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。
Summary
- Standard Error
The standard error of a random variable $X$ is defined by $$SE(X)=\sqrt{E((X-E(X))^2)}$$ $SE$ measures the rough size of the chance error in $X$: roughly how far off $X$ is from $E(X)$. - Standard Deviation
The standard deviation of a list of numbers is $$SD=\sqrt{E((x-\mu)^2)}$$ where $\mu=E(x)$. $SD$ measures the rough size of the deviations: roughly how far off the numbers are from the average. - $SE$ of the Sum of the Draws
$n$ draws at random with replacement from a box of numbered tickets, the standard error of the sum of the draw is $$SE=\sqrt{\text{number of draws}}\cdot(SD\ \text{of the box})=\sqrt{n}\cdot\sigma$$ where $\sigma=\sqrt{E((x-\mu)^2)}$ - Chebychev‘s Inequality
The probability that $X$ is $k$ or more $SEs$ away from $E(X)$ is at most $\frac{1}{k^2}$, that is $$P(X\ \text{is outside the interval}\ E(X)\pm k\cdot SE(X))\leq\frac{1}{k^2}$$ For instance, $$P(X\ \text{is inside the interval}\ E(X)\pm2\cdot SE(X))\geq1-\frac{1}{2^2}=\frac{3}{4}$$ - De Moivre - Laplace Theorem
Fix any $p$ strictly between $0$ and $1$. As the number of trials $n$ increases, the probability histogram for the binomial distribution looks like the normal curve with mean $\mu=n\cdot p$ and $SD=\sqrt{n\cdot p\cdot(1-p)}$. - Central Limit Theorem
Let $X_1, X_2, \ldots, X_n$ be independent and identically distributed, each with expected value $\mu$ and standard error $\sigma$. Let $S_n=X_1+X_2+\ldots+X_n$. Then for large $n$, the probability distribution of $S_n$ is approximately normal with mean $n\mu$ and standard deviation $\sqrt{n}\sigma$, no matter what the distribution of each $X_i$. - Normal Approximation of Binomial Distribution
$$\mu=n\cdot p, SE=\sqrt{n\cdot p\cdot(1-p)}$$ $$Z_1=\frac{X_1-\mu}{SE}, Z_2=\frac{X_2-\mu}{SE}$$ $$P(X_1\leq X\leq X_2)=\text{Area under the standard normal curve between}\ X_1,X_2 $$ R code:mu = n * p; se = sqrt(n * p * (1 - p)) z1 = (x1 - mu) / se; z2 = (x2 - mu) / se pnorm(z2) - pnorm(z1)
PRACTICE
PROBLEM 1
In 6000 rolls of a die, approximately what is the chance of getting between 950 and 1050 sixes (inclusive)?
Solution
Binomial distribution $n=6000, k=950:1050, p=1/6$: $$P(\text{between 950 and 1050 sixes})$$ $$=\sum_{k=950}^{1050}C_{6000}^{k}(\frac{1}{6})^k\cdot(\frac{5}{6})^{6000-k}\doteq0.9198021$$ R code:
sum(dbinom(x = 950:1050, size = 6000, p = 1/6)) [1] 0.9198021
Alternatively, using Normal Approximation: $$\mu=np=6000\times\frac{1}{6}=1000$$ $$SE=\sqrt{n\cdot p\cdot(1-p)}\doteq28.86751$$ $$Z_1=\frac{950-1000}{SE}, Z_2=\frac{1050-1000}{SE}$$ $$P(\text{between 950 and 1050 sixes})$$ $$=\text{Area under the standard normal curve between}\ Z_1\ \text{and}\ Z_2$$ $$=0.9167355$$ R code:
n = 6000; p = 1/6 mu = n * p; se = sqrt(n * p * (1 - p)) z1 = (950 - mu) / se; z2 = (1050 - mu) / se pnorm(z2) - pnorm(z1) [1] 0.9167355
PROBLEM 2
The “column” bet in roulette pays 2 to 1 and there are 12 chances in 38 to win. Suppose you bet \$1 100 times independently on a column. Find
a) the expected number of times you win
b) the SE of the number of times you win
c) the expected value of your net gain
d) the $SE$ of your net gain
e) the chance that you come out ahead
Solution
2a) $$E(\text{times of win})=100\times\frac{12}{38}\doteq31.57895$$
2b) $$SE=\sqrt{n\cdot p\cdot(1-p)}=\sqrt{100\times\frac{12}{38}\times\frac{26}{38}}\doteq4.648295$$
2c) $$E(\text{net gain})=100\times(2\times\frac{12}{38}+(-1)\times\frac{26}{38})\doteq-5.263158$$ Alternatively, Let $W$ be the number of wins and $X$ the net gain. Then $$X=2\cdot W-1\cdot(100-W)=3\cdot W-100$$ $$E(X)=3\cdot E(W)-100=3\times31.579895-100=-5.26315$$
2d) Because $SE=\sqrt{n}\sigma$ and $$n=100, \mu=2\times\frac{12}{38}+(-1)\times\frac{26}{38}=-\frac{1}{19}$$ $$\sigma=\sqrt{E((X-\mu)^2)}=\sqrt{(2+\frac{1}{19})^2\times\frac{12}{38}+(-1+\frac{1}{19})^2\times\frac{26}{38}}\doteq1.394489$$ Thus $$SE=\sqrt{n}\sigma\doteq13.94489$$ Alternatively, $$SE(X)=3\cdot SE(W)=3\times4.6483=13.945$$
2e) $X > 0 \Rightarrow W > \frac{100}{3}\Rightarrow W \geq 34$. Binomial distribution $n=100, k=34:100, p=12/38$: $$\sum_{k=34}^{100}C_{100}^{k}\cdot(\frac{12}{38})^k\cdot(\frac{26}{38})^{100-k}\doteq0.3357928$$ R code:
sum(dbinom(x = 34:100, size = 100, p = 12/38)) [1] 0.3357928
PROBLEM 3
Find the normal approximation to the chance of getting 43 heads in 100 tosses of a coin.
