Suppose a random sample of size n is drawn from a population of r objects without replacement. Find the probability that k given objects are included in the sample.
So you are saying that there are 'k' specified objects in the population of 'n' objects and you want the probability that all k of them are in the sample of 'r' objects? That isn't quite what you said initially.
Obviously if r< k the probability is 0. Assuming $\displaystyle r\ge k$, the probability the first object drawn is a specified object is k/n. There are then n- 1 objects left with k-1 of them being the specified objects. The probability that the second object drawn is a specified object is (k-1)/(n-1). Continuing, the probability the first k objects are all specified objects is $\displaystyle \frac{k}{n}\frac{k-1}{n-1}\cdot\cdot\cdot\frac{1}{n-(k-1)}= \frac{k!}{n(n-1)\cdot\cdot\cdot (n-k+1)}$. That denominator can be written as $\displaystyle P^n_k= \frac{n!}{k!}$ so the probability can be written as $\displaystyle \frac{(k!)^2}{n!}$.
That is the probability of drawing all k of the specified objects first then the rest of the k objects. But the same analysis will show that if we assume any other order, we get different fractions but the same numerators and denominators in different orders so the same product. To find the total probability we only have to multiply by the number of different orders of k things in a total of r: $\displaystyle C^r_k= \frac{r!}{k!(r- k)!}$.
made a mistake here
population of 'r' objects.Random sample of 'n' objects selected without replacement.
We have to find the probability that 'k' given objects are there in the sample selected
I have done this
to select 'n' objects randomly from the population of 'r' objects = rCn.
probability that 'k' objects are in the randomly selected 'n' objects = nCk/rCn.
Is this correct?
I am a software Engineer and I am studying this
Thanks
I think it is nCk/rCk. I find these sort of problems easier to solve using a real example as a sanity check. So lets say the population is r =10, sample is n=5 and number of red objects (or round or blue or whatever) is k=3. So the probability of getting the k=3 objects as the first 3 of our sample of n=5 equals 3/10*2/9*1/8 = 1/120. We can see that this is 3!/(10!/7!) = 3!7!/10! which is equal to 1/rCk. So HallsofIvy's equation for the denominator I think should be k!(r-k)!/r! instead of (k!)^{2}/n!. And for the numerator we can get the 3 red objects in our sample of 5 in (5 choose 3) = 10 ways (which is nCk). So nCk/rCk is I beleive the correct answer.