00:01
For this problem to begin we have that the pdf of the binomial distribution is that the probability that x is equal to a particular value, k, is given by n choose k times p to the power of k times 1 minus p to the power of n minus k.
00:26
Now presumably when they're referring to each one of the different terms, you have one term, the second term, and then the third term.
00:35
So for that first term, when we have n choose k, we know that we want to have k successes among n trials.
01:04
So when we have n choose k, n factorial divided by k factorial times n minus k factorial, which actually i'll divide up in a slightly different way.
01:23
So actually we'll have 1 over k factorial times n factorial divided by n minus k factorial.
01:32
So n factorial divided by n minus k factorial is the number of arrangements, or a number of ways to pick k things from n, i'll say k unique things to be specific.
02:09
And then we have that 1 over k factorial, that is included to avoid over counting.
02:20
Since we're treating this as a situation where the order of the events doesn't matter, we're just counting up the number of successes and number of failures.
02:37
So any arrangement of k successes is considered to be equivalent, so to avoid over counting we need to do that division.
02:48
We're dividing by the number of different arrangements of k elements.
02:54
Then for the second term, we know that p, let's see here, we have p is the probability of success.
03:07
How do i want to write this? p is probability of success...