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

$$P(A\cap B)=P(B | A)\cdot P(A)$$ $$P(A\cup B)=P(A)+P(B)-P(A\cap B)$$

PRACTICE PROBLEMS FOR EXERCISE SET 1

1. A bowl contains two Snickers bars, two Mars bars, and one packet of sunflower seeds. I pick two items out of the bowl at random with replacement. Find the chance that

a) I get the sunflower seeds: ___________

b) I get two bars of the same kind: ___________ (Assume that the packet of sunflower seeds is not a bar.)

c) I get two bars of different kinds: ___________

d) I get at least one Snickers bar: ____________

2. Repeat Exercise 1 if the two items are drawn at random without replacement.

3. A die will be rolled five times. Pick one of the options (i)-(iii) to show which of the two events below is more likely.

(A) The same face appears on all five rolls.

(B) An ace (the face with one spot) appears on all five rolls.

(i) A (ii) B (iii) The two events have the same chance.

4. There are 26 letters in the English alphabet; $A, E, I, O$, and $U$ are vowels; the rest are consonants. A box contains the letters $O, R, A, N, G, E, S$. Three letters are drawn at random with replacement. Find the chance that

a) all the letters drawn are consonants

b) not all the letters drawn are vowels

c) at least one of the letters drawn is a vowel

5. Three draws are made at random without replacement from a box which contains 4 red tickets, 4 blue tickets, and 2 green tickets. The conditional chance that all three draws are the same color, given that the first two are red, is

(i) $\frac{4}{10}\times\frac{3}{9}\times\frac{2}{8}$

(ii) $\frac{2}{8}$ (iii) other (specify): ______________

6. A standard deck consists of 52 cards of which 26 are red and 26 black. Two of the red cards are aces; two of the black cards are aces. Two cards are dealt from a well shuffled standard deck (this means “at random without replacement”). Find the chance that

a) each of the two cards is an ace or red

b) both the cards are aces or both are red

7. On the day before the final exam, each of the two TAs (Teaching Assistants) is asked to pick one of the following slots for office hours: 10-11, 11-12, 12-1, 1-2, 2-3, 3-4, and 4-5. The TAs pick at random with replacement from the slots. The chance that the 12-1 hour gets picked is

(i) $\frac{1}{7}+\frac{1}{7}$

(ii) other (specify): _______________

8. A box contains the letters $O, R, A, N, G, E, S$. Letters are drawn one by one at random without replacement until the box is empty. Find the chance that

a) the letter $R$ appears on the last draw: _____________

b) the letter $R$ appears on the last draw, given that a vowel appeared on the first draw:______________

c) vowels appear on the first two draws: ______________

d) there is one vowel and one consonant among the first two letters drawn: _____________

Solutions:

1. Because it is replacement so the total outcome is $5\times5=25$ possibilities.

a) If no sunflower seeds the total possibilities should be $4\times4=16$, so $$P(\text{get the sunflower seeds})=\frac{25-16}{25}=0.36$$

b) The possible pairs of Snickers bars (denotes $S$): $S_1, S_2;\ S_2, S_1;\ S_1, S_1;\ S_2, S_2$, similarly, there are 4 possible pairs of Mars bars. Therefore, $$P(\text{two bars of the same kind})=\frac{4+4}{25}=0.32$$

c) There are 8 possible pairs of different kinds (Mars denotes $M$): $S_1, M_1;\ M_1, S_1;\ S_1, M_2;\ M_2, S_1;\ S_2, M_1;\ M_1, S_2;\ S_2, M_2;\ M_2, S_2$. So $$P(\text{two bars of different kind})=\frac{8}{25}=0.32$$

d) Similar to a), $$P(\text{at least one Snickers bar})=1-P(\text{no Snickers bar})=1-\frac{9}{25}=0.64$$

2. There are $5\times 4=20$ possible pairs if without replacement.

a) $$P(\text{get the sunflower seeds})=1-P(\text{no sunflower seeds})=1-\frac{4\times3}{20}=0.4$$

b) There are 4 possible pairs: $S_1, S_2;\ S_2, S_1;\ M_1, M_2;\ M_2, M_1$. Hence $$P(\text{two bars of the same kind})=\frac{4}{20}=0.2$$

c) It is the same to Exercise 1.c, so $$P(\text{two bars of different kind})=\frac{8}{25}=0.32$$

d) There are 6 possible pairs without Snickers bars (seeds denotes $Se$): $M_1, M_2;\ M_2, M_1;\ M_1, Se;\ Se, M_1;\ M_2, Se, Se, M_2$. Therefore $$P(\text{at least one Snickers bar})=1-P(\text{no Snickers bar})=1-\frac{6}{20}=0.7$$

3. $$P(A)=(\frac{1}{6})^{5}\times6=(\frac{1}{6})^{4},\ P(B)=(\frac{1}{6})^{5}$$. Therefore, $P(A)>P(B)$.

