加州大学伯克利分校Stat2.2x Probability 概率初步学习笔记: Section 1 The Two Fundamental Rules (1.5-1.6)

Stat2.2x Probability（概率）课程由加州大学伯克利分校（University of California, Berkeley）于2014年在edX平台讲授。

PDF笔记下载（Academia.edu）

Summary

Bayes Theorem $$P(A_i|B)=\frac{P(B|A_i)\cdot P(A_i)}{\sum_{j}P(B|A_j)\cdot P(A_j)}$$ where $$P(B)=\sum_{j}P(B|A_j)\cdot P(A_j)$$

GRADED EXERCISE SET 2

PROBLEM 1

Cards are dealt from a well shuffled standard deck. Find the chance that:

1A the first card is not a heart

1B the 10th card is the ace of spades

1C the 10th card is the ace of spades, given that the 50th card is the ace of diamonds

1D the 32nd card is red, given that the 51st and 52nd cards are red

1E the 51st card is red, given that the 32nd and 52nd cards are red

Solution

1A) $$P(\text{first card is not heart})=1-P(\text{first card is heart})=1-\frac{13}{52}=\frac{3}{4}$$

1B) Each card is randomly drawn thus the chance of a specific card is equivalent: $$P(\text{10th is ace of spades})=\frac{1}{52}$$

1C) There are 51 cards except the given card ace of diamonds, thus $$P(\text{10th is ace of spades}\ |\ \text{50th is ace of diamonds})=\frac{1}{51}$$

1D) $$P(\text{32nd is red}\ |\ \text{51st and 52nd are red})=\frac{26-2}{52-2}=\frac{12}{25}$$

1E) $$P(\text{51st is red}\ |\ \text{32nd and 52nd are red})=\frac{26-2}{52-2}=\frac{12}{25}$$

PROBLEM 2

A person is selected at random from a population that has the following characteristics: 65% of the people are men; the others are women 12% of the men are smokers 7% of the women are smokers

2A Find (correct to four decimal places) the chance that the selected person is a non-smoker, given that the person is a man.

2B Find (correct to four decimal places) the chance that the selected person is a woman who smokes.

2C Find (correct to four decimal places) the chance that the selected person is a smoker.

2D Given that the selected person is a smoker, what is the chance that the person is a woman? Please give your answer correct to four decimal places.

2E Find (correct to four decimal places) the chance that the selected person is a man or a non-smoker.

Solution

2A) $$P(\text{non-smoker}\ |\ \text{man})=1-0.12=0.88$$

2B) $$P(\text{woman & smoker})=(1-0.65)\times0.07=0.0245$$

2C) $$P(\text{smoker})=P(\text{smoker & man})+P(\text{smoker & woman})$$ $$=0.65\times0.12+(1-0.65)\times0.07=0.1025$$

2D) $$P(\text{woman}\ |\ \text{smoker})=\frac{P(\text{woman & smoker})}{P(\text{smoker})}=\frac{0.35\times0.07}{0.1025}=0.2390244$$

2E) $$P(\text{man or non-smoker})=P(\text{man})+P(\text{non-smoker})-P(\text{man & non-smoker})$$ $$=0.65+(1-0.1025)-0.65\times0.88=0.9755$$

PROBLEM 3

I am with a group of five of my friends. A box contains 8 pieces each of milk chocolate, white chocolate, and dark chocolate. The box is passed around the six of us, with each person taking 4 pieces. Assume that each person chooses at random without replacement from the available pieces. I am the last person to whom the box is passed. Find the chance that I pick 4 dark chocolates.

Solution

It it no difference between "I" am the first and "I" am the last. $$P=\frac{C_{8}^{4}}{C_{24}^{4}}=\frac{35}{5313}=0.006587615$$ $C_{n}^{m}=\frac{n!}{m!\cdot(n-m)!}$ and its R function is $\text{choose(n, m)}$.

PROBLEM 4

An urn contains 10 white balls and 5 blue balls. Draws are made repeatedly from the urn as follows. On each draw, a ball is drawn and its color noted; then it is replaced in the urn along with 3 more balls of its color. For example, if you draw a white ball the first time, then you will put it back in the urn along with 3 more white balls. So you will end up with 13 white balls and 5 blue balls.

4A Find the chance that the first ball drawn is blue.

4B Find the chance that the second ball drawn is blue, given that the first ball drawn is blue.

4C Find the chance that the second ball drawn is blue.

4D Find the chance that the first ball drawn is blue, given that the second ball drawn is blue.

Solution

4A) $$P(\text{first is blue})=\frac{5}{10+5}=\frac{1}{3}$$

4B) If the first is blue then it would be 10 white balls and 8 blue balls before the second draw. Hence $$P(\text{second is blue}\ |\ \text{first is blue})=\frac{8}{10+8}=\frac{4}{9}$$

4C) $$P(\text{second is blue})=P(\text{second is blue & first is white})+P(\text{second is blue & first is blue})$$ $$=P(\text{first is white})\cdot P(\text{second is blue}\ |\ \text{first is white})$$ $$+P(\text{first is blue})\cdot P(\text{second is blue}\ |\ \text{first is blue})$$ $$=\frac{10}{15}\times\frac{5}{13+5}+\frac{5}{15}\times\frac{8}{10+8}=\frac{1}{3}$$

4D) $$P(\text{first is blue}\ |\ \text{second is blue})=\frac{P(\text{first & second are blue})}{P(\text{second is blue})}=\frac{\frac{5}{15}\times\frac{8}{10+8}}{\frac{1}{3}}=\frac{4}{9}$$