# Quod Erat Demonstrandum

## 2013/02/07

### To 5E

Filed under: NSS — johnmayhk @ 5:09 下午
Tags:

Q.1

(a) John 被抽出玩快活角遊戲之機會；
(b) John 成為快活角遊戲主角之機會。

(a)

method 1: multiplication

P(John joins)

= P(JNNNN or NJNNN of NNJNN or NNNJN or NNNNJ)

= P(JNNNN)+P(NJNNN)+P(NNJNN)+P(NNNJN)+P(NNNNJ)

= $\frac{1}{40}\frac{39}{39}\frac{38}{38}\frac{37}{37}\frac{36}{36}+\frac{39}{40}\frac{1}{39}\frac{38}{38}\frac{37}{37}\frac{36}{36}+\frac{39}{40}\frac{38}{39}\frac{1}{38}\frac{37}{37}\frac{36}{36}+\frac{39}{40}\frac{38}{39}\frac{37}{38}\frac{1}{37}\frac{36}{36}+\frac{39}{40}\frac{38}{39}\frac{37}{38}\frac{36}{37}\frac{1}{36}$

= $5\times \frac{1}{40}\frac{39}{39}\frac{38}{38}\frac{37}{37}\frac{36}{36}$

= $\frac{5}{40}$

= $\frac{1}{8}$

method 2: combination

no. of all possible outcomes = $_{40}C_5$

to count the no. of favourable outcomes, since John must be a member in the 5 participants, the remaining 4 are from 39 people, that is we can select 4 people out of 39; thus,

no. of favourable outcomes = $_{39}C_4$

P(John joins)

= $_{39}C_4/_{40}C_5$

= $\frac{39\times 38\times 37\times 36}{4!}\div \frac{40\times 39\times 38\times 37\times 36}{5!}$

= $\frac{5}{40}$

= $\frac{1}{8}$

(b)

method 1

P(John dies)

= P(John joins & John dies)

= $\frac{1}{8}\times \frac{1}{5}$

= $\frac{1}{40}$

method 2

P(John dies) = $\frac{1}{40}$

Q.2
P.6.69 question 30

(a)

P(△)
= P(3 上 1 右)
= P(上上上右)+P(上上右上)+P(上右上上)+P(右上上上)
= $4\times \frac{1}{2}\frac{1}{2}\frac{1}{2}\frac{1}{2}$

(c)
Hint:

Let X be the score. Then X may be 3, 4 or 0.
Let
a = P(X is 3)
b = P(X is 4)
c = P(X is 0)
Then, the expected value of X is $3\times a+4\times b+0\times c$.

Q.3
P.6.68 question 25

The following table shows the departure times of routes 6C and 6F buses from terminus A from 6:00 a.m. to 7:30 a.m.

Jane arrives at terminus A every day between 6:00 a.m. and 7:00 a.m. at random. She gets on a route 6C or 6F bus whichever departs earlier.
Find the probability that Jane gets on a route 6C bus at least once in three successive days.

Jane 可在 6 時至 7 時這 60 分鐘內的任何一刻到車站。

P(gets on a route 6C bus at least once in three successive days)

= 1 – P(can’t get on a route 6C bus in three successive days)

= $1-(1-\frac{1}{5})^3$