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805+ (Hard)|   Probability|      
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BrentGMATPrepNow
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A box contains 7 balls, and each ball is exactly 1 color 07 Mar 2023, 11:04
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95% (hard)

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42% (02:15) correct 58% (02:03) wrong based on 130 sessions

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A box contains 7 balls, and each ball is exactly 1 color (i.e., there are no multi-colored balls). If two balls are randomly selected without replacement, what is the probability that both selected balls will be blue?

(1) The probability is greater than 0.5 that both selected balls will be yellow.
(2) The probability is less than 0.25 that the first ball selected will be blue.
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D
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A box contains 7 balls, and each ball is exactly 1 color 07 Mar 2023, 11:24
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BrentGMATPrepNow
A box contains 7 balls, and each ball is exactly 1 color (i.e., there are no multi-colored balls). If two balls are randomly selected without replacement, what is the probability that both selected balls will be blue?

(1) The probability is greater than 0.5 that both selected balls will be yellow.
(2) The probability is less than 0.25 that the first ball selected will be blue.
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We know that number of balls in the box = 7

Statement 1

(1) The probability is greater than 0.5 that both selected balls will be yellow.

Let the number of yellow balls = y

\(\frac{^yC_2 }{ ^7C_2} > \frac{1}{2}\)

\(\frac{y*(y-1) }{ 7*6} > \frac{1}{2}\)

\(y*(y-1) > 21\)

Let's try out some values

y = 4; 4 * 3 = 12
y = 5; 5 * 4 = 20
y = 6; 6 * 5 = 30
y = 7; 7 * 6 = 42

So the number of yellow balls is either 6 or 7.

If the number of yellow balls is either 6 or 7, the maximum number of blue balls = 1.

Hence, the probability that both selected balls will be blue = 0

The statement is sufficient.

Statement 2

(2) The probability is less than 0.25 that the first ball selected will be blue.

Let the number of blue balls = b

\(\frac{b }{ 7} < \frac{1}{4}\)

\(b < \frac{7}{4}\)

Therefore the number of blue balls is either 1 or 0.

Hence, the probability that both selected balls will be blue = 0

This statement is also sufficient.

Option D
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A box contains 7 balls, and each ball is exactly 1 color 10 Mar 2023, 08:29
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BrentGMATPrepNow
A box contains 7 balls, and each ball is exactly 1 color (i.e., there are no multi-colored balls). If two balls are randomly selected without replacement, what is the probability that both selected balls will be blue?

(1) The probability is greater than 0.5 that both selected balls will be yellow.
(2) The probability is less than 0.25 that the first ball selected will be blue.
Show more

Target question: What is the probability that both selected balls will be blue?

Statement 1: The probability is greater than 0.5 that both selected balls will be yellow.
Let's test some possible case:
case i: All 7 balls are yellow. In this case, p(both selected balls are yellow) = 1, which is greater than 0.5.
case ii: 6 of the 7 balls are yellow. In this case, p(both selected balls are yellow) = 6/7 x 5/6 = 5/7 ≈ 0.71, which is greater than 0.5.
case iii: 5 of the 7 balls are yellow. In this case, p(both selected balls are yellow) = 5/7 x 4/6 = 20/42, which is LESS THAN 0.5

Since we are told p(both selected balls are yellow) > 0.5, then there are EITHER 6 yellow balls OR 7 yellow balls in the box.
This means there is AT MOST 1 blue ball in the box.
If there is AT MOST 1 blue ball in the box, then p(both selected balls are blue) = 0
Since we can answer the target question with certainty, statement 1 is SUFFICIENT

Statement 2: The probability is less than 0.25 that the first ball selected will be blue.
Let's test some possible case:
case i: None of the 7 balls are blue. In this case, p(first selected ball is blue) = 0, which is less than 0.25.
case ii: 1 of the 7 balls is blue. In this case, p(first selected ball is blue) = 1/7 = 0.14, which is less than 0.25.
case iii: 2 of the 7 balls are blue. In this case, p(first selected ball is blue) = 2/7 = 0.28, which is GREATER THAN 0.25.
So, there are EITHER 0 blue balls in a box OR 1 blue ball in the box.
In both possible cases, p(both selected balls are blue) = 0
Since we can answer the target question with certainty, statement 2 is SUFFICIENT

Answer: D
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A box contains 7 balls, and each ball is exactly 1 color 22 Jul 2026, 21:12
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How are we supposed to assume there are only two colors of balls in the box?
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A box contains 7 balls, and each ball is exactly 1 color 22 Jul 2026, 22:09
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How are we supposed to assume there are only two colors of balls in the box?
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We do not need to assume that there are only two colors.

Statement (1) implies that at least 6 balls are yellow, so at most 1 ball can be blue.

Statement (2) implies that fewer than 1.75 balls are blue, so there can be only 0 or 1 blue ball.

In either case, selecting 2 blue balls is impossible.
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A box contains 7 balls, and each ball is exactly 1 color 30 Jul 2026, 08:03
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Hi tsvruko,

Great catch, and you're right to be suspicious: the question never says the box holds only blue and yellow balls. There could be red, green, anything. The good news is that the solutions don't actually need that assumption - it just isn't spelled out clearly. Let me show why the color mix outside of blue never matters.

The target question only cares about how many blue balls there are. As soon as we can pin down that blue ≤ 1, P(both blue) = 0, no matter what the other balls are.

Statement (2) is the cleanest version of this. P(first ball is blue) = b/7 < 0.25 gives b < 1.75, so b = 0 or 1. That's a direct fact about blue balls - the other 6 or 7 balls could be any assortment of colors, and it changes nothing. With at most 1 blue ball, you can't draw 2 blue, so the probability is 0. Sufficient.

Statement (1) reaches the same place indirectly. "P(both yellow) > 0.5" forces y = 6 or 7. That leaves at most 1 non-yellow ball total - and blue is one kind of non-yellow. So blue ≤ 1 again, and P(both blue) = 0. The leftover ball being "yellow vs. something else" was never assumed; what's assumed is only that almost everything is yellow.

Here's the check that makes it concrete - try to build a case that satisfies statement (1) yet gives a nonzero blue probability:

- 6 yellow + 1 blue - only 1 blue - both-blue prob = 0
- 6 yellow + 1 red - 0 blue - both-blue prob = 0
- 7 yellow - 0 blue - both-blue prob = 0

Every allowed case lands on 0, extra colors and all. That's exactly what "sufficient" means: the answer stays fixed across every scenario the statement permits.

So the takeaway: you don't assume two colors - you just notice each statement caps the blue count, and once blue ≤ 1, the other colors are irrelevant.

Answer: D

tsvruko
How are we supposed to assume there are only two colors of balls in the box?
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