MATH 431 第一章作业

1.12
a) Let $X_i$ be the event that we need i rolls and $A$ be the event of at most 3 trials, So $A=X_1 \cup X_2\cup X_3$ , which $X_1,X_2,X_3$ are obviously disjoint, so$$P(A)=P(X_1)+P(X_2)+P(X_3)= \frac{1}{6} + \frac{5}{6} \cdot \frac{1}{6}+\frac{5}{6} \cdot \frac{5}{6} \cdot \frac{1}{6}= \frac{91}{216}$$
b) Let $B$ be the event of even number needed, so $$\begin{align*}
P(B) &= P(X_2) + P(X_4) + \cdots = \sum_{i=1}^{\infty} P(2i) \
&= \sum_{i=1}^{\infty} \left(\frac{5}{6}\right)^{2i-1} \cdot \frac{1}{6} \
&= \frac{5}{36} \sum_{i=0}^{\infty}\left(\frac{25}{36}\right)^i = \frac{5}{11}
\end{align*}$$
1.14

Assume $P(AB)<0.1$, we get $$P(A\cup B)=P(A)+P(B)-P(AB)>1$$
which disobeys the rules of a probability measure.

Assume $P(AB)>0.4$ ,we get
$$P(A)=P(A\cap B)+P(A\cap B^c)>0.4$$
which could not satisfy this problem assumption.

So $0.1\leq P(AB)\leq 0.4$ .

1.16

for five times, $X\in {-5,-3,-1,1,3,5}$, for $X=\frac{i+5}{2}$ , we have $$P(X)=\binom{5}{i}\cdot0.5^5=\frac{15}{(5-i)!\cdot i!\cdot 4}$$

1.18
We could choose 4 words including 2 words with 4 length,one with 3 length and one with 5 length.
So X=2,3 or 4.
$P(X=i)=\frac{#X=i}{16}$,so $P(X=3)=\frac{3}{16}$,$P(X=4)=\frac{5+3}{16}=\frac{1}{2},$$P(X=5)=\frac{5}{16}$

1.32
Let the event be A, so$$P(A)=\frac{#A}{#\Omega}$$
$#\Omega=\binom{52}{5}$ and $#A=13\times1\times1\times\binom{4}{3}\times12\times1\binom{4}{2}$
So $P(A)=\frac{6}{4165}$

1.34

Let A be the event and assume the distance from (0,0) is r.
So $$P(A)=P(r<\frac{2}{3})=\frac{\pi (2/3)^2}{\pi\cdot 1^2}=\frac{4}{9}$$
1.36
a)
$$P(X\in(a,b))=\frac{#A}{#\Omega}=\frac{b-a}{1}=b-a$$
b) Draw lines of y=x+0.25 and y=x-0.25, let S be the surrounded area.
So
$$P(|X-Y|<\frac{1}{4})=\frac{S}{S_\Omega}=\frac{7}{16}$$

1.40

$$#\Omega=4^4=256$$
Let G = {exactly 2 green balls}, R = {exactly 2 red balls}, Y = {exactly 2 yellow balls}, W = {exactly 2 white balls}, A be the event we want.
So
$$
\begin{align*}
#A &= #G + #Y+ #R + #W - #(GY) - #(GR) - #(GW) - #(YR) - #(YW) - #(RW) \
&= 54+54+54+54 - 6 \times 6 \
&= 180
\end{align*}$$
So $P(A)=\frac{180}{256}=\frac{45}{64}$