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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)

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  • $\begingroup$ Well, CDF of Binomial distribution shows the probability that the number of successes is less than or equal to a specific value. So, your CDF will show the probability that you have 0 successes, at most 1 success, and at most 2 successes (which, of course, covers all sample space). You can calculate all of it from the PMF you circled. $\endgroup$
    – Evgeny
    Mar 1, 2020 at 1:06

1 Answer 1

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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.

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

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

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

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  • $\begingroup$ Thank you for the explanation! But I still don't get why possible values for r are specifically from 0 to 2? $\endgroup$ Mar 1, 2020 at 1:07
  • $\begingroup$ Thank you so much, now I understand it. How do we get the value of s then? 3/2 $\endgroup$ Mar 1, 2020 at 1:23
  • $\begingroup$ You want to work out the value of $F$ for all values between 0 and $n$. They just picked $\frac 32$ as an example. $\endgroup$
    – tomi
    Mar 1, 2020 at 1:28

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