- A person may be able to snap his/her fingers. Likewise, a person may be able to whistle. Let
*S*be the event that a person can snap his/her fingers. Let*W*be the event that a person can whistle. We have

P(*S*) = 0.68 and P(*W*) = 0.79

- What is the probability of the largest possible intersection of
*S*and*W*?

Since P(S) < P(W), then the largest possible intersection is

Therefore,

0.68 [Answer]

- What is the probability of the smallest possible intersection of
*S*and*W*?

Solution:

The probability is between 0 and 1 (inclusive)

Multiply by -1

Add 1.47 to both sides

The probability cannot exceed 1.

Hence, the probability of the smallest possible intersection of *S* and *W is 0.47.*

*Answer: 0.47*

- It is known that P(
*S*∩^{C}*W*) = 0.12. What does^{C}*S*∩^{C}*W*represent? (Write this out in words). What is P(^{C}*S*∪*W*) using P(*S*∩^{C}*W*) = 0.12?^{C}

Solution:

*S ^{C}*∩

*W*represents all elements in the Universal set that are not in the union of S and W.

^{C}

= 1 – 0.12

= **0.88** [Answer]

- Using the information in part
**c)**, determine P(*S*∩*W*).

[Answer]

- Draw a Venn diagram using the information in parts
**c)**and**d)**.

- Using the information in part
**c)**, determine P(*S*|^{C}*W*).

= 0.25316 [Answer]

- At the dog park on a Sunday afternoon, 35% of the dogs are purebred and the rest are mixed breeds (i.e. “mutts”). 40% of the dogs are aggressive. At the same time we know that 75% of the purebred dogs are aggressive in their encounters with any other dog.
- What is the probability that a dog is aggressive and purebred?

Let A = event that a dog is purebred

B = event that a dog is aggressive

Given:

Find:

Solution:

**0.2625** [Answer]

- What is the probability that a mixed breed dog is not aggressive?

Side Work:

Hence,

= or 0.788462 [Answer]

- We will roll 2 fair dice and are interested in the following events:

Event J: both dice show the same number

Event K: the sum of the dice is 6

Event L: the second die is a 4

- Write out all the elements in each event J, K and L and find P(J), P(K), P(L).

J = {(1,1), (2,2),(3,3),(4,4),(5,5),(6,6)}

K = {(1,5), (2,4), (3,3), (4,2), (5,1)}

L = {(1, 4), (2,4), (3,4), (4,4), (5,4), (6,4)}

n(J) = 6

n(K) = 5

n(L) = 6

There are 36 possible outcomes when two fair dice are rolled. There are 6 outcomes for the first roll and 6 outcomes for the second roll. Multiplying, there are 6 × 6 =36 outcomes in all.

Hence,

P(J) = 6/36 or 1/6

P(K) = 5/36

P(L) = 6/36 or 1/6

**b)**Find P(J ∩ L)

Solution:

Hence,

[Answer]

- Find P(K ∪ L)

Solution:

K = {(1,5), (2,4), (3,3), (4,2), (5,1)}

L = {(1, 4), (2,4), (3,4), (4,4), (5,4), (6,4)}

Hence, or [Answer]

- Is Event J independent from event K? Explain and show your work.

Solution:

If , then events J and K are independent, otherwise not independent.

J = {(1,1), (2,2),(3,3),(4,4),(5,5),(6,6)}

K = {(1,5), (2,4), (3,3), (4,2), (5,1)}

This shows that

Therefore, Event J is NOT independent from event K .

- Is Event J independent from event L? Explain and show your work.

Solution:

J = {(1,1), (2,2),(3,3),(4,4),(5,5),(6,6)}

L = {(1, 4), (2,4), (3,4), (4,4), (5,4), (6,4)}

Hence,

and from part “a”.

This shows that

Therefore, Event J is independent from event L .

- Is Event K independent from event L? Explain and show your work.

Solution:

K = {(1,5), (2,4), (3,3), (4,2), (5,1)}

L = {(1, 4), (2,4), (3,4), (4,4), (5,4), (6,4)}

Hence,

and from part “a”.

This shows that

Therefore, Event K is NOT independent from event L .

- Is event L mutually exclusive from event K? Why or why not?

which is not an empty set.

Therefore, event L is NOT mutually exclusive from event K.

- What is the probability that the sum of the dice is at least 10?

Solution:

Let T = the sum of the dice is at least 10

There are 6 elements in T.

Hence, or [Answer]

- Given that Event J has occurred, what is the probability that the sum is less than 6?

Solution:

Let A = event that sum is less than 6

A =

J = {(1,1), (2,2),(3,3),(4,4),(5,5),(6,6)}

n(A) = 10

P(A) = 10/36 = 5/18

= [Answer]

- Three components are connected to form a system as shown in the diagram below. In order for the whole system to function, component 1 must function and at least one of other components must function. Assume that each component functions or not independently of the others. The experiment consists of determining the condition of each component. Let
*S*(success) denote a functioning component and*F*(failure) denote a nonfunctioning component. For example,*SSF*, is the outcome that component 1 functioned, component 2 functioned and component 3 failed.

