# Probability problem

Amith2006

## Homework Statement

A Company manufactures bulb. 4 percent of it are defective. A Supervisor selects 2 bulbs at random. What is the probability that atleast one of the bulb is not defective?

## The Attempt at a Solution

probability of selecting atleast one non defective bulb = 1 - probability of selecting 2 defective bulbs
= 1 - C[4,2]/C[100,2]
= 824/825
IS IT RIGHT?

exk
Your problem doesn't state how many bulbs were manufactured total so you can't assume it's 100, hence your solution is wrong.

I think the following might work better for you:
Let X= # of defective bulbs. Then you want to find P(X=<1)=P(X=0)+P(X=1)
P(X=0)=.96^2 P(X=1)=2*.04*.96 P(X=<1)=0.9984

mistermath
What Exk is saying is that the probability of picking a defective item is 4%. Even if you pick a defective item, what's the chance of the next one being defective? 4%! But if you limit yourself to 100 total bulbs, then as you pick bulbs, the probabilities are changing. So in a way, in this particular problem, you have an "infinite" number of bulbs to work with.

What is weird is that both ways have an error associated with it but from the wording it sounds like the book is testing the infinite # of bulbs idea.

exk
Actually there are specifically 2 bulbs that you are dealing with so when calculating the probabilities you have C(2,0) and C(2,1) which I didn't put explicitly, but are in the calculations I mentioned.

Homework Helper
Hi Amith2006! probability of selecting a tleast one non defective bulb = 1 - probability of selecting 2 defective bulbs

So far, so good.

Now, since we can assume that there are a very large number of bulbs, the probabilities for the two bulbs are independent, and so we can multiply them:

probability of selecting 2 defective bulbs

= (probability of selecting 1 defective bulb)(probability of selecting another defective bulb). Staff Emeritus I've found another bulb! 