W01 Homework B

Due date: Tuesday 1/20, 11:59pm

Events and outcomes

01

03

Events - descriptions to sets

You are modeling quality assurance for cars coming off an assembly line. They are either good (G) or broken (B). You watch 4 cars come off and record their status as a sequence of these letters, for example ‘GGBG’.

Determine the sets defined by the events having the following descriptions:

(a) “third car is broken”

(b) “all cars have the same status”

(c) “at least one car is broken”

(d) “no consecutive cars have the same status”

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Solution

Solutions - 5010-03

(a)

$$S_1=\{abcd\in {\{G,B\}}^4:c=B\}$$

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(b)

$$S_2=\{abcd\in {\{G,B\}}^4:a=b=c=d\}$$

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(c)

$$S_3=\{abcd\in {\{G,B\}}^4:abcd\ne GGGG\}$$

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(d)

$$S_4=\{a_1{a}_2{a}_3{a}_4\in {\{G,B\}}^4:a_i\ne a_{i+1},i\in \{1,2,3\}\}$$Link to original

Probability models

02 $★$

06

Researcher’s degree

Of 1000 researchers at a research laboratory, 375 have a degree in mathematics, 450 have a degree in computer science, and 150 of the researchers have a degree in both fields. One researcher’s name is selected at random.

(a) What is the probability that the researcher has a degree in mathematics, but not in computer science?

(b) What is the probability that the researcher has no degree in either mathematics or computer science?

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Solution

Solutions - 5020-06

(a)

Use $P[M]=P[M\cap C]+P[M\cap C^c]$ to isolate $P[M\cap C^c]$:

$$\begin{array}{c}P[M\cap C^c]=P[M]-P[M\cap C]\\ \\ \gg \gg 0.375-0.15\gg \gg 0.225\end{array}$$

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(b)

Apply De Morgan’s law and inclusion-exclusion:

$$\begin{array}{c}P[M^c\cap C^c]=1-P[M\cup C]\\ \\ \gg \gg 1-(0.375+0.45-0.15)\gg \gg 0.325\end{array}$$Link to original

03

02

Inclusion-exclusion reasoning

Your friend says: “according to my calculations, the probability of $A$ is $0.5$ and the probability of $B$ is $0.7$, but the probability of $A$ and $B$ both happening is only $0.001$.”

You tell your friend they don’t understand probability. Why?

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Solution

Solutions - 5020-02

(1) State the inclusion-exclusion principle:

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

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(2) Derive a contradiction:

Given $P[A]=0.5$ and $P[B]=0.7$, we have $P[A]+P[B]=1.2$.

Since $P[A\cup B]\le 1$, it follows that $P[A\cap B]\ge 0.2$.

Therefore $P[A\cap B]=0.001$ is impossible.

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Conditional probability

04 $★$

02

Conditioning - two dice, at least one is 5

Two dice are rolled, and at least one is a 5.

What is the probability that their sum is 10?

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Solution

Solutions - 5030-02

(1) Set up the conditional probability formula:

$$P[D_1+D_2=10|D_1=5\cup D_2=5]=\frac{P[(D_1+D_2=10)\cap (D_1=5\cup D_2=5)]}{P[D_1=5\cup D_2=5]}$$

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(2) Count favorable and total outcomes:

There is $1$ combination where at least one die is a $5$ and the sum is $10$: $(\mathrm{5,5})$.

There are $2\cdot 6-1=11$ combinations where at least one die is a $5$.

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(3) Evaluate:

$$\begin{array}{c}P[D_1+D_2=10|D_1=5\cup D_2=5]=\frac{1/36}{11/36}\\ \\ \gg \gg \frac{1}{11}\end{array}$$Link to original

05

03

Conditioning - two dice, differing numbers

Two dice are rolled, and the outcomes are different.

What is the probability of getting at least one 1?

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Solution

Solutions - 5030-03

(1) Set up the conditional probability formula:

$$P[D_1=1\cup D_2=1|D_1\ne D_2]=\frac{P[(D_1=1\cup D_2=1)\cap (D_1\ne D_2)]}{P[D_1\ne D_2]}$$

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(2) Count favorable and total outcomes:

There are $2\cdot 6-1=11$ combinations with at least one $1$. Excluding $(\mathrm{1,1})$ gives $10$ combinations where the outcomes also differ.

There are $36-6=30$ combinations where the two dice differ.

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(3) Evaluate:

$$\begin{array}{c}P[D_1=1\cup D_2=1|D_1\ne D_2]=\frac{10/36}{30/36}\\ \\ \gg \gg \frac{1}{3}\end{array}$$Link to original

06 $★$

06

Conditioning relation

Suppose you know $P[A\cap B]=0.036$ and $P[A|B]=0.18$ and $P[B|A]=0.60$.

Calculate $P[A]$ and $P[B]$ and $P[A\cup B]$.

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Solution

Solutions - 5030-06

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

$$\begin{array}{c}P[B|A]=\frac{P[A\cap B]}{P[A]}\gg \gg P[A]=\frac{P[A\cap B]}{P[B|A]}\\ \\ \gg \gg \frac{0.036}{0.60}\gg \gg 0.06\end{array}$$

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

$$\begin{array}{c}P[A|B]=\frac{P[A\cap B]}{P[B]}\gg \gg P[B]=\frac{P[A\cap B]}{P[A|B]}\\ \\ \gg \gg \frac{0.036}{0.18}\gg \gg 0.2\end{array}$$

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

$$\begin{array}{c}P[A\cup B]=P[A]+P[B]-P[A\cap B]\\ \\ \gg \gg 0.06+0.2-0.036\gg \gg 0.224\end{array}$$Link to original