Quod Erat Demonstrandum

2013/02/07

To 5E

Filed under: NSS — johnmayhk @ 5:09 下午
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純粹答學生提問,高手見諒。

Q.1
班中 40 人,其中一人是 John。為玩快活角遊戲,先隨意抽出 5 人。再在 5 人中隨意抽出 1 人做主角。求

(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。

johnmayhk-hc

(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

若然抽人的過程是隨機,即 John 或其他同學地位上無異,被抽出之機會均等,故

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.

johnmayhk-2013-02-05
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.

略解

先找某天 Jane 乘坐 6C 巴士之機會。

Jane 可在 6 時至 7 時這 60 分鐘內的任何一刻到車站。

那麼,到達時間是甚麼才可乘坐 6C 巴士?

因每 10 分鐘開出一班車,我們便考慮由 6:00 開始,每 10 分鐘之情況:

如果在 6:00-6:02 到,她會乘坐 6F 巴士;
如果在 6:02-6:10 到,她會乘坐 6C 巴士;
這 10 分鐘內,有 8 分鐘代表她可以乘坐 6C 巴士。

如果在 6:10-6:12 到,她會乘坐 6F 巴士;
如果在 6:12-6:20 到,她會乘坐 6C 巴士;
這 10 分鐘內,有 8 分鐘代表她可以乘坐 6C 巴士。

如此類推。

故 P(某天 Jane 乘坐 6C 巴士) = \frac{8}{10}=\frac{1}{5}

而要求的概率可以考慮:

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

Read also

https://johnmayhk.wordpress.com/2012/02/13/core-math-problem-probability/

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