Cumulative distribution function of a discrete random variable

Question

Can someone please explain how we got the pmf of the X ~ Bin (2, 2/3) ? (I understand that it is the Binomial distribution but where did we get the 0, 1, 2 from)

Answer 1

The Binomial distribution is used to model the situation when we perform a trial a certain number of times. We call one outcome of the trial a 'success' and the alternative to this is a 'failure.' We then count the number of trials. The Binomial distribution gives us the probability of a particular number of successes.

Usually $n$ is the number of trials, $p$ is the probability of success, $1-p$ or $q$ is the probability of failure. $r$ is used for a particular number of successes.

Example: Janice plays 4 games of chess against Jack. She has a probability of 0.8 of winning a game.

In this example $n=4$ and $p=0.8$

Janice can win 0 games, 1 game, 2 games, 3 games or 4 games. These correspond to $r=0, r=1, r=2, r=3, r=4$.

The probability that Janice wins 3 games is given by $P(X=3)$. To find this we substitute $n=4,p=0.8, r=3$ into the formula $P(X=r)={}_nC_rp^r(1-p)^{n-r}$

$P(X=3)={}_4C_30.8^30.2^1=0.4096$

For $X$ ~ $Bin(n,p)$ we have $P(X=r)={}_nC_rp^r(1-p)^{n-r}$

In your case the possible values of $r$ are 0, 1, 2.

$P(X=0)={}_2C_0(\frac23)^0(\frac13)^2=1 \times 1 \times \frac 19=\frac 19$

$P(X=1)={}_2C_1(\frac23)^1(\frac13)^1=2 \times \frac 23 \times \frac 13=\frac 49$

$P(X=2)={}_2C_2(\frac23)^2(\frac13)^0=1 \times \frac 49 \times 1=\frac 49$