1) We are creating a new card game with a new deck. Unlike the normal deck that has 13 ranks (Ace through King) and 4 Suits (hearts, diamonds, spades, and clubs), our deck will be made up of the following.

Each card will have:
i) One rank from 1 to 16.
ii) One of 5 different suits.

Hence, there are 80 cards in the deck with 16 ranks for each of the 5 different suits, and none of the cards will be face cards! So, a card rank 11 would just have an 11 on it. Hence, there is no discussion of "royal" anything since there won't be any cards that are "royalty" like King or Queen, and no face cards!

The game is played by dealing each player 5 cards from the deck. Our goal is to determine which hands would beat other hands using probability. Obviously the hands that are harder to get (i.e. are more rare) should beat hands that are easier to get.

i) How many different ways are there to get a flush (all cards have the same suit, but they don't form a straight)?
Hint: Find all flush hands and then just subtract the number of straight flushes from your calculation above.
The number of ways of getting a flush that is not a straight flush is
DO NOT USE ANY COMMAS

What is the probability of being dealt a flush that is not a straight flush?
Round your answer to 7 decimal places.

j) How many different ways are there to get a straight that is not a straight flush (again, a straight flush has cards that go in consecutive order like 4, 5, 6, 7, 8 and all have the same suit. Also, we are assuming there is no wrapping, so you cannot have the ranks be 14, 15, 16, 1, 2)?
Hint: Find all possible straights and then just subtract the number of straight flushes from your calculation above.
The number of ways of getting a straight that is not a straight flush is  
DO NOT USE ANY COMMAS

What is the probability of being dealt a straight that is not a straight flush?
Round your answer to 7 decimal places.

2) Given the following information, answer questions a - d.
P(A)=0.48P(A)=0.48
P(B)=0.41P(B)=0.41
A and B are independent.
Round all answers to 5 decimal places as needed

a) Find P(A∩B).

b) Find P(A∪B).
c) Find P(A∣B).  

d) Find P(B∣A).  

Given the following information, answer questions e - g.
P(A)=0.48P(A)=0.48
P(B)=0.41P(B)=0.41
A and B are dependent.
P(A|B) = 0.14
Round all answers to 5 decimal places as needed

e) Find P(A∩B).     

f) Find P(A∪B)

g) Find P(B∣A).

Suppose approximately 80% of all marketing personnel are extroverts, whereas about 70% of all computer programmers are introverts. (Round your answers to three decimal places.)

(a) At a meeting of 15 marketing personnel, what is the probability that 10 or more are extroverts?


What is the probability that 5 or more are extroverts?


What is the probability that all are extroverts?


(b) In a group of 4 computer programmers, what is the probability that none are introverts?


What is the probability that 2 or more are introverts?


What is the probability that all are introverts?

Suppose approximately 75% of all marketing personnel are extroverts, whereas about 60% of all computer programmers are introverts. (Round your answers to three decimal places.)

At a meeting of 15 marketing personnel, what is the probability that 10 or more are extroverts?

What is the probability that 5 or more are extroverts?

What is the probability that all are extroverts?

(b) In a group of 5 computer programmers, what is the probability that none are introverts?

What is the probability that 3 or more are introverts?

What is the probability that all are introverts?

Suppose approximately 75% of all marketing personnel are extroverts, whereas about 55% of all computer programmers are introverts. (Round your answers to three decimal places.)

(a) At a meeting of 15 marketing personnel, what is the probability that 10 or more are extroverts?

What is the probability that 5 or more are extroverts?

What is the probability that all are extroverts?

(b) In a group of 5 computer programmers, what is the probability that none are introverts?

What is the probability that 3 or more are introverts?

What is the probability that all are introverts?

Suppose approximately 80% of all marketing personnel are extroverts, whereas about 60% of all computer programmers are introverts. (Round your answers to three decimal places.)
(a) At a meeting of 15 marketing personnel, what is the probability that 10 or more are extroverts?


What is the probability that 5 or more are extroverts?


What is the probability that all are extroverts?


(b) In a group of 5 computer programmers, what is the probability that none are introverts?


What is the probability that 3 or more are introverts?


What is the probability that all are introverts?

