Show that $E(X)=np$ when $X$ is a binomial random variable.

[Hint: First express $E(X)$ as a sum with lower limit $x=1$. Then factor out $np$, let $y=x-1$ so that the sum is from $y=0$ to $n-1$, and show that the sum equals $1$.]
 

The variance is $V(X)=np(1-p)=npq$, and the Standard Deviation (SD) of $X$ is $\sigma_X=\sqrt{npq}$ (where $q=1-p$).

The proof can be seen in Question 32. (Where the variance is denoted by $D(X)$.)

(a) Show that $b(x;n,1-p)=b(n-x;n,p)$.

(b) Show that $B(x;n,1-p)=1-B(n-x-1;n,p)$. [Hint: At most $x$ $S$'s is equivalent to at least $(n-x)$ $F$'s.]

 

Twenty percent of all telephones of a certain type are submitted for service while under warranty. Of these, $60\%$ can be repaired whereas the other $40\%$ must be replaced with new units. If a company purchases ten of these telephones, what is the probability that exactly two will end up being replaced under warranty?

[中文]

$20\%$的某種類型的電話在保修期間需要維修。其中$60\%$可以修好，其餘$40\%$必須更換新裝置。如果一家公司購買了其中十部手機，那麼在保修期內，正好有兩部電話被替換的可能性是多少？

[法语]

Vingt pour cent de tous les téléphones d'un certain type sont soumis au service pendant la garantie. De ce nombre, $60\%$ peuvent être réparés alors que les $40\%$ restants doivent être remplacés par de nouvelles unités. Si une entreprise achète dix de ces téléphones, quelle est la probabilité que exactement deux finissent par être remplacés sous garantie?
 

A company that produces fine crystal knows from experience that $10\%$ of its goblets have cosmetic flaws and must be classified as "seconds."

• (a) Among six randomly selected goblets, how likely is it that only one is a second?
• (b) Among six randomly selected goblets, what is the probability that at least two are seconds?
• (c) If goblets are examined one by one, what is the probability that at most five must be selected to find four that are not seconds?
   

Definition. Given a binomial experiment consisting of $n$ trials, the binomial random variable(二项随机变量) $X$ associated with this experiment is defined as

\[
X=\text{the number of S's among the}\ n\ \text{trials}
\]

Here we use S to denote the outcome H(heads) and F to denote the outcome T(tails).

We write $X\sim\mathrm{Bin}(n,p)$ to indicate that $X$ is a binomial rv based on $n$ trials with success probability $p$.

We denote the pmf of a binomial rv $X$ by $b(x;n,p)$. The cdf is denoted by
\[
P(X\leqslant x)=B(x;n,p)=\sum_{y=0}^{x}b(y;n,p)\quad x=0,1,2,\ldots,n.
\]

Here

\[
b(x;n,p)=\begin{cases}
\binom{n}{x}p^x(1-p)^{n-x},&x=0,1,2,\ldots,n\\
0,&\text{otherwise}.
\end{cases}
\]

Calculate the following probabilities:

• (a) $B(4; 10, .3)$
• (b) $b(4; 10, .3)$
• (c) $b(6; 10, .7)$
• (d) $P(2\leqslant X\leqslant 4)$ when $X\sim\mathrm{Bin}(10,.3)$
• (e) $P(2\leqslant X)$ when $X\sim\mathrm{Bin}(10,.3)$
• (f) $P(X\leqslant 1)$ when $X\sim\mathrm{Bin}(10,.7)$
• (g) $P(2 < X < 6)$ when $X\sim\mathrm{Bin}(10,.3)$
   

Suppose $E(X)=5$ and $E[X(X-1)]=27.5$. What is

• $E(X^2)$ ? [Hint: $E[X(X-1)]=E[X^2-X]=E(X^2)-E(X)$.]
• $V(X)$ ?
• The general relationship among the quantities $E(X)$, $E[X(X-1)]$, and $V(X)$.

Using the definition of variance, prove that
\[
V(aX+b)=a^2\cdot\sigma_X^2
\]

[Hint] With $h(X)=aX+b$, $E[h(X)]=a\mu+b$ where $\mu=E(X)$.

The expected value of $X$ measures where the probability distribution is centered. We will use the variance of $X$ to measure the amount of variability in (the distribution of) $X$.

Let $X$ have pmf $p(x)$ and expected value $\mu$. Then the variance of $X$ ($X$ 的方差), denoted by $V(X)$ or $\sigma_X^2$, or just $\sigma^2$, is defined by
\[
V(X):=\sum_{D}(x-\mu)^2\cdot p(x)=E\bigl[(X-\mu)^2\bigr]
\]
Prove that
\[
V(X)=E(X^2)-(E(X))^2.
\]

i.e.,

\[V(X)=\sigma^2=\biggl[\sum_{D}x^2\cdot p(x)\biggr]-\mu^2\]
 

Let $X$ have pmf
\[
p(x)=\begin{cases}
\frac{k}{x^2}, & x=1,2,3,\ldots\\
0, & \text{otherwise}
\end{cases}
\]
where $k$ is chosen so that $\sum_{x=1}^{\infty}\frac{k}{x^2}=1$. So the expected value of $X$ is
\[
\mu=E(X)=\sum_{x=1}^{\infty}x\cdot\frac{k}{x^2}=k\sum_{x=1}^{\infty}\frac{1}{x}
\]
What is the value of $k$?

Answer: The value of $k$ is equal to $\frac{6}{\pi^2}$.

Please see the Question 20, or search $\frac{\pi^2}{6}$ in this site.
 

Starting at a fixed time, we observe the gender of each newborn child at a certain hospital until a boy ($B$) is born. Let $p=P(B)$, assume that successive births are independent, and define the rv $X$ by $X=$ number of births observed. Then
\[
\begin{aligned}
p(1)&=P(X=1)=P(B)=p,\\
p(2)&=P(X=2)=P(GB)=P(G)\cdot P(B)=(1-p)p,\\
\end{aligned}
\]
and
\[
p(3)=P(X=3)=P(GGB)=P(G)\cdot P(G)\cdot P(B)=(1-p)^2 p.\\
\]
Continuing in this way, write the general formula for the pmf $p(x)$. And compute the following

• the cdf $F(x)$.
• $E(X)$
   

Remark. Here rv stands for random variable(随机变量), pmf stands for probability mass function (or probability density function 概率密度函数), and cdf stands for cumulative distribution function(累积分布函数).