A box has coins HH, TT, and a fair coin (HT). One coin is chosen uniformly at random and flipped. Given the first flip is Head, what is the probability the next flip (same coin) is Head?
Solution (Bayes):
Let \(C\in\{\text{HH},\text{TT},\text{F}\}\),
each with prior \(1/3\).
\(P(H\mid \text{HH})=1,\;P(H\mid
\text{TT})=0,\;P(H\mid \text{F})=1/2\).
Posterior: \[ P(\text{HH}\mid H)=\frac{(1/3)\cdot 1}{(1/3)\cdot 1 + (1/3)\cdot (1/2)}=\frac{2}{3},\quad P(\text{F}\mid H)=\frac{1}{3}. \]
Therefore \[ P(\text{next }H\mid H)=\tfrac{2}{3}\cdot 1+\tfrac{1}{3}\cdot \tfrac{1}{2}=\tfrac{5}{6}. \]
ans_q1 <- 5/6
ans_q1## [1] 0.8333333We interpret the diagram as C — (A ∥ B) — D in
series, with component reliabilities: \(P(C)=0.8,\;P(A)=0.95,\;P(B)=0.7,\;P(D)=0.9\).
The middle block is parallel: \(P(A\cup
B)=1-(1-0.95)(1-0.7)=0.985\).
pC <- 0.8; pA <- 0.95; pB <- 0.7; pD <- 0.9
p_parallel <- 1 - (1 - pA)*(1 - pB) # A || B
p_sys <- pC * p_parallel * pD
p_sys## [1] 0.7092So, \(P(\text{system works}) = 0.8 \times 0.985 \times 0.9 = 0.7092\).
Let \(q_S=1-p_{\text{sys}}\) and \(q_D=1-p_D=0.1\). Because series blocks are independent, if \(D\) fails, the system fails. Thus \[ P(D\text{ fails}\mid S\text{ fails}) =\frac{P(D\text{ fails}\cap S\text{ fails})}{P(S\text{ fails})} =\frac{P(D\text{ fails})}{q_S} =\frac{0.1}{1-0.7092}\approx 0.3439. \]
qS <- 1 - p_sys
ans_q2b <- (1 - pD)/qS
ans_q2b## [1] 0.343879Four gardeners: Oliver (miss 0.3 ⇒ show 0.7), Walter (miss 0.1 ⇒ show 0.9), Tom (miss 0.2 ⇒ show 0.8), and Jerry who comes exactly when Tom comes. Assume independence among Oliver, Walter, Tom; Jerry mirrors Tom. Let \(X\) be the number who show up. Find \(f_X(x)\).
Let \(O\sim \text{Bern}(0.7)\), \(W\sim \text{Bern}(0.9)\), and a pair \(T_2\in\{0,2\}\) with \(P(T_2=2)=0.8\), \(P(T_2=0)=0.2\). Then \(X=O+W+T_2\).
pO <- 0.7; pW <- 0.9; pT2 <- 0.8 # prob of {Tom & Jerry both present}=2
# Distribution of O+W:
pS0 <- (1-pO)*(1-pW)
pS1 <- pO*(1-pW) + (1-pO)*pW
pS2 <- pO*pW
# Mix with T-pair:
pmf <- c(
`0` = 0.2*pS0,
`1` = 0.2*pS1,
`2` = 0.2*pS2 + 0.8*pS0,
`3` = 0.8*pS1,
`4` = 0.8*pS2
)
pmf## 0 1 2 3 4
## 0.006 0.068 0.150 0.272 0.504sum(pmf)## [1] 1Thus the pmf: - \(P(X=0)=0.006\)
- \(P(X=1)=0.068\)
- \(P(X=2)=0.150\)
- \(P(X=3)=0.272\)
- \(P(X=4)=0.504\)
A discrete rv with pmf \(f(x)=\dfrac{c}{4^x}\) for \(x=0,1,2,\dots\)
\[ 1=\sum_{x=0}^\infty \frac{c}{4^x}=c\sum_{x=0}^\infty \left(\frac{1}{4}\right)^x =c\cdot \frac{1}{1-1/4}=\frac{4}{3}c \ \Rightarrow\ c=\frac{3}{4}. \]
c_q4 <- 3/4
c_q4## [1] 0.75\[ F(2.5)=\sum_{x=0}^{2}\frac{3}{4\cdot 4^x} =\frac{3}{4}\left(1+\frac{1}{4}+\frac{1}{16}\right) =\frac{3}{4}\cdot \frac{21}{16} =\frac{63}{64}=0.984375. \]
F_2_5 <- (3/4)*(1 + 1/4 + 1/16)
F_2_5## [1] 0.984375A continuous rv with pdf \(f(x)=c(x-1)\) for \(1\le x<3\), else \(0\).
