W02 Homework B

Due date: Tuesday 1/27, 11:59pm

Conditional probability

01

04

Multiplication - drawing two hearts

Two cards are drawn from a standard deck (without replacement).

(a) What is the probability that both are hearts?

(b) What is the probability that both are 4?

Link to original

Solution

Solutions - 5030-04

Let $C_1$ be the outcome of the first card, $C_2$ be the outcome of the second card, and $H$ denote “hearts.”

(a)

Apply the multiplication rule for sequential draws without replacement:

$$\begin{array}{c}P[C_1=H\cap C_2=H]=P[C_1=H]\cdot P[C_2=H|C_1=H]\\ \\ \gg \gg \frac{13}{52}\cdot \frac{12}{51}\gg \gg \frac{1}{17}\end{array}$$

---

(b)

Apply the multiplication rule:

$$\begin{array}{c}P[C_1=4\cap C_2=4]=P[C_1=4]\cdot P[C_2=4|C_1=4]\\ \\ \gg \gg \frac{4}{52}\cdot \frac{3}{51}\gg \gg \frac{1}{221}\end{array}$$Link to original

02

09

Syntax errors vs. logic errors, Part A

A computer program may contain a syntax error or a logic error or both types of errors. The probability that a program has both types of error is 0.16. The probability that a program has a syntax error given that it has a logic error is 0.4. The probability that a program has a logic error given that it has a syntax error is 0.5.

Find the probability that a particular program has at least one type of error.

Link to original

Solution

Solutions - 5030-09

Let $A$ = syntax error, $B$ = logic error.

(1) Solve for $P[B]$ using $P[A|B]$:

$$\begin{array}{c}P[A|B]=\frac{P[A\cap B]}{P[B]}\gg \gg P[B]=\frac{P[A\cap B]}{P[A|B]}\\ \\ \gg \gg \frac{0.16}{0.4}\gg \gg 0.4\end{array}$$

---

(2) Solve for $P[A]$ using $P[B|A]$:

$$\begin{array}{c}P[B|A]=\frac{P[A\cap B]}{P[A]}\gg \gg P[A]=\frac{P[A\cap B]}{P[B|A]}\\ \\ \gg \gg \frac{0.16}{0.5}\gg \gg 0.32\end{array}$$

---

(3) Apply inclusion-exclusion to find $P[A\cup B]$:

$$\begin{array}{c}P[A\cup B]=P[A]+P[B]-P[A\cap B]\\ \\ \gg \gg 0.32+0.4-0.16\gg \gg 0.56\end{array}$$Link to original

Bayes’ Theorem

03

02

Bayes’ Theorem - Inferring die from roll

A bag contains one 4-sided die, one 6-sided die, and one 12-sided die. You draw a random die from the bag, roll it, and get a 4.

What is the probability that you drew the 6-sided die?

Link to original

Solution

Solutions - 5040-02

(1) Define events:

Let $D_4$, $D_6$, $D_{12}$ = event of drawing each die.

$P[4|D_4]=\frac{1}{4}$, $$ $P[4|D_6]=\frac{1}{6}$, $$ $P[4|D_{12}]=\frac{1}{12}$, $$ and $$ $P[D_4]=P[D_6]=P[D_{12}]=\frac{1}{3}$.

---

(2) Apply Bayes’ Theorem:

$$\begin{array}{c}P[D_6|4]=\frac{P[4|D_6]P[D_6]}{P[4]}\\ \\ \gg \gg \frac{P[4|D_6]P[D_6]}{P[4|D_4]P[D_4]+P[4|D_6]P[D_6]+P[4|D_{12}]P[D_{12}]}\\ \\ \gg \gg \frac{\frac{1}{6}\cdot \frac{1}{3}}{\frac{1}{4}\cdot \frac{1}{3}+\frac{1}{6}\cdot \frac{1}{3}+\frac{1}{12}\cdot \frac{1}{3}}\\ \\ \gg \gg \frac{1}{3}\end{array}$$Link to original

Independence

04

04

Syntax errors vs. logic errors, Part B

A computer program may contain a syntax error or a logic error or both types of errors. The probability that a program has both types of error is 0.16. The probability that a program has a syntax error given that it has a logic error is 0.4. The probability that a program has a logic error given that it has a syntax error is 0.5.

Are the events “program has a syntax error” and “program has a logic error” independent? Justify your answer.

Link to original

Solution

Solutions - 5050-04

Let $A$ = syntax error, $B$ = logic error.

(1) Find $P[A]$ and $P[B]$:

$$P[B]=\frac{P[A\cap B]}{P[A|B]}=\frac{0.16}{0.4}=0.4P[A]=\frac{P[A\cap B]}{P[B|A]}=\frac{0.16}{0.5}=0.32$$

---

(2) Check the independence condition:

$$P[A]\cdot P[B]=0.32\cdot 0.4=0.128\ne 0.16=P[A\cap B]\Rightarrow \text{NOT independent}$$Link to original

05

02

Pairwise independent, not mutually independent: three coin flips

Flip a coin three times in sequence. Label events like this:

• $A$ – exactly one heads among first and second flips
• $B$ – exactly one heads among second and third flips
• $C$ – exactly one heads among first and third flips

Verify that $A,B,C$ are pairwise independent but not actually mutually independent.

Link to original

Solution

Solutions - 5050-02

(1) Compute individual probabilities:

$P[A]=\frac{1}{2}$, $$ $P[B]=\frac{1}{2}$, $$ $P[C]=\frac{1}{2}$.

