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

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

by swerve » Wed Nov 28, 2018 9:14 am
A box contains red and blue balls only. If there are 8 balls in total, how many red balls are in the box?

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.

The OA is A

Source: Magoosh
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by Jay@ManhattanReview » Thu Nov 29, 2018 12:32 am
swerve 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.

The OA is A

Source: Magoosh
Let's take each statement one by one.

Say there are n red balls; thus, there are (8 - n) blue balls.

We have to find out the value of n.

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

=> n/8 * (n - 1)/7 = 5/14
n(n - 1) = 20

We see thatn and (n - 1) are two consectuive integers, thus, we can write 20 as 5*4, where = 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.

=> n/8 * (8 - n)/7 = 15/56
n(8 - n) = 15 = 5*3

We see that n can be 5 or 3. No unique value of n. Insufficient.

The correct answer: A

Hope this helps!

-Jay
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by [email protected] » Thu Nov 29, 2018 2:33 pm
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

Final Answer: A

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Rich
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