4. a) $$P(\text{all three letters are consonants})=(\frac{4}{7})^3=\frac{64}{343}$$

b) $$P(\text{not all the letters are vowels})=1-P(\text{all the letters are vowels})=1-(\frac{3}{7})^3=\frac{316}{343}$$

c) $$P(\text{at least one letter is vowel})=1-P(\text{no letter is vowel})=1-(\frac{4}{7})^3=\frac{279}{343}$$

5. $$P(\text{all same color | first two red})=\frac{P(\text{first two red & all same color})}{P(\text{first two red})}$$ $$=\frac{P(\text{all of three are red})}{P(\text{first two red})}=\frac{\frac{4}{10}\times\frac{3}{9}\times\frac{2}{8}}{\frac{4}{10}\times\frac{3}{9}}=\frac{1}{4}$$

6. a) $$P(\text{each of the two cards is ace or red})=\frac{28}{52}\times\frac{27}{51}=\frac{63}{221}$$

b)$$P(\text{both are ace or both are red})$$ $$=P(\text{both are ace})+P(\text{both are red})-P(\text{both are red ace})$$ $$=\frac{4}{52}\times\frac{3}{51}+\frac{26}{52}\times\frac{25}{51}-\frac{2}{52}\times\frac{1}{51}=\frac{55}{221}$$

7. $$P(\text{12-1 hour gets picked})=1-P(\text{12-1 hour does not get picked})$$ $$=1-(\frac{6}{7})^2=\frac{13}{49}$$

8. a) It appears equally on any draw, so $$P(\text{R appears on the last draw})=\frac{1}{7}$$

b) $$P(\text{R appears on the last draw | the first draw is a vowel})$$ $$=\frac{P(\text{the first is a vowel & the last is R})}{P(\text{the first draw is a vowel})}=\frac{\frac{3}{6}\times\frac{1}{7}}{\frac{3}{7}}=\frac{1}{6}$$

c) $$P(\text{the first two are vowels})=\frac{3}{7}\times\frac{2}{6}=\frac{1}{7}$$

d) $$P(\text{one consonant & one vowel on the first two draws})$$ $$=P(\text{first is vowel & second is consonant})+P(\text{first is consonant & second is vowel})$$ $$=\frac{3}{7}\times\frac{4}{6}+\frac{4}{7}\times\frac{3}{6}=\frac{4}{7}$$

GRADED EXERCISE SET 1

PROBLEM 1

There are six chairs in a row. Person A picks one of the chairs at random. Then Person B picks a chair at random from the five chairs that remain. Find the chance that they pick chairs that are next to each other.

Solution

The total number of possibilities is $6\times5=30$. And the possibilities of paired chairs are $5\times2=10$. Therefore $$P(\text{two paired chairs})=\frac{10}{30}=\frac{1}{3}$$

PROBLEM 2

A standard deck consists of 52 cards, distributed as follows: There are 13 cards in each of 4 suits: hearts, diamonds, spades, and clubs. Hearts and diamonds are red. Spades and clubs are black. In each suit there are 13 ranks: Ace, 2, 3, 4, 5, 6, 7, 8, 9, 10, Jack, Queen, King. In Exercises 2A through 2E, assume that cards are dealt from a well-shuffled deck; that is, cards are dealt at random without replacement.

2A The "face" cards are those that have a person’s face on them: Jack, Queen, or King. One card is dealt. Find the chance that it is a face card.

2B One card is dealt. Find the chance that its color is black and its rank is an even number.

2C One card is dealt. Find the chance that its color is red or its rank is an even number.

2D A poker hand (5 cards) is dealt. Find the chance that the cards are all hearts.

2E A poker hand (5 cards) is dealt. Find the chance that the cards are all of the same suit. [This is sometimes called a “flush.”]

