probablility question?

Question

How do you approach and solve this question? Thanks!

the probablity that a machine produces a defective item is 0.01. each item is checked as it is produced. assume that these are independant trials and compute the probablility that atleast 100 items must be checked to find one that is defective.

Answer 1

Let r = the number of successes. In this case, r = 1, the first success. When r = 1, this is a special case of the negative binomial, called the geometric probability distribution. And the probability of x trials given p = 0.01 is:
P(x) = p(1 - p)^(x - 1)
For x = 100, p(100) = 0.01(0.99)^(99) = 0.0037
***Edit***
Note: Galaxy's solution was on the right track, but I think his probability is for finding 1 defective during any one of 100 trials, not the last trial.
***Edit(2)***
I'm glad that , at least, "Wantingtoknow" agrees with me.

Answer 2

Hmm. I think previous responders are answering different questions.

Having to check "at least 100 items to find one that is defective" means, to me, that the first 99 were OK.

Your scenario follows a geometric distribution with n=100 and p=0.01. The PMF is:

P(X=x) = (1-p)^(x-1) p

Since,
the probability that the first 99 items were OK is:

0.99^99

and the probability that the next item is deffective is:

0.01^1

Your desired probability is:

0.99^99 * 0.01 = 0.004 (approx.)

Answer 3

First, you're able to be able to desire to locate all the achievable outcomes of rolling 2 cube. they're: one million and one million one million and a pair of one million and 3 one million and four one million and 5 one million and 6 2 and a pair of two and 3 2 and four 2 and 5 2 and 6 3 and 3 3 and four 3 and 5 3 and 6 4 and four 4 and 5 4 and 6 5 and 5 5 and 6 6 and 6 Now, of those 21 techniques, in common terms rolling a one million and 3, 2 and four, 3 and 5 or a 4 and 6 can provide 2 once you subtract. So, our possibility of four/21.

Answer 4

Actually, the binomial calculations could be tedious. A normal distribution is used to approximate this binomial event frequently when n ( the number of samples ) is large.

How large is 'large'? As a rule of thumb n>25 usually. In other words, for n>25, the binomial distribution (n,p,q) can be approximated by a normal distribution whose mean and variance are as follows :

Mean = pq
Variance = npq

Where,

p = probablity for the event to "happen" or sucess.
q = 1- p = the probability of the event to NOT happen or Fail.

I'll leave the 'calculations; to you.

Hope this sheds some light.

Answer 5

q=0.01 that a given item is defective and p=0.99 that a given item is OK. If 100 items are sampled, then prob(all are OK)= 0.99^100=0.366.

Answer 6

i would use binomial distribution.

p = 100C1 * (.01) ^1 (.99)^99

i think there is also some normal distribution approximation to binomial distribution calculations, but offhand i cant recall it.