\[ 1=\int_{1}^{3} c(x-1)\,dx = c\cdot \frac{(x-1)^2}{2}\Big|_{1}^{3} = c\cdot \frac{4}{2}=2c\ \Rightarrow\ c=\tfrac{1}{2}. \]
c_q5 <- 1/2
c_q5## [1] 0.5\[ P(1\le X<2)=\int_{1}^{2}\tfrac{1}{2}(x-1)\,dx =\tfrac{1}{2}\cdot \frac{(x-1)^2}{2}\Big|_{1}^{2} =\frac{1}{4}. \]
p_1_2 <- 1/4
p_1_2## [1] 0.25\[ F(x)= \begin{cases} 0, & x<1,\\\\ \displaystyle \int_{1}^{x}\tfrac{1}{2}(t-1)\,dt =\frac{(x-1)^2}{4}, & 1\le x<3,\\\\ 1, & x\ge 3. \end{cases} \]
\[ E[X]=\int_{1}^{3} x\cdot \tfrac{1}{2}(x-1)\,dx =\tfrac{1}{2}\int_{1}^{3} (x^{2}-x)\,dx =\tfrac{1}{2}\Big(\frac{x^{3}}{3}-\frac{x^{2}}{2}\Big)\Big|_{1}^{3} =\frac{7}{3}\approx 2.333333. \]
EX <- 7/3
EX## [1] 2.333333\[ E[X^{2}]=\tfrac{1}{2}\int_{1}^{3}(x^{3}-x^{2})\,dx =\tfrac{1}{2}\Big(\frac{x^{4}}{4}-\frac{x^{3}}{3}\Big)\Big|_{1}^{3} =\frac{17}{3}. \] \[ \mathrm{Var}(X)=E[X^{2}]-(E[X])^{2} =\frac{17}{3}-\Big(\frac{7}{3}\Big)^{2} =\frac{2}{9}\approx 0.222222. \]
VarX <- 2/9
c(EX=EX, VarX=VarX)## EX VarX
## 2.3333333 0.2222222\[ \mathrm{Var}(Y)=25\,\mathrm{Var}(X)=\frac{50}{9}\approx 5.555556. \]
VarY <- 25*VarX
VarY## [1] 5.555556A 10-item multiple-choice test; each item has 4 options. The student can eliminate one wrong option and guesses among the remaining 3, so \(p=1/3\) for each item. Passing is \(\ge 8\) correct. Compute the probability of passing.
Let \(X\sim\text{Binomial}(n=10,p=1/3)\). \[ P(X\ge 8)=\sum_{k=8}^{10} \binom{10}{k}\Big(\frac{1}{3}\Big)^k\Big(\frac{2}{3}\Big)^{10-k} =\frac{201}{3^{10}}\approx 0.003406. \]
n <- 10; p <- 1/3
ans_q6 <- pbinom(7, size=n, prob=p, lower.tail = FALSE)
c(exact_fraction = "201 / 3^10", numeric = ans_q6)## exact_fraction numeric
## "201 / 3^10" "0.00340395264949449"There are 8 dogs and 6 cats (total 14). Four pets are chosen without replacement.
Let \(X\) = number of dogs chosen. Then \(X\sim \text{Hypergeometric}(N=14,K=8,n=4)\): \[ P(X=k)=\frac{\binom{8}{k}\binom{6}{4-k}}{\binom{14}{4}}, \quad k=0,1,2,3,4, \] with \(\binom{14}{4}=1001\).
N <- 14; K <- 8; n <- 4
den <- choose(14,4)
k <- 0:4
pmf_q7 <- choose(8,k)*choose(6,4-k)/den
data.frame(k, P=pmf_q7)## k P
## 1 0 0.01498501
## 2 1 0.15984016
## 3 2 0.41958042
## 4 3 0.33566434
## 5 4 0.06993007Values: - \(P(0)=15/1001\)
- \(P(1)=160/1001\)
- \(P(2)=420/1001\)
- \(P(3)=336/1001\)
- \(P(4)=70/1001\)
p_atleast1 <- 1 - pmf_q7[1]
p_atleast1## [1] 0.985015EX_q7 <- n*(K/N)
EX_q7## [1] 2.285714Three identical fair coins are tossed simultaneously until all three show the same face (HHH or TTT). Each trial succeeds with \(p=2/8=1/4\).
Let \(T\) be the number of tosses (trials) until first success; \(T\sim\text{Geometric}(p=1/4)\) with support \(\{1,2,\dots\}\).
p <- 1/4
P_more_than_3 <- (1-p)^3
P_more_than_3## [1] 0.421875E_T <- 1/p
E_T## [1] 4Let \(X_1,X_2,X_3 \overset{iid}{\sim}\text{Exp}(\beta=6)\) (scale). Find \(P(\text{at least one }X_i>8)\).
For \(\text{Exp}(\beta)\), \(F(x)=1-e^{-x/\beta}\) for \(x>0\).
\[ P\Big(\max_i X_i>8\Big)=1-P(X_1\le 8, X_2\le 8, X_3\le 8) =1-\left(1-e^{-8/6}\right)^3. \]
beta <- 6; x <- 8
ans_q9_exact <- 1 - (1 - exp(-x/beta))^3
c(numeric = ans_q9_exact)## numeric
## 0.6006567Numerically, \(e^{-8/6}\approx 0.2636\), so the probability is about 0.601.