Consider n independent trials of a success failure...

Consider n independent trials of a "success-failure" experiment (with probability of success being p). Let X be the random variable that denotes the total number of "successes" in n trials. a) What are the values that the random variable X takes? b) Write down the expression which calculates the probability for X = k "successes": P(X = k) c) Provided E(X) = np; show that Var(X) = np(1-p)

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00:01 For part a if we are considering that if we're considering the values that the random variable x can take well.

00:07 We know that x is going to be giving us the number of times or the number of successes in n trials so x can take on any value from 0 to 1 to 2 all the way up to n we can't have more than n successes in n trials, but we can have 0 successes in n trials but we also cannot have 0 .5 successes then for part b the expression that calculates the probability for x equals k successes so the thing that we need to recognize here is that this is an example of what's called a binomial random variable the pmf for a binomial random variable is that probability that x equals k is going to be n choose k x p to the power of n or pardon me p to the power of k times 1 minus p the power of n minus k or for n choose k.

01:14 I'll note that that's n factorial divided by k factorial times n minus k factorial and then we multiply by p to the power of k times 1 minus p to the power of n minus k as usual and then for part c the let's see here, so we are provided that the expected value of x is equal to n times p? so we have that the variance of x should be equal to the expected value of x squared minus the square of the expected value of x so the key thing that we're going to have to do here then is figure out what the expected value of x squared is? now what we'll do is let x i be the result of the ith trial so we have that x i is going to be equal to 1 if it's a success or 0 if it's a failure so we have that the expected value of x i is just going to be equal to p and the expected value of x i squared is equal to p as well since x i squared is just going to be equal to x i since x i is just 1 or 0 now we know that x the number of successes is going to be the sum of each one of the x eyes all the way up to x n so that then means that x squared b x 1 plus x 2 plus all the way up to plus x n all squared which we can then expand out it would become x 1 squared plus x 2 squared plus dot dot dot up to plus x n squared plus 2 times x 1 x 2 plus 2 times x 1 x 3 and so on as we go through multiplying each term as we go along so what that will give us is that the expected value of x squared is going to be equal to n times e of x 1 squared plus 2 times e of x 1 times x 2 now we know that the expected value of each trial squared is still going to be just equal to p so we have e of x squared is going to be n times p and then we'd have plus 2 times e times x 1 times x 2 but since x 1 and x 2 are independent, that's just going to be equal to 2 times e of x 1 times e of x 2 which then becomes just p squared so we get n p plus 2 p squared...

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