- What outcomes are contained in event
*G*that exactly two of the three components are functioning?

Answer: G = {FSS, SFS, SSF}

- What outcomes are contained in the event
*H*that at least two of the components are functioning?

Answer: H = {FSS, SFS, SSF, SSS}

- Assume that each outcome in the experiment is equally likely. What is the probability that the whole system is functioning?

Let W = event that the whole system is functioning.

W = {SSF, SFS, SSS}

There are 8 possible outcomes in the sample space.

Hence P(W) =

or 0.375 [Answer]

- Given that the system did not function, what is the probability that component 1 was functioning?

Let W = event that the whole system is functioning.

W = {SSF, SFS, SSS}

W^{C} = {FFF, SFF, FSF, FFS, FSS}

Let A = component 1 was functioning

A = {SFF, SFS, SSF, SSS}

[Answer]

- Suppose that with further investigation, it is known that component 1 fails 20% of the time, and components 2 fails 10% of the time and component 3 fails 5% of the time. Using this information, what is the probability that the whole system functions?

Let W = event that the whole system is functioning.

W = {SSF, SFS, SSS}

P(W) = (1-0.2)((1-0.1)(0.05) + (1-0.2)(0.1)(1-0.05) + (1-0.2)(1-.1)(1-0.05) = **0.796** [Answer]

- At a buffet dinner for 105 people, there were 3 choices for vegetables—artichokes, broccoli and cauliflower. Diners may have more than one type of vegetable or none at all. Mary Meediun recorded the vegetable choices of the diners that evening.
- 15 took cauliflower and broccoli
- 75 took broccoli or cauliflower
- 28 took artichokes
- 12 took cauliflower and artichokes
- 8 took all three
- 17 did not take any vegetables
- 33 took broccoli only
- 22 took two or more vegetables

- Draw a Venn diagram for this information—label with number of diners in each section of the Venn diagram.

- How many diners took only cauliflower?

20 diners took only cauliflower

- What is the probability that someone took only cauliflower?

Solution:

Let A = event that a diner took artichokes

B = event that a diner took broccoli

C = event that a diner took cauliflower

There are 105 diners in all.

Therefore, or [Answer]

- What is the probability that someone took broccoli?

or [Answer]

- What is the probability that someone did not take cauliflower?

or [Answer]

- What is the probability that someone took broccoli and artichokes?

[Answer]

- What is the probability that someone took cauliflower or artichokes?

or [Answer]

- What is the probability that someone took artichokes or took no vegetables?

Let D = event that someone took artichokes or took no vegetables

or [Answer]

- Of those taking broccoli, what is the probability that they also took cauliflower?

[Answer]

- At the Anytown International Airport, there is a coffee kiosk, which sells regular and decaffeinated (i.e. decaf) coffee in sizes small, medium and large. The small size coffee is purchased 30% of the time, medium size coffee 37% of the time and large size coffee the rest of the time. 62% of the small coffees, 84% of the medium coffees, and 75% of the large coffees sold are regular.
- Draw a tree diagram to illustrate the above information.

**[The tree diagram is on the next page]**

- If a customer is randomly chosen, what is the probability that he/she purchases a large coffee?

From the Tree Diagram, the probability is equal to **0.33**.

[Subtract the sum (0.3 + 0.37) from 1]

Answer: 0.33

- What is the probability that a customer purchases a coffee that is medium and regular?

From Tree Diagram,

Probability = (0.37)(0.84) = 0.3108 [Answer]

- What is the probability that a customer purchases decaf coffee?

From Tree Diagram,

Probability = (0.3)(0.38) + (0.37)(0.16) + (0.33)(0.25) = **0.2557 **[Answer]

- Given that a customer purchased decaf coffee, what is the probability that it was a small size?

Let D = event that a customer purchased decaf coffee

S= event that a customer purchased a small size coffee.

= 0.445835 [Answer]

- Is the size of the coffee purchased independent from the type (regular or decaf)? Explain your answer mathematically.

Let D = event that a customer purchased decaf coffee

S= event that a customer purchased a small size coffee.

P(S) = 0.3 and P(S|D) = 0.445835

Therefore,

This shows that the size of the coffee purchased is NOT independent from the type (regular or decaf).

- Suppose that one die is rolled and you observe the number of dots facing up. The sample space for this experiment is Ω = {1, 2, 3, 4, 5, 6}. The following table provides 5 different potential probability assignments to the possible outcomes.

