A time article sounded warning bells aout fast growing in vitro fertilization industry, which caters to infertile couples desperate to have children. The program charged $7000 per attempt. We will use an average success rate of 1 success in 10 attempts. Suppose 4 attempts (28000) is the maximum couples are prepared to pay and that they will try until successful uptill that maximum.

It would appear we could use a geometric probability for this one.

This is given by P(x)=p\cdot (1-p)^{x-1}

A geometric is used when:

1. a trial is repeated until a success occurs.

2. the trials are independent

3. the probability is constant for each trial.

So, the prob. A pregnancy occurs on the first attempt knowing that the chance is 1/10 is

P(1)=.10(.90)^{0}=.10

second attempt:

P(2)=.10(.90)^{1}=.09

third attempt:

P(3)=.10(.90)^{2}=.081

fourth attempt:

P(4)=.10(.90)^{3}=.0729

Therefore, the prob. A pregnancy occurs on the first, second, third, or fourth attempt would be
P(1)+P(2)+P(3)+P(4)

The prob. It does not occur would be the complement of this result.

The mean or expected value for the geometric distribution is given by {\mu}=\frac{1}{p}

So, showing you the geometric probability wasn't helpful? Don't worry, you will not find me helping you anymore.

One does not feel like helping much when one goes to the trouble to write up a post and the OP does not appreciate it, but gives a 'not helpful'.

does not find this helpful : I find binomial distribution could be the right approach in solving the problem.

The fact is that you cannot go around using the binomial distribution here.

You can use the binomial only in the case where you have the first attempt successful, out of 1 attempt.

If you use the binomial distribution for the second attempt, you are directly
Including the case that they succeed in the first attempt and failed in the second attempt, which is not what the question is looking for.

You are looking for any of these cases:
S
FS
FFS
FFFS

Where I'm denoting F to be fail and S to be success. And this is exactly what galactus gave you. Maybe you don't know the name of this distribution, or the formula for it, but I'm almost certain that you have had such similar problems before.

Your not helpful comment given to galactus is wrong because the post above is correct.

It is only my view that I thought would be OK to share, but if you are offended, I am sorry buddy.
I DO APPRECIATE YOUR HELP TO A GREAT EXTENT.

Thanks mate. I will take it on board and be very careful in any more comments I make.