Data Sufficiency

knight247 wrote:There are 12 balls in an urn, out of which 4 balls are picked up at random. Is the probability of all the four balls being red greater than 1/33 ?
(1)If two balls are picked up, the probability of both being red is 5/33
(2)There are 7 blue balls.

Statement 1: From this information we can calculate the number of red balls and hence we can determine whether the required probability is greater than 1/33 or not.

Sufficient

Statement 2: As there are 7 blue balls, there are (12 - 7) = 5 red balls at max.
Hence, maximum required probability = (5/12)*(4/11)*(3/10)*(2/9) = 1/99
As the maximum probability is less than 1/99, the required probability will be always less than 1/99

Sufficient

The correct answer is D.

In case anyone has doubt with the explanation in statement 1:

• Say, the number of red balls in the urn = n
  Hence, the probability of the first ball being red = n/12
  And, the probability of the second ball being red = (n - 1)/11
  
  So, (n/12)*((n - 1)/11) = 5/33
  --> n(n - 1) = 20
  --> n = 5
  
  Now we can calculate the probability of first four balls being red and we can compare that with 1/33.

Anurag@Gurome wrote:

Statement 2: Doesn't provide any relevant information.

Not sufficient

You can't dismiss Statement 2 so quickly here. If 7 of the balls are blue, then we have at most 5 red balls. If we have 5 red balls, the maximum we could have, the probability we pick 4 red balls in 4 selections is:

(5/12)(4/11)(3/10)(2/9) = 1/99

That's clearly less than 1/33, so we have enough information to be certain the answer to the question is *no* and Statement 2 is also sufficient. The answer here is D.

Ian Stewart wrote:You can't dismiss Statement 2 so quickly here.
...
The answer here is D.

Thanks Ian.
Edited the reply.

knight247 wrote:There are 12 balls in an urn, out of which 4 balls are picked up at random. Is the probability
of all the four balls being red greater than 1/33 ?

(1) If 2 balls are picked up, the probability of both being red is 5/33
(2) There are 7 blue balls

We are given that there are 12 balls in an urn, out of which 4 balls are picked up at random. We need to determine whether the probability of selecting 4 red balls is greater than 1/33.

Statement One Alone:

If 2 balls are picked up, the probability of both being red is 5/33

We can let r = the number of red balls and create the following equation:

(r/12) x (r - 1)/11 = 5/33

(r^2 - r)/132 = 5/33

33(r^2 - r) = 132 x 5

(r^2 - r) = 4 x 5

(r^2 - r) = 20

r^2 - r - 20 = 0

(r - 5)(r + 4) = 0

r = 5 or r = -4

Since r must be positive, we see that there are 5 red balls in the urn and thus there are 7 blue balls. Thus, we have enough information to answer the question.

Statement Two Alone:

There are 7 blue balls

Since there are 7 blue balls, there are at most 5 red balls. Even if all 5 remaining balls are red, the probability that the 4 chosen balls are red is:

(5/12) x (4/11) x (3/10) x (2/9) = 1/(11 x 9) = 1 / 99.

If there are fewer than 5 red balls in the urn, the probability that all 4 chosen balls are red is even smaller. Thus, the probability that all 4 chosen balls are red is definitely less than 1/33.

Once again, we have enough information to answer the question.

Answer: D