior Manager  Joined: 07 Aug 2011
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GMAT 1: 630 Q49 V27 A box contains red and blue balls only. If there are 8 balls in total  [#permalink]

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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.

(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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1
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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GMAT 1: 670 Q44 V38 Re: A box contains red and blue balls only. If there are 8 balls in total  [#permalink]

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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..
Manager  Joined: 06 Mar 2014
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A box contains red and blue balls only. If there are 8 balls in total  [#permalink]

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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.

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.

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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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1
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

Thanks,
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GMAT 1: 660 Q48 V33 GMAT 2: 740 Q50 V40 A box contains red and blue balls only. If there are 8 balls in total  [#permalink]

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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.

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

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Math Expert V
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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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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 [ 36.72 KiB | Viewed 3347 times ]

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Intern  B
Joined: 11 Apr 2014
Posts: 19
Re: A box contains red and blue balls only. If there are 8 balls in total  [#permalink]

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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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Joined: 22 Aug 2013
Posts: 1409
Location: India
Re: A box contains red and blue balls only. If there are 8 balls in total  [#permalink]

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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.
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Posts: 13732
Re: A box contains red and blue balls only. If there are 8 balls in total  [#permalink]

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_________________ Re: A box contains red and blue balls only. If there are 8 balls in total   [#permalink] 23 Nov 2019, 21:54
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