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simultaneous probability

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by faysal204 » Wed Jun 30, 2010 3:33 pm
A box contains 10 light bulbs, fewer than half of which are defective. Two bulbs are to be drawn simultaneously from the box. If n of the bulbs in box are defective, what is the value of n?

(1) The probability that the two bulbs to be drawn will be defective is 1/15.

(2) The probability that one of the bulbs to be drawn will be defective and the other will not be defective is 7/15.

OA is D.

Here,the question asks that the two bulbs be drawn simultaneously(bulbs are not drawn one after another).So how
do we calculate this type of probability?
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Source: — Data Sufficiency |
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by jeffedwards » Wed Jun 30, 2010 4:22 pm
Don't worry about the simultaneous stuff, just solve it as normal. You can eye these, but I wrote out the long way to solve, since that seems to be your main question

A- Sufficient
(x/10)*((x-1)/(10-1))
(x^2-x)/90
x^2-x-6=0
x= 3 or -2
so x= 3

B- Sufficient
Slightly trickier
(x/10)*((10-x)/(10-1)) + ((10-x)/10)*(x/(10-1)) = 7/15
((10x-x^2)/90) + ((10x-x^2)/90) = 7/15
(-2x^2+20x)/90 = 7/15
-2x^2+20x-41=0
-2(x^2-10x+21) = 0
-2(x-7)(x+3) = 0

So x could be 3 or 7. Since we know that it is less than 5 the answer is 3
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