W01 Homework A

Due date: Thursday 1/15, 11:59pm

Events and outcomes

01

01

Sample space - roll a die, flip a coin

A normal 6-sided die is cast, and then a coin is flipped. All results are recorded.

(a) Define a sample space for this experiment.

(b) How many possible events are there?

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Solution

Solutions - 5010-01

(a)

(1) Count outcomes:

Since there are 6 possible results of rolling a die and $2$ possible results of a coin flip, the sample space has $6\cdot 2=12$ elements.

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(2) Describe the sample space $\mathrm{\Omega }$ using set builder notation:

$$\mathrm{\Omega }=\{(i,j):i\in \{1,\dots ,6\},j\in \{H,T\}\}$$

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(b)

The number of events equals the number of subsets of $\mathrm{\Omega }$, i.e. $|𝒫(\mathrm{\Omega })|$.

$$|𝒫(\mathrm{\Omega })|=2^{|\mathrm{\Omega }|}=2^{12}=4096$$Link to original

02

02

Sample space - roll a die then flip coin(s)

A normal 6-sided die is cast. If the result is even, flip a coin two times; if the result is odd, flip a coin one time. All results are recorded.

(a) Define a sample space for this experiment.

(b) How many possible events are there?

Link to original

Solution

Solutions - 5010-02

(a)

(1) Divide the sample space into two disjoint sets:

Let ${\mathrm{\Omega }}_1$ be the sample space where the die result is even, and ${\mathrm{\Omega }}_2$ where it is odd. Then $\mathrm{\Omega }={\mathrm{\Omega }}_1\cup {\mathrm{\Omega }}_2$.

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(2) Describe ${\mathrm{\Omega }}_1$:

There are $3$ even numbers on a die, and $2$ possible results of each of $2$ coin flips, so $|{\mathrm{\Omega }}_1|=3\cdot 2^2=12$.

$${\mathrm{\Omega }}_1=\{(i,j,k):i\in \{2,4,6\},j,k\in \{H,T\}\}$$

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(3) Describe ${\mathrm{\Omega }}_2$:

There are $3$ odd numbers on a die and $2$ possible results of $1$ coin flip, so $|{\mathrm{\Omega }}_2|=3\cdot 2=6$.

$${\mathrm{\Omega }}_2=\{(i,j):i\in \{1,3,5\},j\in \{H,T\}\}$$

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(4) Combine to describe $\mathrm{\Omega }$:

Since ${\mathrm{\Omega }}_1$ and ${\mathrm{\Omega }}_2$ are disjoint, $|\mathrm{\Omega }|=12+6=18$.

$$\mathrm{\Omega }={\mathrm{\Omega }}_1\cup {\mathrm{\Omega }}_2=\{(i_1,j_1),(i_2,j_2,k):i_1\in \{1,3,5\},i_2\in \{2,4,6\},j_1,j_2,k\in \{H,T\}\}$$

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(b)

$$|𝒫(\mathrm{\Omega })|=2^{|\mathrm{\Omega }|}=2^{18}=262144$$Link to original

Probability models

03

01

Venn diagrams - set rules and Kolmogorov additivity

Suppose we know three probabilities of events: $P[A]=0.4$, $P[B]=0.3$, and $P[A\cap B]=0.1$.

Calculate: $P[A\cup B]$, $P[A^c]$, $P[B^c]$, $P[A\cap B^c]$, and $P[{(A\cap B)}^c]$.

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Solution

Solutions - 5020-01

(1) Apply inclusion-exclusion:

$$\begin{array}{c}P[A\cup B]=P[A]+P[B]-P[A\cap B]\\ \\ \gg \gg 0.4+0.3-0.1\gg \gg 0.6\end{array}$$

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(2) Compute complements:

$$\begin{array}{c}P\left[A^c\right]=1-P[A]\\ \\ \gg \gg 1-0.4\gg \gg 0.6\end{array}$$ $$\begin{array}{c}P\left[B^c\right]=1-P[B]\\ \\ \gg \gg 1-0.3\gg \gg 0.7\end{array}$$

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(3) Use $P[A]=P[A\cap B]+P[A\cap B^c]$ to find $P[A\cap B^c]$:

$$\begin{array}{c}P[A\cap B^c]=P[A]-P[A\cap B]\\ \\ \gg \gg 0.4-0.1\gg \gg 0.3\end{array}$$

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(4) Compute the complement of $A\cap B$:

$$\begin{array}{c}P\left[{(A\cap B)}^c\right]=1-P[A\cap B]\\ \\ \gg \gg 1-0.1\gg \gg 0.9\end{array}$$Link to original

04

04

At least two heads from three flips

A coin is flipped three times.

What is the probability that at least two heads appear?

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Solution

Solutions - 5020-04

(1) Describe the sample space:

$$\mathrm{\Omega }={\{H,T\}}^3\gg \gg |\mathrm{\Omega }|=2^3=8$$

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(2) Count favorable outcomes and compute probability:

The sequences with at least two heads are $\{HHT,HTH,THH,HHH\}$, giving $4$ favorable outcomes.

$$P[\text{at least two heads}]=\frac{4}{8}=\frac{1}{2}$$Link to original

Conditional probability

05

01

Conditioning

Student test-passing rates, by year:

	1st year	2nd year	3rd year	4th year
Pass	0.155	0.340	0.255	0.160
Fail	0.025	0.040	0.015	0.010

What is the likelihood that a randomly chosen 4th-year student passed the test?

What about for 1st-year students?

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Solution

Solutions - 5030-01

(1) Apply conditional probability to find $P[\text{pass}|\text{4th}]$:

$$\begin{array}{c}P[\text{pass}|\text{4th}]=\frac{P[\text{pass}\cap \text{4th}]}{P[\text{4th}]}\\ \\ \gg \gg \frac{0.160}{0.170}\gg \gg \approx 0.9412\end{array}$$

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(2) Repeat for 1st-year students:

$$\begin{array}{c}P[\text{pass}|\text{1st}]=\frac{P[\text{pass}\cap \text{1st}]}{P[\text{1st}]}\\ \\ \gg \gg \frac{0.155}{0.180}\gg \gg \approx 0.8611\end{array}$$Link to original