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 Post subject: GMAT Probability (Data Sufficiency)
PostPosted: Sat Mar 24, 2012 5:36 pm 
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Joined: Sun May 30, 2010 3:15 am
Posts: 424
A jar has 10 marbles, a mix of red and white. Two marbles are randomly chosen from the jar. If b is the probability that both will be red, is b > 1/3?
(1) Less than 1/2 of the marbles in the jar are white.
(2) The probability that 1 white marble and 1 red marble will be chosen together is 7/15.

A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D. EACH statement ALONE is sufficient.
E. Statements (1) and (2) TOGETHER are NOT sufficient.
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I understood the solution, however, it seems too time consuming. I tried another solution, like this:

r + w = 10 --> w = 10 – r

out of statement (2),
(r/10) × (w/9) = (7/15) --> substituting w,
(r/10) × ((10 – r)/9) = (7/15)
r² – 10r + 42 = 0

This is where I struggled. This equation has no solution:
r = (-(-10) +-(√(-10)² -(4 × 1 × 42)))/2
i would have to take the square root of a negative number.

My question is: where is my mistake? Was there any other solution for this problem besides try all scenarios?


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 Post subject: Re: Probability, data sufficiency.
PostPosted: Sat Mar 24, 2012 6:03 pm 
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Joined: Sun May 30, 2010 2:23 am
Posts: 498
Quote:
r + w = 10 --> w = 10 – r
This is a very good approach you use.

Quote:
out of statement (2),
(r/10) × (w/9) = (7/15)
Here is a mistake you've made. The left side of the equality defines the probability of the following event:
"We choose two marbles, one at a time. The first marble is red. The second marble is white."

But the question and statement (2) deal with the event "two marbles are chosen at the same time. One of them is red, another one is white."

Try to feel the difference. ... Now, if to convert "two marbles are chosen at the same time" into "we choose two marbles, one at a time", then "one of them is red, another one is white" consists of two possibilities
- "the first one is red, the second one is white"
- "the first one is white, the second one is red".
Therefore your equation should be the following:
(r/10)(w/9) + (w/10)(r/9) = (7/15)

Simplify it first, then plug in w = 10 – r and you'll get two solutions (w = 7, r = 3 and w = 3, r =7).


Alternatively, you can deal with selected pairs (imagine we numbered marbles and select two at once). For statement (2) there will be w × r such pairs (one white and one red). The total number of possible pairs will be 2-combinations of 10 elements (10 C 2 = 10!/(8! × 2!) = 45).
The equation will be the same wr/45 = 7/15.


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