Solution

2A) $$P(\text{face cards})=\frac{12}{52}=\frac{3}{13}$$

2B) $$P(\text{black & even number})=\frac{10}{52}=\frac{5}{26}$$

2C) $$P(\text{red or even number})=P(\text{red})+P(\text{even number})-P(\text{red & even number})$$ $$=\frac{26}{52}+\frac{20}{52}-\frac{10}{52}=\frac{9}{13}$$

2D) $$P(\text{all hearts in a hand})=\frac{13}{52}\times\frac{12}{51}\times\frac{11}{50}\times\frac{10}{49}\times\frac{9}{48}\doteq0.0004951981$$

2E) Very similar to 2D, there are 4 suits here. $$P(\text{flush})=\frac{52}{52}\times\frac{12}{51}\times\frac{11}{50}\times\frac{10}{49}\times\frac{9}{48}\doteq0.001980792$$

PROBLEM 3

A die is a six-sided figure; each side shows a number of spots; the numbers of spots on the six sides are 1, 2, 3, 4, 5, and 6. Unless the problem says otherwise, you can assume that the die is fair. That is, every time the die is rolled, each face has chance 1/6 of appearing, regardless of what happens on other rolls.

3A A die is rolled three times. Find the chance that the face with 4 spots appears at least once.

3B A die is rolled three times. Find the chance that three different faces appear.

Solution

3A) $$P(\text{4 spots appears at least once})=1-P(\text{4 spots doesn‘t appear})=1-(\frac{5}{6})^4=\frac{91}{216}$$

3B) $$P(\text{three different faces appear})=\frac{6}{6}\times\frac{5}{6}\times\frac{4}{6}=\frac{5}{9}$$

PROBLEM 4

There are 5 people in a room. Make the following simplifying assumptions about their birthdays: Ignore leap years; assume 365 days in the year. For each person, assume that the chance he/she is born on a specified day is 1/365. Assume that the chance that someone is born on a specified day is not affected by other people’s birthdays.

4A Find the chance that the 5 people have 5 different birthdays.

4B Find the chance of a "match": that is, at least two people in the room have the same birthday.

4C Repeat 4B for a room in which there are 400 people.

Solution

4A) $$P(\text{5 different birthdays})=\frac{365}{365}\times\frac{364}{365}\times\frac{363}{365}\times\frac{362}{365}\times\frac{361}{365}\doteq0.9728644$$

4B) $$P(\text{at least two people have the same birthday})=1-P(\text{5 different birthdays})$$ $$=1-\frac{365}{365}\times\frac{364}{365}\times\frac{363}{365}\times\frac{362}{365}\times\frac{361}{365}\doteq0.02713557$$

4C) Note that $400>365$ so the chance is $1$.

PROBLEM 5

There are 26 letters in the English alphabet. Letters are picked one by one at random, so that each letter has the same chance of appearing as any other letter, regardless of which letters have appeared or not appeared.

5A Six letters are picked. Find the chance that the sequence that appears is $RANDOM$, in that order.

5B Six letters are picked. Find the chance that they can be arranged to form the word $RANDOM$.

Solution

5A) $$P(\text{ordered RANDOM})=(\frac{1}{26})^6=\frac{1}{165765600}$$ Or $$P(\text{ordered RANDOM})=\frac{1}{26^6}$$

5B) $$P(\text{unordered RANDOM})=\frac{6}{26}\times\frac{5}{26}\times\frac{4}{26}\times\frac{3}{26}\times\frac{2}{26}\times\frac{1}{26}\doteq2.330732\times10^{-6}$$ Or $$P(\text{unordered RANDOM})=\frac{6!}{26^6}$$

PROBLEM 6

A two-day conference is going to be held in a city. The leading newspaper in the city says there is a 10% chance of rain on the first day of the conference and a 40% chance of rain on the second day.

6A For your answer, pick just one option, please: it should be the best of the available options, based on the information given in the problem. If the newspaper’s figures are correct, the chance that it rains in the city during the conference is at least__________

6B For your answer, pick just one option, please: it should be the best of the available options, based on the information given in the problem. If the newspaper’s figures are correct, the chance that it rains in the city during the conference is at most__________

Solution

$P_i$ denotes the $i\text{th}$ day rains.

6A) $$P(\text{rains during the conference})=P(\text{first day rains or second day rains})$$ $$=P_1\cup P_2\ge\text{max}\{P_1,P_2\}=0.4$$ Therefore, the chance that it rains in the city during the conference is at least $40\%$.

6B) $$P(\text{rains during the conference})=P(\text{first day rains or second day rains})$$ $$=P_1\cup P_2=P_1+P_2-(P_1\cap P_2)\le P_1+P_2=0.5$$ Therefore, the chance that it rains in the city during the conference is at most $50\%$.

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 (1.1-1.4)