In the Blade Runner universe, replicants are bioengineered androids that are virtually identical to humans. The “Voight-Kampff” test is designed to distinguish replicants from humans based on their emotional response to test questions. The test designers guarantee an accuracy rate of 90%. In other words, they guarantee that if a replicant is subjected to the test, then the test will correctly label them as a replicant with probability q = 90%. With the remaining probability, the test incorrectly labels the replicant as a human. Similarly, if a human is subjected to the test, then they will be correctly labelled as human with probability q = 90%, and with the remaining probability they will be incorrectly labelled as a replicant. A subject, Leon, is suspected to be a replicant. Your prior probability that Leon is a replicant equals p = 75% and with the remaining probability 1 − p = 25% you suspect Leon is a human. (a) What is the probability that if Leon takes the Voight-Kampff test, the test will label him as a replicant? (b) Leon is subjected to the Voight-Kampff test, and the test labels Leon as a replicant. What is your posterior probability about whether Leon is a replicant or not? (c) Another subject, Deckard, is also suspected to be a replicant, and your prior probability is that Deckard is a replicant with probability p1 = 10% and human with probability 1 − p1 = 90%. Deckard takes the test, and is labelled as a human. What is your posterior probability about Deckard?

Probability of getting 3 in a single toss of die =

aw of addition: page 168

1. In a manufacturing line there are several operations to produce a component. The probability of getting a surface finish defect on operation-10 is 2/3and probability of getting a dimensional defect on operation-20 is 5/9.After manufacturing finished components,

1. what is the probability of getting a surface finish defect?

2. what is the probability of NOTgetting at least one surface finish defect?

Probability of an event A is P(A) and its complement is Ā. P(A)+P(Ā)=1  or P(Ā) = 1 - P(A)

3. what is the probability of getting both type of defects in the finished component?

If A and B are independent events, the probability that both A and Bwill occur is

P(AB) = P(A∩B) = P(A) x P(B)

4. probability of either surface finish defect ordimensional defect in the finished product is

P(A U B) = P(A) + P(B) – P(A∩B) =

Law of multiplication:

2. In one shift, the probability of tool change on operation 10, P(A) is 0.6. The probability of tool change on operation 20, P(B) is 0.5. The probability of tool change on both operations 10 and 20, P(A∩B) is 0.3.

Note: The tool change on operations are independent.

P(A) = 0.6, P(B) = 0.5 and P(A∩B) = 0.3

Independent:P(AB) = P(A∩B) = P(A) x P(B)

The probability of tool change on both operations 10 and 20

P(A∩B)  = P(A) x P(B) =

The probability of tool change on eitheroperation 10 or 20

P(A) or  P(B) = P(A U B) =

3. The probability of three independent machines on an assembly line with failures are as follows:

P(A failing) = 0.15, P(B failing) = 0.05, P(C failing) = 0.10

1. What is the probability of all three machine work?

Independent failures:

P(A failing) = 0.15,  P(A NOT failing) =

P(B failing) = 0.05, P(B NOT failing) =

P(C failing) = 0.10, P(C NOT failing) =

Probability of all three machines work =

2. What is the probability of the assembly line fail (NOT work)?

Terrific Thai Food Truck wants to know about its waiting line. Currently, they use a single-server, single-phase system when serving customers. Based on historical evidence, the average number of customers arriving per hour is 80 and is described by a Poisson distribution. They estimate that they can serve 130 customers per hour with service times that follow an exponential distribution. Being that TTFT customers have infinite patience and are willing to wait in rain, sleet, and snow for the chance to get the best Thai food in town, we don’t see any balking or reneging. Being that Terrific Thai is located in Louisville, Kentucky… a city with a large population… customers come from an infinite population. The owners would like you to help them solve some thorny operational problems and figure out some system performance measures.

Compute each the following for this queueing system:

1. Average utilization.

2. Average time in the system.

3. Average time in the queue.

4. Average number of people in the system.

5. Average number of people in the queue.

6. The probability of there being more than 3 people in the system.

Each of 12 refrigerators of a certain type has been returned to a distributor because of an audible, high-pitched, oscillating noise when the refrigerators are running. Suppose that 7 of these refrigerators have a defective compressor and the other 5 have less serious problems. If the refrigerators are examined in random order, let X be the number among the first 6 examined that have a defective compressor.

(a) Calculate P(X = 4) and P(X ≤ 4).

(b) Determine the probability that X exceeds its mean value by more than 1 standard deviation.

(c) Consider a large shipment of 400 refrigerators, of which 40 have defective compressors. If X is the number among 10 randomly selected refrigerators that have defective compressors, describe a less tedious way to calculate (at least approximately) P(X ≤ 2) than to use the hypergeometric pmf. We can approximate the hypergeometric distribution with the (choose one: negative binomal, geometricm or binomal) distribution if the population size and the number of successes are large. Here n = (blank) and p = M/N = (blank). Approximate P(X ≤ 2) using that method.