Solution
Normal Approximation: $$\mu=100\times0.5=50, SE=\sqrt{n\cdot p\cdot(1-p)}=\sqrt{100\times0.5\times0.5}=5$$ $$Z_1=\frac{42.5-50}{5}, Z_2=\frac{43.5-50}{5}$$ $$P(\text{getting 43 heads in 100 tosses of a coin})\doteq0.02999328$$ R code:
n = 100; p = 1/2 mu = n * p; se = sqrt(n * p * (1 - p)) z1 = (42.5 - mu) / se; z2 = (43.5 - mu) / se pnorm(z2) - pnorm(z1) [1] 0.02999328
Binomial distribution (exact value): $$C_{100}^{43}\times(\frac{1}{2})^{100}\doteq0.03006864$$ R code:
dbinom(x = 43, size = 100, p = 1/2) [1] 0.03006864
Therefore the normal approximation is excellent.
EXERCISE 4
PROBLEM 1
A random variable $W$ has the probability distribution
value 1 2 3 4
probability 0.5 0.25 0.125 0.125
(For those of you who are interested, this is the geometric $p=0.5$ “killed” at 4. $W$ is the number of times I toss a coin if I follow this rule: I’ll toss the coin till I get the first head, but I’ll stop after 4 tosses even if I haven’t got a head by that time.)
1A Find $E(W)$
1B Find $SE(W)$
Solution
1A) $$E(W)=1\times0.5+2\times0.25+3\times0.125+4\times0.125=1.875$$
1B) $$SE(W)=\sqrt{E[(W-E(W))^2]}$$ $$=\sqrt{(1-1.875)^2\times0.5+(2-1.875)^2\times0.25+(3-1.875)^2\times0.125+(4-1.875)^2\times0.125}$$ $$\doteq1.053269$$ R code:
v = 1:4; p = c(.5, .25, .125, .125) mu = sum(v * p) sqrt(sum((v - mu) ^ 2 * p)) [1] 1.053269
PROBLEM 2
A true-false test consists of 20 questions, each of which has one correct answer: true, or false. One point is awarded for every correct answer, but one point is taken off for each wrong answer. Suppose a student answers every question by guessing at random, independently of other questions. Let $S$ be the student’s score on the test.
2A Find $E(S)$
2B Find $SE(S)$
2C Find $P(S=0)$ without using a large-sample approximation.
Solution
2A) This is very similar to the net gain, $$E(S)=20\times(1\times\frac{1}{2}+(-1)\times\frac{1}{2})=0$$
2B) $S$ is the sum score, $$\mu=1\times\frac{1}{2}+(-1)\times\frac{1}{2}=0$$ $$SE(S)=\sqrt{n}\sigma=\sqrt{20\times((1-0)^2\times\frac{1}{2}+(-1-0)^2\times\frac{1}{2})}\doteq4.472136$$
2C) $S=0$ means there are 10 correct answers and 10 incorrect answers, binomial distribution $n=20, k=10, p=\frac{1}{2}$, $$P(S=0)=C_{20}^{10}\times(\frac{1}{2})^{20}\doteq0.1761971$$ R code:
dbinom(x = 10, size = 20, prob = 1/2) [1] 0.1761971
PROBLEM 3
A die is rolled 60 times.
3A Find the expected number of times the face with 6 spots appears.
3B Find the $SE$ of the number of times the face with 6 spots appears.
3C Find the normal approximation to the chance that the face with six spots appears 10 times.
3D Find the exact chance that the face with six spots appears 10 times.
