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W03 - HW Solutions

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03-01 - Rolling two dice

Two dice are rolled. Find the probabilities of the following events:

  • , the event that the sum is 10
  • , the event that the sum is 12
  • , the event that the two numbers are equal

Solution

  1. & Consider the total number of outcomes.
    • Since there are two dice being rolled, there are 36 total outcomes.
  2. & Consider the total number of desired outcomes for .
    • There are 3 total desired outcomes, , , (6, 4).
  3. & Use formula to find .
  4. & Consider the total number of desired outcomes for .
    • There is only 1 desired outcome, .
  5. & Use formula to find .

  6. & Consider the total number of desired outcomes for .

    • There are 6 total desired outcomes: , , , , ,
  7. & Use formula to find .

03-02 - Binomial - Repeated coin flips

A coin is flipped times and the sequence of results recorded as an outcome.

  • (a) How many possible outcomes have exactly 3 heads?
  • (b) How many possible outcomes have at least 3 heads?

Solution (a)

  1. & Out of trials, we choose of them to be heads. Thus, (b)
  2. & Out of 7 trials, we choose at least 3 of them to be heads.
    • Using summation notation, we get

03-03 - Multinomial - Colored marbles in a line

How many ways are there to line up 10 colored marbles (2 red, 3 white, 5 blue), assuming you cannot distinguish marbles of the same color?

Solution

  1. & Use the multinomial coefficient.
    • Use that .
    • We have three bins, , , and .
    • Thus,

03-04 - Multinomial - Many rolls of a die

Roll a die 100 times.

  • (a) What is the probability that you rolled exactly 16 ones and 17 twos? Hint: use three bins. What are the bins?
  • (b) Using summation notation, write down a formula for the probability of rolling exactly 25 ones and at least 50 twos.

For this problem, use “desired outcomes over total outcomes” (simple counting), not repeated trials theory (next section).

Solution (a)

  1. & Consider total number of outcomes.
    • You roll a die times, so there are total outcomes.
  2. & Consider total number of desired outcomes.
    • We choose 16 of the 100 to be ones.
    • We choose 17 of the remaining 100 - 16 = 84 to be twos.
    • The other can be any of . So, there are remaining outcomes.
  3. & Set up formula. (b)
  4. & Consider total number of desired outcomes.
    • We choose 25 of the 100 to be ones.
    • Out of the remaining 100 - 25 = 75 rolls, we choose at least 50 to be two.
    • The remaining rolls, where , can be any of , , , . So, there are remaining outcomes.
  5. & Set up formula.

03-05 - Independent trials - At least 45 good paper clips

For a paper clip production line, of the paper clips come off good, and come off broken.

You buy a box of 50 paper clips from this line. What is the probability that at least 45 of them are good?

Solution

  1. &&& State the formula for a binomial distribution.
  2. & State parameters of binomial distribution.
    • since you buy paper clips.
    • ranges from to .
    • .
    • .
  3. & Use summation notation to find .

03-06 - Geometric wait time - Takes 10 rolls to get 6

A fair die is rolled until a six comes up. What are the odds that it takes at least 10 rolls?

Solution

  1. & Describe the situation.
    • Let .
    • If it takes at least rolls for a to come up, then the first rolls resulting in a number that is not a six.
  2. & Compute relevant probabilities.
    • The probability that you roll a six is .
    • The probability that you do not roll a six is .
  3. & Set up formula.
  4. &&& Use the formula for geometric series to evaluate the sum.

03-07 - Intersection accidents

Suppose that the odds of an accident occurring on any given day at the intersection of Ivy and Emmett is 0.05.

  • (a) What are the odds of a perfect week? (No accidents.)
  • (b) What are the odds of exactly 2 accidents in 30 days?
  • (c) What are the odds of the first accident occurring after day 4 and by day 10?

Solution (a)

  1. & The odds of a perfect week follows a binomial distribution.
    • since there are days to account for.
    • We choose of them to have accidents. (b)
  2. & The odds of exactly 2 accidents in 30 days follows a binomial distribution.
    • since there are 30 days to account for.
    • We choose of these days to have accidents. (c)
  3. && Describe the situation.
    • We want four perfect days before an accident occurs.
    • Over the course of the next six days, we want at least one accident occurring.
    • Since these are independent trials, multiplying the individual probabilities together will yield the desired value.
  4. & Set up the first condition.
    • since we want to account for 4 days.
    • We choose of these days to have accidents.
  5. & Set up second condition.
    • since we now want to account for the next six days.
    • We choose of these days to have accidents, where ranges from through .
  6. & Multiply the two expressions.

03-08 - Guessing on a test

Your odds of getting any given exam question right are 80. The exam has 4 questions, and you need to answer 3 correctly to pass.

  • What is the probability that you pass?
  • After finishing the exam, you are sure that you got the second question right. Now, what are the odds that you pass?

