More Exercises

Section 4.9 More Exercises

Checkpoint 4.9.1. Conditional Basic computation.

Given \(P(A) = 0.43, P(B) = 0.72,\) and \(P(A \cap B) = 0.29\text{,}\) determine

\(\displaystyle P(A \cup B)\)
\(\displaystyle P(B | A)\)
\(\displaystyle P(A | B)\)
\(\displaystyle P(A^c \cap B^c)\)

Use the Theorems and Definitions provided to complete the first three. For the fourth part, it might be useful to use DeMorgan's Laws from set theory to rewrite \(A^c \cap B^c = (A \cup B)^c\) and then use a theorem.

Checkpoint 4.9.2. Gender vs University Major.

The table below classifies students at your university according to gender and according to major.

Table 4.9.3. Gender vs Major

Enrollment	Male	Female	Totals
STEM	420	510	930
Business	320	270	590
Other	610	710	1320
Totals	1350	1490	2840

Determine the following:

\(\displaystyle P( \text{STEM major} )\)
\(\displaystyle P( \text{STEM | Female} )\)
\(\displaystyle P( \text{Female | STEM} )\)
\(\displaystyle P( \text{Female | Not STEM} )\)

Checkpoint 4.9.4. Mean Tough Teacher.

You are in a probability and statistics class with a teacher who has predetermined that only one student can make an A for the course. To be "fair", he places a number of slips of paper in a bowl equal to the number of students in the course with one of the slips having an A designation. Students in the course each can pick once randomly from the bowl and without replacement to see if they can get the lucky slip. Determine the following:

If there are 15 students in your course, determine the probabilities of getting an A in the course if you pick first and if you pick last.
Since the teacher likes you the most, she will give you the option of deciding whether to pick at any position. If so, determine the position that would give you the best likelihood of getting the A slop.
Suppose again that the teacher was feeling more generous and decided instead to allow for two A's. Determine how that changes your likelihood of winning and on what position you would like to choose.
Continue as above except that only one slip does not have an A on it.
Discuss how your choice is affected by the number of students in the course or the number of A slips included.

Solution

Using the normal equally-likely definition, \(P(\text{first}) = \frac{1}{15}\text{.}\)

To get the A on the last pick requires that all of the previous picks to be something else. You don't get the opportunity to pick the A if it has already been selected. So, if L stands for losing (not getting the A), then

\begin{align*} P(\text{last}) & = P(\text{LLLLLLLLLLLLLLA}) \\ & = \frac{14 \cdot 13 \cdot 12 \cdot 11 \cdot 10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}{15 \cdot 14 \cdot 13 \cdot 12 \cdot 11 \cdot 10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2} = \frac{1}{15}. \end{align*}

Therefore, it is the same probability of getting the A whether you pick first or last. In general, to win on the kth pick gives

\begin{align*} P(\text{kth}) & = P(\text{LL...LA})\\ & = \frac{14 \cdot 13 \cdot ... \cdot (15-k) \cdot 1}{15 \cdot 14 ... \cdot (16-k) \cdot (15-k)} = \frac{1}{15} \end{align*}

Hence, it is the same probability regardless of when you get to pick.

If there are two A's possible, then the options for person k in include either receiving the first of the two slips or the second. The probability for determining the first of the two is computed in a manner similar to above except that there is one more A and one less other.

\begin{align*} P(\text{kth as first}) & = P(\text{LL...LA})\\ & = \frac{13 \cdot 12 \cdot ... \cdot (15-k) \cdot 2}{15 \cdot 14 ... \cdot (16-(k+1)) \cdot (16-k)} = \frac{2 \cdot (15-k)}{15 \cdot 14} \end{align*}

The probability of getting the second A means exactly one of the previous k-1 selections also picked the other A. There are k-1 ways that this could happen. Computing for one of the options and multiplying by k-1 gives

\begin{align*} P(\text{kth as second}) & = P(\text{LL...LAA})\\ & = (k-1) \cdot \frac{13 \cdot 12 \cdot ... \cdot (15-k) \cdot 2 \cdot 1}{15 \cdot 14 ... \cdot (16-k) \cdot (15-k)} = \frac{2 \cdot (k-1)}{15 \cdot 14}. \end{align*}

Adding these two together gives

\begin{gather*} P(\text{getting an A when there are two}) = \frac{2 \cdot (15-k) + 2 \cdot (k-1)}{15 \cdot 14}\\ = \frac{28}{15 \cdot 14} = \frac{2}{15}. \end{gather*}

