light bulbs

Mar 2010
110
0
Suppose you have 100 light bulbs and one of them is defective. If you pick out two light bulbs at random (either borh at the same time, or first one, then another, without replacing the first light bulb), what is the probability that one of your chosen light bulbs is defective?

i got an answer, but it seemed too easy

1/100 + (1/100)(1/99)

is this right?
 
May 2010
13
1
isnt it just 1/50?

1 - P(Not being defective) = 1 - (99/100).(98/99) = 1/50
 

Plato

MHF Helper
Aug 2006
22,455
8,631
Consider how we could get one good and one bad: \(\displaystyle GB\text{ or }BG\).
What are those probabilities: \(\displaystyle \frac{99}{100}\frac{1}{99} +\frac{1}{100} \frac{99}{100}=? \)
 
Dec 2009
3,120
1,342
Suppose you have 100 light bulbs and one of them is defective. If you pick out two light bulbs at random (either borh at the same time, or first one, then another, without replacing the first light bulb), what is the probability that one of your chosen light bulbs is defective?

i got an answer, but it seemed too easy

1/100 + (1/100)(1/99)

is this right?
You have summed the probabilities of getting the bad one first
and getting the bad one first followed by a "specific" good one.

You need the bad one first or 2nd, which is BG+GB

\(\displaystyle \frac{1}{100}\ \frac{99}{99}+\frac{99}{100}\ \frac{1}{99}\)

which follows from ..... pick one bulb then another.
After the first bulb is picked there are 99 to choose from.
 
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