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by sana.noor » Mon Jul 22, 2013 12:05 pm
Alan is a bartender, and someone has removed all the labels from the bottles. Some bottles contain liquor, which has alcohol, and the rest contain mixers, which do not have any alcohol. If Alan randomly selects three bottles and mixes a drink, what is p, the probability that the drink will contain no alcohol?

(1) There are twice as many bottles of liquor as there are of mixers.
(2) 1/80 < p < 1/50
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Source: — Data Sufficiency |

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by [email protected] » Mon Jul 22, 2013 1:33 pm
Hi sana.noor,

This DS question focuses on probability math, but there's a "question behind the question" hidden within it:

The original prompt asks about the probability of mixing 3 bottles and having NO ALCOHOL.

The Q behind the Q asks "what's the probability of randomly grabbing 3 "mixers"? Let's call L the number of Liquor Bottles and M the number of Mixer Bottles.

Fact 1 tells us that there's twice as many liquor bottles as mixer bottles. TESTing values, we could have:

L = 2
M = 1
In this scenario, there is a 0% chance of getting 3 mixers.

L = 6
M = 3
Here, there's clearly a chance that's GREATER THAN 0%

Inconsistent answers = INSUFFICIENT


Fact 2 gives us a range of possibilities, but not enough info to give us a specific answer, so...

Fact 2 is INSUFFICIENT.


Combining facts you'll have to do some math to prove the possibilities. Since there is a range (from Fact 2), we have to TEST some possibilities (using the info from Fact 1) to see what falls within the range:

If
L = 6
M = 3
The odds of getting 3 mixers is 3/9 x 2/8 x 1/7 = 1/84 which is NOT possible, given the info in Fact 2

If
L = 8
M = 4
The odds of getting 3 mixers is 4/12 x 3/11 x 2/10 = 1/55 which IS POSSIBLE

At this point, notice how the odds are INCREASING? What do you think will happen when we raise the numbers?

If
L = 10
M = 5
The odds of getting 3 mixers is 5/15 x 4/14 x 3/13 = 2/91 which is about 1/45, which is NOT POSSIBLE

There's ONLY ONE ANSWER that fits both Facts:
L = 8
M = 4

Final Answer: C

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