Outcome | #1 | #2 | #3 | #4 | #5 |

1 | 1/6 | 0.1 | 0.2 | 1/2 | 1/16 |

2 | 1/6 | 0.15 | 0.15 | 1/4 | 1/8 |

3 | 1/6 | 0.4 | 0.2 | -1/4 | 1/4 |

4 | 1/6 | 0.05 | 0.2 | 1/2 | 0 |

5 | 1/6 | 0.1 | 0.1 | -1/8 | 7/16 |

6 | 1/6 | 0.2 | 0.2 | 1/8 | 1/8 |

- Which of the assignments #1-#5 are legitimate probability assignments? Explain your answers.

# 4 is NOT a legitimate probability assignment because some of the probabilities are negative (-1/4 and -1/8).

A valid probability value is between 0 and 1 (inclusive).

#3 is also NOT a legitimate probability assignment because the total probability exceeds 1 which is not valid (total =0.2+0.15+0.2+0.2+0.1+0.2 =1.05 > 1). The total probabilities of all outcomes in the sample space should be equal to 1.

#1, #2, and #5 are legitimate probability assignments. All probability assignments are between 0 and 1 (inclusive) and total probabilities of all outcomes in each assignment (#1, #2, and #5) is one.

- Let
*A*= die comes up odd;*B*= die comes up at least 2;*C*= die comes up at most 4;*D*= die comes up 5. Determine the probability of each of these four events using each of the legitimate probability assignments (i.e. only use the assignments you said were legitimate in part a).

A = {1,3,5}

B = {2,3,4,5,6}

C = {1,2,3,4}

D = {5}

Assignment #1

P(A) = (1/6) + (1/6) + (1/6) = ½

P(B) = (1/6) + (1/6) + (1/6) + (1/6) + (1/6) = 5/6

P(C) = (1/6) + (1/6) + (1/6) + (1/6) = 4/6 or 2/3

P(D) = 1/6

Assignment #2

P(A) = 0.1 + 0.4 + 0.1 = 0.6

P(B) = 0.15 + 0.4 + 0.05 + 0.1 + 0.2 = 0.9

P(C) = 0.1 + 0.15 + 0.4 + 0.05 = 0.7

P(D) = 0.1

Assignment #5

P(A) = (1/16) + (1/4) + (7/16) = 12/16 or 3/4

P(B) = (1/8) + (1/4) + 0 + (7/16) + (1/8) = 15/16

P(C) = (1/16) + (1/8) + (1/4) + 0 = 7/16

P(D) = 7/16

- Determine the probability that the die comes up even using each of the legitimate probability assignments in the table.

Let *E* = die comes up even

E = {2, 4, 6}

Assignment #1

P(E) = (1/6) + (1/6) + (1/6) = 1/2

Assignment #2

P(E) = 0.15 + 0.05 + 0.2 = 0.4

Assignment #5

P(E) = (1/8) + 0 + (1/8) = 1/4

- If the die is balanced, which probability assignment would be appropriate? Why?

If the die is balanced, then each outcome has an equal chance of occurring.

Therefore, probability assignment #1 would be appropriate.

- Using probability assignment #2 and the events defined in part b) above, find P(
*A*∩*B*), P(*B*∩*C*), P(*B*∪*C*), P(*A*∩^{c}*D*), P(*A*∪*C*)^{c}, P(*D*|*B*), and P(*A*∪ (*B*∩*C*)).

Solution:

A = {1,3,5}

B = {2,3,4,5,6}

C = {1,2,3,4}

D = {5}

Outcome | #2 |

1 | 0.1 |

2 | 0.15 |

3 | 0.4 |

4 | 0.05 |

5 | 0.1 |

6 | 0.2 |

[Answer]

[Answer]

[ from part ‘b’ P(B) = 0.9]

= [Answer]

[Answer]

- JJ’s Video Game store sells used games. For Pokémon fans, the store has used copies of Pokémon Pearl, Diamond and Ruby for sale. 22% of the Pokémon games sold are Pearl, 36% are Diamond and the rest are Ruby. Annie knows that counterfeit (i.e. fake) games are commonly sold in stores used. She has done some research on this and has found that a used Diamond game will be fake with a probability of 0.18, a used Pearl game will be fake with a probability of 0.1 and a used Ruby game will be fake with a probability of 0.26. Find the probability that a customer buying a used Pokémon game at this store will purchase a counterfeit Pokémon game.

Solution:

Let A = event that a customer purchased used Pokémon Pearl game.

B = event that a customer purchased used Pokémon Diamond game.

C = event that a customer purchased used Pokémon Ruby game.

D = event that a customer purchased counterfeit Pokémon game.

Given: P(A) = 0.22

P(B) = 0.36

P(C) = 0.42

P(D|B) = 0.18

P(D|A) = 0.1

P(D|C) = 0.26

Find: P(D)

Tree Diagram:

From above tree diagram, we have

P(D) = (0.22)(0.1) + (0.36)(0.18) + (0.42)(0.26) = **0.196** [Answer]