If you take 5 balls at random from a box with M-white balls and N-balls, the number of white balls will be a random quantity. Then write the law of its distribution

Question

help me with this

Answer 1

I assume there are M white balls and N balls which are not white. This is a two-state system, because each ball is either white or it's not white. There are M+N balls altogether. The probability of selecting a white ball is M/(M+N) and the probability of picking a ball which is not white is N/(M+N).

This is a binomial distribution:

(M/(M+N)+N/(M+N))^T. The number of trials is denoted by T. 

I'm assuming that the question implies a binomial distribution, rather than a hypergeometric distribution. The question doesn't state how the selection of a ball is made. If a ball is removed and not replaced the distribution is hypergeometric; but if the ball is replaced, its colour noted, then replaced it's binomial. A binomial distribution implies that at every trial there are always M+N balls to select from. In the hypergeometric distribution as each ball is removed without replacement, the number of balls in the set decreases by 1. 

The distribution function will have two parameters: (1) T the number of trials, that is, the number of times a ball is selected at random from the set and replaced; (2) the number of white balls, w, we are trying to find the probability of capturing.

If P(T,w) means the probability of finding w white balls in T trials, then:

P(T,w)=^TC_w(M/(M+N))^w(N/(M+N))^T-w. ^TC_w is the number of combinations of w white balls in the mixed set of balls, and it's equal to T!/(w!(T-w)!). This can also be written:

P(T,w)=^TC_wM^wN^T-w/(M+N)^T

Let's look at an example. M=8, N=16, so M/(M+N)=8/24=⅓ and N/(M+N)=16/24=⅔.

Let T=10 and w=3. So we have 24 balls, 8 of which are white. We select a ball and note its colour, then put it back in the set. We select 10 balls in this way and at the end of the exercise we note that out of 10 only 3 are white.

P(10,3)=120×⅓³×⅔⁷=5120/19683=0.26 or 26% approx. (Note that we might have expected about 33% because one in three balls in the set is white, so we perhaps conclude logically that about 3 in 10 balls is white and 3/10 is about 33%. The probability function says it's more like 26% (about a quarter).)