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by Joy Shaha » Tue Jul 19, 2016 6:35 am
Q. A box contains 10 light bulbs, and 2 of the 10 light bulbs are defective are defective.there 10 bulbs taken randomly from the box. Find the probability that there is exactly one defective bulb out of the 3 bulbs taken from the box?
A.112/720;
B.128/1000;
C.2/10;
D.336/720;
E.6/10
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by Brent@GMATPrepNow » Wed Jul 20, 2016 6:04 am
Joy Shaha wrote:Q. A box contains 10 light bulbs, and 2 of the 10 light bulbs are defective are defective.there 10 bulbs taken randomly from the box. Find the probability that there is exactly one defective bulb out of the 3 bulbs taken from the box?

A. 112/720
B. 128/1000
C. 2/10
D. 336/720
E. 6/10
One approach is to use COUNTING TECHNIQUES

P(exactly one defective bulb) = (# of ways to select one defective bulb)/(TOTAL # of outcomes)

As always start with the DENOMINATOR
TOTAL # of outcomes
Let's pretend that the 10 bulbs are all unique
# of ways to select any 3 of them = 10C3 = 120

# of ways to select one defective bulb
We can select 1 of the 2 defective bulbs in 2 ways.
We can select 2 of the 8 non-defective bulbs in 8C2 (28) ways.
So, # of ways to select one defective bulb = (2)(28) = 56

So, P(exactly one defective bulb) = 56/120
= [spoiler]7/15[/spoiler]

NOTE: On a REAL GMAT question, the answer choices are given in SIMPLEST TERMS.
Here, 336/720 = 7/15, so the correct answer is D

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Brent
Brent Hanneson - Creator of GMATPrepNow.com
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by Matt@VeritasPrep » Wed Jul 20, 2016 11:23 pm
p(exactly one defective bulb) = p(only first bulb is defective) + p(only second bulb is defective) + p(only third bulb is defective)

Since each of the three probabilities on the right is the same, we can find the probability of one of them (say "only first bulb is defective"), then multiply that by three.

If only the first bulb is defective, we have

(First bulb defective) * (Second not) * (Third not)

or

(2/10) * (8/9) * (7/8)

or

7/45

Multiplying this by 3 gives us 21/45, or 7/15, which is equivalent to D.
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