find the mean value of x if The probability distribution of a discrete random variable x is given

Question

The probability distribution of a discrete random variable x is $$f (x)= \begin{pmatrix}3 \\ x \end{pmatrix} (1/4)^x (3/4)^{3-x} $$ Find the mean value of x. Construct a cumulative distribution function for f (x). i find out$$ P(X=o) = 0.421875$$ $$ P(X=1) = 0.421875$$ $$ P(X=2) = 0.140625$$ $$ P(X=3) = 0.015625$$ put the value in Binomial distribution.

Answer 1

This discrete random variable follows Binomial distribution, i.e. $$f(x)=\binom{n}{x}p^x(1-p)^{n-x}$$ where $n$ is the number of trials and $p$ is the proportion. What this distribution mean is that when there are $n$ trials and every trial follows $p$ proportion(success rate), the probability that $x$ successes occur is $f(x)$. And for Binomial distribution, the expect(mean) value and variance are $$E(X)=np$$ $$Var(X)=np(1-p)$$ respectively.

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From above, you can know how to calculate the mean value. For cumulative distribution function, it is just the the sum of probability of first k outcomes, which can be expressed as $$F(k;n,p)=P(X\le \lfloor k\rfloor)=\sum_{i=0}^{\lfloor k \rfloor} \binom{n}{i}p^i(1-p)^{n-i}$$ where $\lfloor k\rfloor$ is the largest integer that is less than or equal to $k$. For this problem, I think this form is enough because of the small $n$ that equals to 3.

Answer 2

"The essence of the probability argument is that we have 3 independent trials, and on each trial, outcome 1 occurs with probability p and some other outcome with probability 1−p. help me for finding mean value"

I would say that x = {0,1}, P(0) = 1-p, P(1) = p, E(x) =0*(1-p)+1*p = p

When $X = x_1+x_2+\cdots + x_n$, then $E(X) = E(x_1)+E(x_2)+\cdots+E(x_n)=np$ (generally apply to both dependent and independent experiments).