MHB Consider the following probability mass function of a random variable x

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The provided probability mass function, $p(x) = p^x (1-p)^{1-x}$ for $x=0.1$ and $0$ otherwise, is not valid as it does not sum to one. Since $x$ only takes the value of 0.1, $p(0.1)$ should equal 1, which is not achievable with the given function. Consequently, the mean of the random variable $X$ is 0.1, and the variance is 0, as there is no deviation from the mean. This indicates that the function fails to meet the criteria for a proper probability mass function.
dylbester
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$p(x) ={p}^{x}*{(1-p)}^{1-x}$
for $x=0.1$
$0$ otherwise

Where $p$ is such that $0<=p<=1$

Question:
Find the mean and variance of $X
 
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dylbester said:
$p(x) ={p}^{x}*{(1-p)}^{1-x}$
for $x=0.1$
$0$ otherwise

Where $p$ is such that $0<=p<=1$

Question:
Find the mean and variance of $X$

Hi dylbester! Welcome to MHB! (Smile)

That doesn't seem to be a proper probability mass function.
The sum of $p(x)$ for all values of $x$ is supposed to be $1$.
Since there is only one value for $x$, which is $0.1$, that should mean that $p(0.1)=1$, but that is not the case for any choice for $p$.

Anyway, if there is only one value for $x$, the mean will have to be that value.
And the deviation of that mean can only be $0$, so the variance will then have to be $0$.
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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