For example, if k = 5,

\begin{gather*} P(\text{5th as first}) = P(\text{LLLLA}) \\ = \frac{13 \cdot 12 \cdot 11 \cdot 10 \cdot 2}{15 \cdot 14 \cdot 13 \cdot 12 \cdot 11} = \frac{20}{15 \cdot 14} \end{gather*}

\begin{gather*} P(\text{5th as second}) = P(\text{LL...LAA}) \\ = 4 \cdot \frac{13 \cdot 12 \cdot \cdot 11 \cdot 2 \cdot 1}{15 \cdot 14 \cdot 13 \cdot 12 \cdot 11} = \frac{8}{15 \cdot 14}. \end{gather*}

Adding these together yields the general result. So, once again, it doesn't matter which pick you use since the likelihood of getting an A is the same for all positions.

Checkpoint 4.9.5. Shared Birthdays.

In this problem, you want to consider how many people are necessary in order to have an even chance of finding two or more who share a common birthday. Toward that end, assuming a year has exactly 365 equally likely 4.3.11 days let r be the number of people in a sample and consider the following:

Determine the number of different outcomes of birthdays when order matters and birthdays are allowed to be repeated.
Determine the number of different outcomes when birthdays are not allowed to be repeated.
Determine the probability that two or more of your r students have the same birthday.
Prepare a spreadsheet with the probabilities found above from r=2 to r=50. Determine the value of r for which this probability is closest to 0.5.
As best as you can, sample two groups of the size found above and gather birthday information. For each group, determine if there is a shared birthday or not. Compare your results with others in the class to check whether the sampling validates that about half of the samples should have a shared birthday group.

Solution

The correct sample size to get past a probability of 0.5 is 23 people. You should justify this numerically by justifying the following probabilities:

Checkpoint 4.9.6. Internet meme solution.

This one is from an internet meme: Two fair 6-sided dice are rolled together and you are told that at least one of the dice is a 6. Given that a 6 will be removed, determine the probability that the other die is a 6.

Solution

In this case, you are presented with an outcome where the possible choices consist of

\begin{equation*} (1,6), (2,6), (3,6), (4,6), (5,6), (6,6), (6,5), (6,4), (6,3), (6,2), (6,1). \end{equation*}

Each of these would satisfy the condition that at least one of the dice is a 6. From this group, the only success that satisfies being a 6, given that another 6 has already been removed, is the (6,6) outcome. Therefore, the conditional probability is 1/11.

It is interesting to note that if the question instead was posed so that one of the dice was a 6 and it was removed, then the probability of the other dice showing a 6 would be 1/6.

Checkpoint 4.9.7. 100 people on an airplane with boarding pass issues.

This is a famous problem. 100 people are in line, boarding an airplane with 100 seats, one at a time. They are in no particular order. The first person has lost his boarding pass, so he sits in a random seat. The second person does the following:

Goes to his seat (the one it says to go to on the boarding pass). If unoccupied, sit in it.
If occupied, find a random seat to sit in.

Everyone else behind him does the same. What is the probability that the last person sits in his correct seat?

Solution

To get the idea, consider what happens with only 2 people, then only 3. Generalize.

The answer is 1/2. To obtain this, you can define recursively the probability that the kth person sits in their own set as f(k). Consider the first traveler's and your seats. Then you get the following cases:

P(first guy sits in his own seat and you sit in yours) = \(\frac{1}{k} \cdot 1\)
P(first guy sits in your seat and you do not sit in yours) = \(\frac{1}{k} \cdot 0\)
P(other k-2 travelers make their choices) = \((k-2) \frac{1}{k} f(k-1)\)

\begin{equation*} f(k) = 1/k + 0 + (k-2)/k f(k-1) \end{equation*}

with f(2) = 1/2.

For example,

f(3) = 1/3 + f(2)/3 = 1/3 + 1/6 = 1/2.
f(4) = 1/4 + 2/4 f(3) = 1/4 + 1/2 1/2 = 1/2.
f(5) = 1/5 + 3/5 1/2 = 1/2.
f(6) = 1/6 + 4/6 1/2 = 1/2.

Etc.

Checkpoint 4.9.8. Basic Independence Calculations.

Given \(P(A) = 0.43, P(B) = 0.72, \) and \(P(A \cap B) = 0.31\text{,}\) verify that A and B are not independent.

