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The Probability Game Week 6, Wednesday. Teams This game affects your Quiz 4 Grade First place:+4 points Second place: +3 points Third place: +2 points.

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Presentation on theme: "The Probability Game Week 6, Wednesday. Teams This game affects your Quiz 4 Grade First place:+4 points Second place: +3 points Third place: +2 points."— Presentation transcript:

1 The Probability Game Week 6, Wednesday

2 Teams

3 This game affects your Quiz 4 Grade First place:+4 points Second place: +3 points Third place: +2 points Fourth place: +1 point Fifth place: +0 points Last place: -1 point ** You can’t earn more than 10/10 on quiz 4 **

4 Study These Questions Most of these questions are similar to the questions you’ll find on midterm #2!!!

5 Rules There are 7 questions, and they will appear on the screen one at a time. For each question, the group will discuss the solution, and the Team Captain will record a single answer and hand the paper into me. We will discuss the solution as a group, and the paper will be handed back for the next question to be answered. The order of winners is decided by the number of correct answers each team provides.

6 Question #1 Question: Which of these probability assignments are possible? Assignment A XP[X] 025% 1 2 3 Assignment B XP[X] 120% 230% 3 420% Assignment C XP[X] 0100% 10% 2 3 Assignment D XP[X] 0-5% 15% 2100% 30% Assignment E XP[X] 010% 130% 240% 350% Answer: A, B, C

7 Question #2 Question: In a standardized test 80% of the students pass. Three students take the test. What is the probability that: (a) All three pass (b) None pass (c) At least one passes Answer: (a) P[1p and 2p and 3p] = P[1p]*P[2p]*P[3p] =.8*.8*.8 = 51% (b) P[1f and 2f and 3f] = P[1f]*P[2f]*P[3f] =.2*.2*.2 =.8% (c) P[At least one passes] = 1 – P[nobody passes] = 99.2%

8 Question #3 Question: Let X represent the number of fights at “Tiny Tavern” during a week. The following lists the probability model for X: (a) Find P[more than 2 fights this week] (b) Find P[6 fights this week] Answer: (a) P[more than 2 fights this week] = P[3]+P[4]+P[5] = 10% (b) P[6 fights this week] = 0% X012345 P[X]50%30%10%?3%2%

9 Question #4 Question: Let X represent the number of fights at “Tiny Tavern” during a week. The following lists the probability model for X: Calculate Answer: X012345 P[X]50%30%10%?3%2% X012345 P[X]50%30%10%?3%2% xP[x]0.3.2.15.12.1 E[X] = 0 +.3 +.2 +.15 +.12 +.1 =.87

10 Question #5 Question: Administration wants to find out if UA students felt appreciated on Student Appreciation Day. To do this they randomly select students and surveyed them about the experience. Suppose the data are gathered and summarized in the table below. (a) Find P[Appreciated | Grads] and (b) P[Appreciated]. (c) Can you conclude independence? Felt Appreciated? Yes No Undergrad 100 250 Grad 50 125 Answer: P[A | G] = 50/175 = 28.6%. P[A] = 150/525 = 28.6%. (equal therefore indep.)

11 Question #6 Question: For the following two situations, list all of the outcomes in the sample space and tell whether the outcomes are equally likely: (a) Flip a coin until you observe a tail or have four tosses. (b) Toss the coin four times and record the number of tails. Answer (a): {T, HT, HHT, HHHT, HHHH}. These outcomes are not equally likely – for example, the probability of flipping a tails on the first try is 50%, but the probability of flipping four heads in a row is much smaller. Answer (b): {0, 1, 2, 3, 4}. These outcomes are not equally likely. This is a binomial distribution with n=4 and p=50%.

12 Question #7 Question: (a) How many binomial distributions exist? (b) What two things do you need to know in order to identify the exact binomial distribution needed to solve a given problem? (c) A telemarketer makes 1,000 phone calls per day. The probability that he will make a sale to a given person is 0.5%. Assume each person he calls is independent of the others. Define a random variable, X, as the number of sales made in a given day. Is this binomial? If so, what is the mean and variance of X? Answer (a): Infinite. For every n and p, there is a unique binomial distribution. Answer (b): n (number of trials) and p (probability of success) Answer (c): Binomial. E[X] = np = 1000(.005) = 5. VAR[X] = npq = 1000(.005)(.995) = 4.98


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