3E Find the normal approximation to the chance that the face with six spots appears 9, 10, or 11 times.
3F Find the exact chance that the face with six spots appears 9, 10, or 11 times.
Solution
3A) $$E(\text{6 spots appears})=60\times\frac{1}{6}=10$$
3B) $$SE(\text{6 spots appears})=\sqrt{60\times\frac{1}{6}\times(1-\frac{1}{6})}\doteq2.886751$$
3C) $$Z_1=\frac{9.5-10}{SE}, Z_2=\frac{10.5-10}{SE}$$ Computing in R:
mu = 10; se = sqrt(60 * 1/6 * 5/6) z1 = (9.5 - mu) / se; z2 = (10.5 - mu) / se pnorm(z2) - pnorm(z1) [1] 0.1375098
3D) Binomial distribution $n=60, k=10, p=\frac{1}{6}$: $$C_{60}^{10}\times(\frac{1}{6})^{10}\times(\frac{5}{6})^{50}\doteq0.1370131$$ R code:
dbinom(x = 10, size = 60, prob = 1/6) [1] 0.1370131
3E) $$Z_1=\frac{8.5-10}{SE}, Z_2=\frac{11.5-10}{SE}$$ Computing in R:
mu = 10; se = sqrt(60 * 1/6 * 5/6) z1 = (8.5 - mu) / se; z2 = (11.5 - mu) / se pnorm(z2) - pnorm(z1) [1] 0.3966682
3F) Binomial distribution $n=60, k=9:11, p=\frac{1}{6}$: $$\sum_{k=9}^{11}C_{60}^{k}\cdot(\frac{1}{6})^{k}\cdot(\frac{5}{6})^{60-k}\doteq0.3958971$$ R code:
sum(dbinom(x = 9:11, size = 60, prob = 1/6)) [1] 0.3958971
PROBLEM 4
According to genetic theory, plants of a particular species have a 25% chance of being red-flowering, independently of other plants. Find the normal approximation to the chance that among 10,000 plants of this species, more than 2400 are red-flowering.
Solution
Normal approximation: $$p=0.25, n=10000$$ $$\mu=np, SE=\sqrt{np(1-p)}, Z=\frac{2400.5-\mu}{SE}$$ Computing in R:
n = 10000; p = 0.25 mu = n * p; se = sqrt(n * p * (1 - p)) z = (2400.5 - mu) / se 1 - pnorm(z) [1] 0.989215
Binomial distribution $$\sum_{k=2401}^{10000}C_{10000}^{k}\cdot(0.25)^k\cdot(0.75)^{10000-k}$$ R code:
sum(dbinom(x = 2401:10000, size = 10000, prob = 0.25)) [1] 0.9894525
PROBLEM 5
A random number generator draws at random with replacement from the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. In 5000 draws, the chance that the digit 0 appears fewer than 495 times is closest to
Solution
Normal approximation: $$n=5000, p=0.1$$ $$\mu=np, SE=\sqrt{np(1-p)}, Z=\frac{494.5-\mu}{SE}$$ Computing in R:
mu = n * p; se = sqrt(n * p * (1 - p)) z = (494.5 - mu) / se pnorm(z) [1] 0.3977125
Binomial distribution $$\sum_{k=0}^{494}C_{5000}^{k}\cdot(0.1)^k\cdot(0.9)^{5000-k}$$ R code:
sum(dbinom(x = 0:494, size = 5000, prob = 0.1)) [1] 0.3999814
EXERCISE 5
PROBLEM 1
The durations of phone calls taken by the receptionist at an office are like draws made at random with replacement from a list that has an average of 8.5 minutes (that‘s 8 minutes and 30 seconds) and an $SD$ of 3 minutes. Approximately what is the chance that the total duration of the next 100 calls is more than 15 hours?
Solution
Central Limit Theorem: $$\mu=8.5, SD=3, SE=\sqrt{n}\cdot SD=30$$ $$Z=\frac{900-850}{30}$$ Computing in R:
z = (900 - 850) / 30 1 - pnorm(z) [1] 0.04779035
PROBLEM 2
A multiple choice test consists of 100 questions. Each question has 5 possible answers, only one of which is correct. Four points are awarded for each correct answer, and 1 point is taken off for each wrong answer. Suppose you answer all the questions by guessing at random, independently of all other questions.
2A In order to score more than 30 points, you have to get more than ________ answers right. Fill in the blank with the smallest correct whole number.
2B What is the chance that you get more than 30 points?
Solution
2A) Let $x$ be the number of correct answers, we have $$4x+(-1)\cdot(100-x) > 30\Rightarrow x > 26$$ Therefore you have to get more than 26 answers right.