---

(2) Verify pairwise independence:

$A\cap B$ occurs for $\{THT,HTH\}$, so $P[A\cap B]=\frac{1}{4}=P[A]P[B]$ $✓$

$B\cap C$ occurs for $\{TTH,HHT\}$, so $P[B\cap C]=\frac{1}{4}=P[B]P[C]$ $✓$

$A\cap C$ occurs for $\{HTT,THH\}$, so $P[A\cap C]=\frac{1}{4}=P[A]P[C]$ $✓$

---

(3) Disprove mutual independence:

$A$, $B$, and $C$ cannot all occur simultaneously, so

$$P[A\cap B\cap C]=0\ne \frac{1}{8}=P[A]P[B]P[C]$$Link to original

Tree diagrams

06

02

Homework part errors

A homework problem has 10 different parts. You submit and are told that 4 of the 10 answers you provided are incorrect, but you are not told which parts are incorrect.

(a) What is the probability you will have gotten the first part correct and the second part incorrect? Draw a tree diagram.

(b) Suppose the 4 errors have occurred in the first 6 parts. In this case, how many possible arrangements are there for these 4 errors?

(c) What is the probability the 4 errors occurred in the first 6 parts?

Link to original

Solution

Solutions - 5060-02

(a)

$$\begin{array}{c}P[\text{1st correct}\cap \text{2nd incorrect}]=\frac{6}{10}\cdot \frac{4}{9}\\ \\ \gg \gg \frac{24}{90}\approx 0.2667\end{array}$$

---

(b)

$$\begin{array}{c}\left(\genfrac{}{}{0px}{}{6}{4}\right)=\frac{6!}{4!2!}\gg \gg 15\end{array}$$

---

(c)

$$\begin{array}{c}P[\text{4 errors in first 6}]=\frac{\left(\genfrac{}{}{0px}{}{6}{4}\right)}{\left(\genfrac{}{}{0px}{}{10}{4}\right)}\\ \\ \gg \gg \frac{15}{210}\gg \gg \approx 0.0714\end{array}$$Link to original

Counting

07

09

Counting passwords

Suppose a password must be created using 5 letters and 6 digits. (There are 26 letters, a-z, and 10 digits, 0-9.) No letter or digit may be repeated.

(a) How many unique passwords can be created if the letters must come first and the digits last?

(b) How many unique passwords can be created if the 5 letters and 6 digits can appear in any order?

Link to original

Solution

Solutions - 5070-09

(a)

Apply the permutation formula (letters first, digits last):

$$\begin{array}{c}{(26)}_5\cdot {(10)}_6=26\cdot 25\cdot 24\cdot 23\cdot 22\cdot 10\cdot 9\cdot 8\cdot 7\cdot 6\cdot 5\\ \\ \gg \gg \mathrm{1,193,512,320,000}\end{array}$$

---

(b)

Multiply by the number of ways to interleave 5 letter positions and 6 digit positions:

$$\begin{array}{c}\left(\genfrac{}{}{0px}{}{11}{5}\right)\cdot {(26)}_5\cdot {(10)}_6\\ \\ \gg \gg 462\cdot \mathrm{1,193,512,320,000}\gg \gg \mathrm{551,402,691,840,000}\end{array}$$Link to original

08

03

Drawing balls of distinct color

A bin contains 3 green and 4 yellow balls. Two balls are drawn out.

What is the probability that they are different colors?

Link to original

Solution

Solutions - 5070-03

(1) Set up the counting argument:

Since order does not matter, the sample space has $\left(\genfrac{}{}{0px}{}{7}{2}\right)$ outcomes. For different colors, choose one of each.

---

(2) Evaluate:

$$\begin{array}{c}P[\text{different colors}]=\frac{\left(\genfrac{}{}{0px}{}{3}{1}\right)\cdot \left(\genfrac{}{}{0px}{}{4}{1}\right)}{\left(\genfrac{}{}{0px}{}{7}{2}\right)}\\ \\ \gg \gg \frac{12}{21}\gg \gg \frac{4}{7}\end{array}$$Link to original

09

02

Wisconsin flag 2 of 3 days

A kindergarten class hangs a random state flag (50 flags) on the wall every day. What is the probability that two days out of three given days have Wisconsin’s flag?

Link to original

Solution

Solutions - 5070-02

(1) Identify the binomial structure:

Each day has probability $\frac{1}{50}$ of showing Wisconsin’s flag, independently of the other days, and probability $\frac{49}{50}$ of showing a different state’s flag. There are $\left(\genfrac{}{}{0px}{}{3}{2}\right)=3$ ways to choose which 2 of the 3 days have the Wisconsin flag.

---

(2) Evaluate:

$$\begin{array}{c}P[\text{2 of 3 days}]=\left(\genfrac{}{}{0px}{}{3}{2}\right)\cdot {\left(\frac{1}{50}\right)}^2\cdot \frac{49}{50}\\ \\ \gg \gg 3\cdot \frac{49}{\mathrm{125,000}}\gg \gg 0.001176\end{array}$$Link to original

Review problems

10

03

Inclusion-exclusion reasoning

Suppose $P[A]=0.4$ and $P[B]=0.7$. Show that $0.1\le P[AB]\le 0.4$.

Link to original

Solution

Solutions - 5020-03

(1) State the inclusion-exclusion principle:

$$P[A\cup B]=P[A]+P[B]-P[A\cap B]$$

---

(2) Derive the lower bound on $P[AB]$:

Since $P[A\cup B]\le 1$:

$$\begin{array}{c}P[A\cap B]=P[A]+P[B]-P[A\cup B]\\ \\ \gg \gg P[A\cap B]\ge 0.4+0.7-1\gg \gg P[AB]\ge 0.1\end{array}$$

---

(3) Derive the upper bound on $P[AB]$:

Since $A\cap B\subseteq A$, we have $P[A\cap B]\le P[A]$.

$$P[AB]\le 0.4$$Link to original