I have trouble understanding binomial variable and distribution and what it signifies .
A random variable is a variable whose domain is the sample space and connects each outcome with a number.I also know the conditions for a random variable to binomial variable but I don't know what it signifies .Does it tell the total probability of the sample space satisfied by it........for eg
A dice is thrown 7 times.What is the chance that head comes up four times .
Here ,n=7. P (success)=1\2 and P (loss)=1/2.
x-number of tines heads comes
P (X=x) -- 7C4 .(1/2)^4.(1/2)^3
What does this P (X=x) signify....the total cases from the sample space showing 4 heads?If it is , is it like the reverse thing because our X used to connect sample space with number and here we are connecting number with a sample space
$X$ is the random variable. In this case, it can take on any value from $0$ to $7$. We're concerned with finding the probability that the random variable takes on the value $x$, which in this case, is $4$.
$P(X = x)$ refers to the probability that the random variable $X$ is equal to a particular value, denoted by $x$.
This probability, as you mentioned, can be obtained from the binomial distribution.
$$P(X=x)=P(X=4)={7\choose{4}}\cdot {1\over{2}}^4 \cdot {1\over{2}}^3 \approx.273$$
Note that $$\sum_{x=0}^7 P(X=x)=1$$
We can verify this.
$$P(X=0)={7\choose{0}}\cdot {1\over{2}}^0 \cdot {1\over{2}}^7=0.0078125$$ $$P(X=1)={7\choose{1}}\cdot {1\over{2}}^1 \cdot {1\over{2}}^6=0.0546875$$ $$P(X=2)={7\choose{2}}\cdot {1\over{2}}^2 \cdot {1\over{2}}^5=0.1640625$$ $$P(X=3)={7\choose{3}}\cdot {1\over{2}}^3 \cdot {1\over{2}}^4=0.2734375$$ $$P(X=4)={7\choose{4}}\cdot {1\over{2}}^4 \cdot {1\over{2}}^3=0.2734375$$ $$P(X=5)={7\choose{5}}\cdot {1\over{2}}^5 \cdot {1\over{2}}^2=0.1640625$$ $$P(X=6)={7\choose{6}}\cdot {1\over{2}}^6 \cdot {1\over{2}}^1=0.0546875$$ $$P(X=7)={7\choose{7}}\cdot {1\over{2}}^7 \cdot {1\over{2}}^0=0.0078125$$
These do indeed sum to $1$.
Consider a more simple example where we flip a coin twice, and let $X$ denote the number of heads we obtain. We can either get $HH$, $HT$, $TH$, or $TT$. Then $X$ can take on the values $0$, $1$, or $2$, with probabilities $.25$, $.5$, and $.25$, respectively. We let $X$ be arbitrary and let $x$ be the specified value.