Binomial + Condition Distribution

[cse63146]

Homework Statement

Let X be a binomial random variable representing the number of successes in n independent Bernoulli trials. Let Y be the number of successes in the first m trials, where m < n . Find the conditional probability distribution of Y given X=x.

Homework Equations

The Attempt at a Solution

So I think I need to use $$f_{Y|X} (Y|X) = \frac{f(x,y)}{f_X (X)}$$

where f(x,y) = _n-mC_xp^xq^n-m-x + _mC_yp^yq^m

am I correct?

 

[mighty2000]

Hello, you are on the right track with your attempt at a solution. However, there are a few things that need to be clarified. First, the formula you are using is for the joint probability distribution of two random variables X and Y, not the conditional probability distribution of Y given X=x. To find the conditional probability distribution, you need to use the formula P(Y=y|X=x) = \frac{P(X=x, Y=y)}{P(X=x)}.

In this case, we are looking for the probability of Y=y given that X=x, so we can rewrite the formula as P(Y=y|X=x) = \frac{P(X=x, Y=y)}{P(X=x|Y=y)P(Y=y)}.

Next, we need to determine the values for each of these probabilities. P(X=x, Y=y) represents the joint probability of X=x and Y=y, which can be calculated using the binomial distribution formula you mentioned: P(X=x, Y=y) = mCxp^x(1-p)^(m-x).

P(X=x|Y=y) represents the probability of X=x given that Y=y, which can be calculated using the conditional probability formula: P(X=x|Y=y) = \frac{P(X=x, Y=y)}{P(Y=y)}.

Lastly, P(Y=y) represents the marginal probability of Y=y, which can be calculated using the binomial distribution formula as well: P(Y=y) = mCyp^y(1-p)^(m-y).

Now, putting it all together, we get the conditional probability distribution of Y given X=x as P(Y=y|X=x) = \frac{mCxp^x(1-p)^(m-x)}{mCyp^y(1-p)^(m-y)}.

I hope this helps clarify the steps you need to take to find the conditional probability distribution of Y given X=x. Let me know if you have any further questions.