2B) Binomial distribution $n=100, k=27:100, p=\frac{1}{5}$: $$P(\text{more than 30 points})=\sum_{k=27}^{100}C_{100}^{k}\cdot(\frac{1}{5})^k\cdot(\frac{4}{5})^{100-k}\doteq0.05583272$$ R code:
sum(dbinom(x = 27:100, size = 100, prob = 1/5)) [1] 0.05583272
Normal approximation: $$n=100, p=\frac{1}{5}, \mu=np=20, SE=\sqrt{np(1-p)}=4$$ $$Z=\frac{26.5-20}{SE}$$ Computing in R:
z = (26.5 - 20) / 4 > 1 - pnorm(z) [1] 0.05208128
This approximation is not sufficient good.
PROBLEM 3
Assume that each person in a population has chance 2/1000 of carrying a particular disease, independently of all other people. Among 1000 people in this population, the number of people that carry the disease [pick all that are correct]
Solution
First, this is binomial distribution. Second, because $p$ is very small so it is right-skewed.
PROBLEM 4
Jack and Jill gamble on a roll of a die (yes, a fair die), as follows. If the die shows 1 or 2 spots, Jack gives Jill $\$1$. If the die shows 5 or 6 spots, Jill gives Jack $\$1$. If the die shows 3 or 4 spots, no money changes hands. Suppose Jack and Jill play this game 400 times. The chance that Jill’s net gain is more than $\$20$ is closest to?
Solution
$$P(\text{Jill wins 1})=P(\text{Jill loses 1})=P(\text{no money changes hands})=\frac{1}{3}$$ $$\mu=1\times\frac{1}{3}+(-1)\times\frac{1}{3}+0\times\frac{1}{3}=0$$ $$SD=\sqrt{(1-0)^2\times\frac{1}{3}+(-1-0)^2\times\frac{1}{3}+(0-0)^2\times\frac{1}{3}}=\sqrt{\frac{2}{3}}$$ $$SE=\sqrt{n}\cdot SD=\sqrt{\frac{800}{3}}, Z=\frac{20-0}{SE}$$ Computing in R:
se = sqrt(800 / 3) z = (20 - 0) / se 1 - pnorm(z) [1] 0.1103357
PROBLEM 5
In roulette, the bet on a “split” pays 17 to 1 and there are 2 chances in 38 to win. The bet on “red” pays 1 to 1 and there are 18 chances in 38 to win. Compare the following two strategies: A: bet $\$1$ on a split, 200 times independently B: bet $\$1$ on red, 200 times independently In what follows, “making more than $\$x$” means having a net gain of more than $\$x$; “losing more than $\$x$” means having a net gain of less than $-\$x$. Compare the chances between A and B that "coming out ahead, winning more than $\$20$, losing more than $\$20$".
Solution
By using Central Limit Theorem.
Let $P_{X0}$ be "coming out ahead" when following strategy $X$. Similarly, $P_{X20^{+}}$ and $P_{X20^{-}}$ denotes wining and losing $\$20$ respectively. Strategy $A$: $$n=200, \mu=200\times(17\times\frac{2}{38}+(-1)\times\frac{36}{38})=-\frac{200}{19}$$ $$SE=\sqrt{n}\cdot SD=\sqrt{200\times[(17-\mu)^2\times\frac{2}{38}+(-1-\mu)^2\times\frac{36}{38}]}$$ Similarly, we can calculate strategy $B$ in the same way. And finally computing in R:
netgain = function(n, prob, value, gain){
mu = n * (sum(prob * value))
se = sqrt(n * sum((value - mu) ^ 2 * prob))
if (gain >= 0){
z = (gain + 0.5 - mu) / se
print(1 - pnorm(z))
} else {
z = (gain - 0.5 - mu) / se
print(pnorm(z))
}
}
netgain(n = 200, prob = c(2/38, 36/38), value = c(17, -1), gain = 0)
[1] 0.4722959 # A
netgain(n = 200, prob = c(18/38, 20/38), value = c(1, -1), gain = 0)
[1] 0.4704632 # B
netgain(n = 200, prob = c(2/38, 36/38), value = c(17, -1), gain = 20)
[1] 0.4224767 # A
netgain(n = 200, prob = c(18/38, 20/38), value = c(1, -1), gain = 20)
[1] 0.4174109 # B
netgain(n = 200, prob = c(2/38, 36/38), value = c(17, -1), gain = -20)
[1] 0.474937 # A
netgain(n = 200, prob = c(18/38, 20/38), value = c(1, -1), gain = -20)
[1] 0.4732785 # B
According to the results above, $$P_{A0} > P_{B0}$$ $$P_{A20^{+}} > P_{B20^{+}}$$ $$P_{A20^{-}} > P_{B20^{-}}$$ That is, $P_A > P_B$ when
- Coming out ahead
- Winning more than $\$20$
- Losing more than $\$20$