Solution

  1. & The probability that you pass follows a binomial distribution.
    • since there are 4 questions to account for.
    • We choose questions to be correct, where or .
  2. && Describe the situation wherein you get the second question right.
    • Now, out of the remaining three questions, you need to answer at least two correctly.
    • This too follows a binomial distribution.
  3. & Set up binomial distribution formula.
    • , since we only need to account for the remaining 3 questions.
    • We choose questions to be correct, where or . $$$P(\text{pass}) = \sum_{k = 2}^{3}\binom{3}{k}(0.8)^{k}(0.2)^{3-k}\approx 0.896$$

03-09 - Reliability for complex process

Consider a process with the following diagram of components in series and parallel:

Use to denote the event that component succeeds.

Suppose the success probabilities per component are given by this chart:

What are the odds of success for the whole process?

Solution

  1. &&& Describe dependencies of components.
    • The series 4, 5 runs parallel to component 3.
    • The components 2, 3, 4, 5, 6 all run in a series.
    • The latter is parallel to component 7
    • The circuit starts at component 1 and ends at component 8
  2. && Find the probability the component 4 and 5 succeeded. (Denote as )
  3. && Find the probability 3, and succeed. Note that these are in parallel. (Denote as )
  4. && Find the probability 2, , and 6 succeed. (Denote as )
  5. && Find the probability 7 and succeed. Note that these are in parallel. (Denote as )
  6. && Lastly, find the probability , , and 8 succeed. (Denote as ).

03-10 - Digit of a real number

Suppose a real number is chosen randomly in the unit interval . Consider the decimal expansion of this number. Let be a random variable giving the first digit after the decimal point. Find the possible values, the PMF, and the CDF of .

  1. & Find the possible values of .
    • After the decimal point, any of the 10 digits can appear, so .
  2. & Find the PMF of .
    • Each digit has a chance of appearing. So, the PMF of is
    Misplaced &0 & \text{otherwise} \end{cases}$$
  3. && Find the CDF of .
    • The CDF of is given as follows.
Misplaced &0 & y < 0 \\ \frac{1}{10}(y+1) & 0 \leq y \leq 9 \\ 1 & y >9 \end{cases}$$ - Listing out the individual cumulative probabilities for each $y$ is also acceptable. #### 03-11 - Gambling with a coin Two players, A and B, are flipping a fair coin together. If it comes up heads, A pays $\$1$ to B, and if it comes up tails, B pays $\$1$ to A. They play five rounds. Let $X$ be a random variable recording A's final winnings. - (a) Describe the set of possible values of $X$. - (b) Describe the PMF and CDF of $X$. **Solution** **(a)** 1. && Describe the situation. - Let $X_{i}$ be A's (not total) winnings after the $i^{\text{th}}$ round. - Since A either loses one dollar or gains one dollar each round, $X_{i} \in \{\pm1\}$. - $X = \sum_{i = 1}^{5}X_{i}$, so $X \in \{-5, -3, 1, 3, 5\}$. **(b)** 1. && Describe the PMF of $X$. - $X = 5$ if and only if all 5 rounds are tails, so $P(X = 5) = \frac{1}{2^{5}} = \frac{1}{32}$. - $X = -5$ if and only if all 5 rounds are heads, so $P(X=-5)=P(X = 5) = \frac{1}{32}$. - $X = 3$ if 4 rounds are tails, and $X = -3$ if 4 rounds are heads. Since the coin is fair, $P(X = 3) = P(X = -3) = \binom{5}{4}(0.5)^{4}(0.5)^{1} = \frac{5}{32}$. - $X = 1$ if 3 rounds are heads, and $X = -1$ if 3 rounds are tails. Since the coin is fair, $P(X = 1) = P(X = -1) = \binom{5}{3}(0.5)^{3}(0.5)^{2} = \frac{10}{32}$. 2. & Define the PMF of $X$. $$p_{X}(x) = \begin{cases} \frac{1}{32} & x = -5 \\ \frac{5}{32} & x = -3 \\ \frac{10}{32} & x = -1 \\ \frac{10}{32} & x = 1 \\ \frac{5}{32} & x = 3 \\ \frac{1}{32} & x = 5 \\ 0 & \text{otherwise} \end{cases}$$ 3. && Define the CDF of $X$. - Note that the jumps in the PDF are the possible discrete values of $X$. $$F_{X}(x) = \begin{cases} 0 & x < -5 \\ \frac{1}{32} & -5 \leq x < -3 \\ \frac{6}{32} & -3 \leq x < -1 \\ \frac{16}{32} & -1 \leq x < 1 \\ \frac{26}{32} & 1 \leq x < 3 \\ \frac{31}{32} & 3 \leq x < 5 \\ 1 & 5\leq x \end{cases}$$