Solution

A and B are independent by definition 4.7.1 provided

\begin{equation*} P(A \cap B) = P(A) P(B). \end{equation*}

Using the provided values, notice that

\begin{equation*} P(A \cap B) = 0.31 \end{equation*}

but

\begin{equation*} P(A)P(B) = 0.43 \cdot 0.72 = 0.3096. \end{equation*}

Since these are not equal (regardless how close) then A and B are not independent.

Checkpoint 4.9.9. Compound events and Independence.

Given A, B, and C are independent events 4.7.1, with \(P(A) = 2/5, P(B) = 3/4,\) and \(P(C) = 1/6\text{,}\) determine:

\(\displaystyle P(A \cap B \cap C)\)
\(\displaystyle P(A^c \cap B^c \cap C)\)
\(\displaystyle P(A \cup B \cup C)\)

Solution

Extending the definition of independent events 4.7.1 gives

\begin{equation*} P(A \cap B \cap C) = \frac{2}{5} \frac{3}{4} \frac{1}{6}. \end{equation*}

By the corollary for independent events, complements also maintain a similar independence. So

\begin{equation*} P(A^c \cap B^c \cap C) \frac{3}{5} \frac{1}{4} \frac{1}{6} . \end{equation*}

To complete the third part, use the inclusion/exclusion result 4.3.9 for dealing with three sets.

Checkpoint 4.9.10. Rolling multiple dice.

For a pair of dice you want to consider the events A = {rolling a 7 or 11} and B = {otherwise}...as in the first roll in the game of craps. Further, for notation purposes let's take ABA (for example) to mean event A occurs on the first roll, event B occurs on the second roll, and event A occurs again on the third roll...in that order only. If you roll the dice 5 times, determine

P(AABBB)
P(BBBAA)
The probability of getting A on exactly two rolls of the dice.

Solution

Successive rollings of a pair of dice are independent events 4.7.1. Therefore,

\begin{equation*} P(AABBB) = P(A)P(A)P(B)P(B)P(B) = \frac{8}{36} \frac{8}{36} \frac{28}{36} \frac{28}{36} \frac{28}{36} \end{equation*}

Similarly for the second part.

For the third part, notice that there will be \(\binom{5}{2}\) ways to rearrange 2 A's and 3 B's but that each of these will have two 8/36's and three 28/36's but just in a different order. Therefore, you will get

\begin{equation*} 10 \cdot \frac{8}{36} \frac{8}{36} \frac{28}{36} \frac{28}{36} \frac{28}{36} \end{equation*}

Checkpoint 4.9.11. Redundancy.

To help "insure" the success of a mission, you propose several redundant components so that the mission is a success if one or more succeed. Supposing that these separate components act independently of each other and that each component has a 75% chance of success, determine:

The probability of failure if you utilize 2 components.
The probability of failure if you utilize 5 components.
The number of components needed to insure that the probability of success is at least 99%.

Checkpoint 4.9.12. Internet Meme redux.

Again, from an internet meme: Two fair 6-sided dice are rolled together and you are told that at least one of the dice is a 6. A 6 is removed and you are presented with the other die. Determine the probability that it is a 6.

Solution

For this setting, notice that the outcomes from each of the two dice are independent of each other. Removing one of the dice, regardless of it's value, does not affect the other. The question in this case does not ask for a conditional probability.

Checkpoint 4.9.13. Single Elimination Tournament.

Consider a n=4 team single-elimination tournament where the teams are "seeded" from 1 (the best team) to 4 (the worst team). For this tournament, team 1 plays team 4 and team 2 plays team 3. The winner of each play each other to determine the final winner. When teams j and k play, set P(j wins) = \(\frac{k}{j+k}\) and similarly for team k. Assuming separate games are independent of each other, determine the probability that team 4 wins the tournament. What about with 8 teams? What about 64 teams?

Solution

P(4 wins) = P(4 beats 1) P(4 beats the winner of the other bracket)

P(4 wins) = (1/5) * P(4 beats 2 | 2 beats 3) + P(4 beats 3 | 3 beats 2)

P(4 wins) = 1/5 [(3/5)(2/6) + (2/5)(3/7)] = 78/1050 = 0.0742

For the other teams:

P(1 wins) = 4/5 [(3/5)(2/3) + (2/5)(3/4) ] = 0.56

P(2 wins) = 3/5 [(4/5)(1/3) + (1/5)(4/6) ] = 0.24

P(3 wins) = 2/5 [(4/5)(1/4) + (1/5)(4/7) ] = 0.1257