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
Cheers,
Brent
