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

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Manager
Joined: 06 Apr 2010
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A box contains red and blue balls only. If there are 8 balls  [#permalink]

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27 Nov 2010, 20:36
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Difficulty:

75% (hard)

Question Stats:

61% (02:03) correct 39% (02:05) wrong based on 201 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.

Manager
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Re: Balls in the box  [#permalink]

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27 Nov 2010, 21:01
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1
udaymathapati 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.

Lets say there are R red balls and 8-R blue balls.
S1: P(Both-Red) = $$\frac{R}{8}*\frac{R-1}{7} = \frac{5}{14}$$
so R(R-1) = 5*4, hence R = 5 (considering only positive value). Sufficient

S2: P(Red-Blue) = $$\frac{R}{8}*\frac{8-R}{7} = \frac{15}{56}$$
so R(8-R) = 15, hence R = 5 or 3. Not Sufficient

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

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28 Aug 2018, 17:49
udaymathapati 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.

I was able to deduce statement 1 to $$R(R-1)/8*7$$ and statement 2 to $$R(8-R)=15$$. I didn't solve it further as I thought R can be calculated but end up getting it wrong

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

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28 Aug 2018, 18:39
A box contains red and blue balls only. If there are 8 balls in total, how many red balls are in the box?
Let the number of red balla be R and blue balla be B

(1) If two balls are randomly selected without replacement, the probability that both balls are red is 5/14
if there are R red balls , the first can be picked in R ways and the next in R-1 ways
Therefore probability = $$\frac{R}{8}*\frac{R-1}{8-1}=\frac{5}{14}.......R(R-1)=\frac{8*7*5}{14}=5*4$$
Thus R =5
Sufficient

(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.
Probability = $$\frac{R}{8}*\frac{B}{8-1}=\frac{15}{56}.......R*B=\frac{8*7*15}{56}=15=5*3$$
5*3 is the only possibility as 5+3=8 but R could be 3 or 5
Insufficient

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

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28 Aug 2018, 19:57
1
chetan2u wrote:
A box contains red and blue balls only. If there are 8 balls in total, how many red balls are in the box?
Let the number of red balla be R and blue balla be B

(1) If two balls are randomly selected without replacement, the probability that both balls are red is 5/14
if there are R red balls , the first can be picked in R ways and the next in R-1 ways
Therefore probability = $$\frac{R}{8}*\frac{R-1}{8-1}=\frac{5}{14}.......R(R-1)=\frac{8*7*5}{14}=5*4$$
Thus R =5
Sufficient

(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.
Probability = $$\frac{R}{8}*\frac{B}{8-1}=\frac{15}{56}.......R*B=\frac{8*7*15}{56}=15=5*3$$
5*3 is the only possibility as 5+3=8
Sufficient

D

chetan2u,
u also made same mistake which I made. Only statement 1 is sufficient. Statement 2 gives, two values for R=3,5.
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Posts: 7686
Re: A box contains red and blue balls only. If there are 8 balls  [#permalink]

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28 Aug 2018, 20:05
ammuseeru wrote:
chetan2u wrote:
A box contains red and blue balls only. If there are 8 balls in total, how many red balls are in the box?
Let the number of red balla be R and blue balla be B

(1) If two balls are randomly selected without replacement, the probability that both balls are red is 5/14
if there are R red balls , the first can be picked in R ways and the next in R-1 ways
Therefore probability = $$\frac{R}{8}*\frac{R-1}{8-1}=\frac{5}{14}.......R(R-1)=\frac{8*7*5}{14}=5*4$$
Thus R =5
Sufficient

(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.
Probability = $$\frac{R}{8}*\frac{B}{8-1}=\frac{15}{56}.......R*B=\frac{8*7*15}{56}=15=5*3$$
5*3 is the only possibility as 5+3=8
Sufficient

D

chetan2u,
u also made same mistake which I made. Only statement 1 is sufficient. Statement 2 gives, two values for R=3,5.

Thanks..
I for some reason read Red balls are more than blue although it is not mentioned. May be I had a similar question earlier.

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

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29 Nov 2018, 16:42
Hi All,

We're told that a box contains red and blue balls only and that there are 8 balls in total. We're asked for the number of 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

To start, since we're selecting balls from the box WITHOUT replacement, the probability that the first ball will be red is X/8 (where X is the number of red balls). If we pull a red ball on the first try, then there will be one FEWER red ball for the second try, so the probability of pulling a red ball then is (X-1)/7.

Thus, the overall probability will be (X)(X-1)/(8)(7) = (X)(X-1)/56. The given fraction is 5/14, which has clearly been reduced. 5/14 = 20/56, so we're looking for a value of X such that (X)(X-1) = 20. There's only one positive value that fits here... X= 5 --> since (5)(4) = 20
Fact 1 is SUFFICIENT

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.

Based on the work that we did in Fact 1, we can look for a similar pattern in Fact 2. The numerator of 15/56 is '15', so we need two numbers that SUM to 8 (one for the number of red balls and one for the number of blue balls) that when multiplied together give us 15. Those would clearly be 3 and 5. However, we do NOT know which one is which. We could have 3 red balls and 5 blue balls OR 5 red balls and 3 blue balls.
Fact 2 is INSUFFICIENT

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Re: A box contains red and blue balls only. If there are 8 balls   [#permalink] 29 Nov 2018, 16:42
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