There are 12 balls in an urn, out of which 4 balls are picked up at ra

You use combinations approach, which is also correct:

\(\frac{C^2_x}{C^2_{12}}=\frac{(\frac{x!}{2!(x-2)!})}{(\frac{12!}{2!10!})}=\frac{x!}{2!(x-2)!}*\frac{2!10!}{12!}=\frac{(x-1)x}{2}*\frac{1}{11*6}=\frac{x(x-1)}{12*11}\).

Notice that this method naturally gives the same exact equation as my solution above.

Hope it helps.

Bunuel
I do not agree with the second statement being sufficient.

Scenario 1:
There are 7 blue balls, and 5 red balls. Your approach is correct.

Scenario 2:
There are 7 blue balls, 2 red balls, 3 green balls. The probability drops to 0. Nowhere is it mentioned that there are only two colours inside the urn.

The stem asks "Is the probability of all the 4 balls being red greater than 1/33?"

The answer to which is a definite NO. Hence the statement 2 is sufficient.

So your example that gives a probability of Zero satisfies the question i.e the probability of drawing 4 red balls is less than 1/33.

Please note this a Yes/No DS question and hence doesn't need to yield a specific answer.

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

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