W01 - HW Solutions

01 - Sample space - roll a die, flip a coin

A normal -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?

Solution (a)

1. & Describe the sample space.
   • Since there are 6 possible results of rolling a die, and possible results of a coin flip, our sample space has elements.
2. & Use set builder notation to describe the sample space .

(b)

1. ! Note that the number of possible events amount to counting how many subsets there are of . In other words, we are asked to compute .
2. & Compute .

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

A normal -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?

Solution (a)

1. && Divide the sample space into two disjoint sets.
   • Denote as the sample space where the result of the die is even.
   • Denote as the sample space where the result of the die is odd.
   • .
2. & Describe .
   • There are even numbers on a die, and possible results of each coin flip. Since coin flips are independent has elements.
3. & Write using set builder notation.
4. & Describe .
   • There are odd numbers on a die, and 2 possible results of a coin flip. has elements.
5. & Write using set builder notation.
6. & Describe .
   • As above, .
   • Since and are disjoint, .

(b)

1. ! Note that the number of possible events amount to counting how many subsets there are of . In other words, we are asked to compute .
2. & Compute .

03 - Events - descriptions to sets

You are modelling 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”

Solution (a)

1. & Since only the third car is broken, and the other three cars can have any status, the relevant set is

(b)

1. & In this case, either all cars are good or all cars are broken. Therefore, the relevant set is

(c)

1. & In this scenario, the only combination not in the relevant set is ‘GGGG’. Therefore,

(d)

1. & In this scenario, given two cars and , . Therefore,

04 - Venn diagrams - sets rules and Kolmogorov additivity

Suppose we know three probabilities of events: , and .

Calculate: , and .

Solution

1. && is computed by directly applying the inclusion-exclusion principle.
2. &
3. &
4. && We can express as . Therefore,
5. &

05 - Inclusion-exclusion reasoning.

Your friend says: “according to my calculations, the probability of is and the probability of is , but the probability of and both happening is only .”

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

Solution

1. & State the inclusion-exclusion principle.
2. & Examine possibilities based on given values.
   • Given that is and is , we have that .
   • Since , we know that .
   • Therefore, .

06 - Inclusion-exclusion reasoning.

Suppose and . Show that .

Solution

1. & State the inclusion-exclusion principle.
2. & Examine the maximum value of .
   • We know that .
   • Given that and , .
3. & Examine the minimum value of .
   • The minimum value of is the maximum of the individual values.
   • .
   • Therefore, .

07 - At least two heads from three flips

A coin is flipped three times.

What is the probability that at least two heads appear?

Solution

1. & Describe the sample size of this experiment.
2. & Find the probability that at least two heads appear.
   • The sequences of flips that contain at least two heads are , , , .
   • We know that , thus

08 - Conditioning - restrict to 4th-year students

What is the likelihood that a randomly chosen 4th year student passed the test? What about for 1st-year students?

Solution

1. & We are asked to compute . Set up the conditional probability formula.
2. & We have from the table that and . Therefore,
3. & For 1st-year students, we have

09 - Conditioning - two dice, at least one if 5

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

What is the probability that their sum is 10?

Solution

1. & Let be the outcome of the first die and be the outcome of the second die. We are asked to compute
2. && Compute individual probabilities.
   • There is only one combination out of the 36 possible combinations of two dice rolls in which at least 1 die rolls a 5 and both sum up to 10 (5, 5).
   • There are combinations of dice rolls in which at least one is a .
3. & Plug into formula.

10 - Conditioning - two dice, differing numbers.

Two dice are rolled, and the outcomes are different.

What is the probability of getting at least one ?

Solution

1. & Let be the outcome of the first die and be the outcome of the second die. We are asked to compute
2. && Compute individual probabilities.
   • There are combinations in which at least one die rolled a . Since one of these combinations is , we have 10 combinations in which the outcomes are unequal.
   • There are combinations in which the outcome of the two dice differ.
3. & Plug into formula.

11 - 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 ?

Solution (a)

1. & Let be the outcome of the first card, be the outcome of the second card, and denote “hearts”. Since there are 52 cards in a standard deck with 13 of them being hearts, we have

(b)

1. & Similarly to part (a), we have

• ! End