问题及解答

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 $X\sim\mathrm{Bin}(10,.3)$
   

(a)

\[
\begin{split}
B(4;10,0.3)&=\sum_{y=0}^{4}b(y;10,0.3)\\
&=b(0;10,0.3)+b(1;10,0.3)+b(2;10,0.3)+b(3;10,0.3)+b(4;10,0.3)\\
&=\binom{10}{0}0.3^0(1-0.3)^{10-0}+\binom{10}{1}0.3^1(1-0.3)^{10-1}+\binom{10}{2}0.3^2(1-0.3)^{10-2}\\
&\quad+\binom{10}{3}0.3^3(1-0.3)^{10-3}+\binom{10}{4}0.3^4(1-0.3)^{10-4}\\
&=0.0282475249+0.1210608210+0.2334744405+0.2668279320+0.2001209490\\
&=0.8497316674
\end{split}
\]

(b)

\[
b(4;10,0.3)=\binom{10}{4}0.3^4(1-0.3)^{10-4}=0.2001209490
\]

(c)

\[
b(6;10,0.7)=\binom{10}{6}0.7^6(1-0.7)^{10-6}=0.2001209490
\]

(d)

\[
\begin{split}
P(2\leqslant X\leqslant 4)&=\sum_{y=2}^{4}b(y;10,0.3)=b(2;10,0.3)+b(3;10,0.3)+b(4;10,0.3)\\
&=\binom{10}{2}0.3^2(1-0.3)^{10-2}+\binom{10}{3}0.3^3(1-0.3)^{10-3}+\binom{10}{4}0.3^4(1-0.3)^{10-4}\\
&=0.2334744405+0.2668279320+0.2001209490\\
&=0.7004233215
\end{split}
\]

(e)

\[
\begin{split}
P(2\leqslant X)&=1-P(X < 2)=1-P(X\leqslant 1)\\
&=1-\sum_{y=0}^{1}b(y;10,0.3)\\
&=1-\bigl[b(0;10,0.3)+b(1;10,0.3)\bigr]\\
&=1-\bigl[0.0282475249+0.1210608210\bigr]\\
&=0.8506916541
\end{split}
\]

(f)  $P(X\leqslant 1)$ when $X\sim\mathrm{Bin}(10,\ .7)$.

\[
\begin{split}
P(X\leqslant 1)&=B(1;10,0.7)=\sum_{y=0}^{1}b(y;10,0.7)\\
&=b(0;10,0.7)+b(1;10,0.7)\\
&=\binom{10}{0}0.7^0(1-0.7)^{10-0}+\binom{10}{1}0.7^1(1-0.7)^{10-1}\\
&=0.0000059049+0.0001377810\\
&=0.0001436859
\end{split}
\]

(g)  $P(2 < X < 6)$ when $X\sim\mathrm{Bin}(10,\ .3)$

\[
\begin{split}
P(2 < X < 6)&=P(3\leqslant X\leqslant 5)\\
&=\sum_{y=3}^{5}b(y;10,0.3)=b(3;10,0.3)+b(4;10,0.3)+b(5;10,0.3)\\
&=\binom{10}{3}0.3^3(1-0.3)^{10-3}+\binom{10}{4}0.3^4(1-0.3)^{10-4}+\binom{10}{5}0.3^5(1-0.3)^{10-5}\\
&=0.2668279320+0.2001209490+0.1029193452\\
&=0.5698682262
\end{split}
\]