Probability Extra Credit Matthew Mucharski PDF

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1. When a dime is tossed four times, there are 16 possible outcomes: HHHH, HTHH, THHH, TTHH, HHHT, HTHT, THHT, TTHT, HHTH, HTTH, THTH, TTTH HHTT, HTTT, THTT, TTTT List the outcomes that constitute each of the following events and then calculate the probability of each event.

a. A = event exactly two heads are tossed Exactly two heads are tossed = {TTHH, HTHT, THHT, HTTH, THTH, HHTT} Total outcomes = 6 Sample size = 16 Therefore, P (event exactly two heads are tossed) = 6/16 or 3/8. Or 37.5% So we can see that there is 37.5% chance of getting exactly two heads when a dime is tossed 4 times.

b. B = event the first two tosses are tails The first two tosses are tails = {TTHH, TTHT, TTTH, TTTT} Total outcomes = 4 Sample size = 16 Therefore, P (event the first two tosses are tails) = 4/16 or 1/4. Or 25% So we can see that there is 25% chance of getting tails in the first two tosses when a dime is tossed 4 times.

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c. C = event the first toss is heads The first toss is heads = {HHHH, HTHH, HHHT, HTHT, HHTH, HTTH, HHTT, HTTT} Total outcomes = 8 Sample size = 16 Therefore, P (event the first toss is heads) = 8/16 or 1/2. Or 50% So we can see that there is 50% chance of getting heads in the first toss when a dime is tossed 4 times.

d. D = event all four tosses come up the same All four tosses come up the same = {HHHH, TTTT} Total outcomes = 2 Sample size = 16 Therefore, P (event all four tosses come up the same) = 2/16 or 1/8 Or 12.5% So we can see that there is 12.5% chance of getting same heads or tails when a dime is tossed 4 times.

e. E = At least one tail At least one tail = {TTTT, HTHH, THHH, TTHH, HHHT, HTHT, THHT, TTHT, HHTH, HTTH, THTH, TTTH, HHTT, HTTT, THTT} 2

Total outcomes = 15 Sample size = 16 Therefore, P (at least one tail) = 15/16 Or 93.75% So we can see that there is 93.75% chance of getting at least one tail when a dime is tossed 4 times.

f. F = At least on head At least one head = {HHHH, HTHH, THHH, TTHH, HHHT, HTHT, THHT, TTHT, HHTH, HTTH, THTH, TTTH, HHTT, HTTT, THTT}

Total outcomes = 15 Sample size = 16 Therefore, P (at least one tail) = 15/16 Or 93.75% So we can see that there is 93.75% chance of getting at least one head when a dime is tossed 4 times.

2. Refer to problem 1. For each of the following events, list the outcomes that comprise the event, and then calculate the probability of each event. a. (not B) The first two tosses are tails = {TTHH, TTHT, TTTH, TTTT} Total outcomes = 4

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Sample size = 16 Therefore, P (event the first two tosses are tails) or P(B) = 4/16 or 1/4 P (not B) = 1 – P(B) = 1 – 1/4 = 3/4 or 75% So we can see that there is 75% chance of not getting tails in the first two tosses when a dime is tossed 4 times. b. (A & B) For A, when exactly two heads are tossed = {TTHH, HTHT, THHT, HTTH, THTH, HHTT} For B, when first two tosses are tails = {TTHH, TTHT, TTTH, TTTT} P (A & B) = {TTHH} Total outcomes = 1 Sample size = 16 Therefore, P (A & B) = 1/16 or 6.25% So we can see that there is 6.25% chance of getting both A & B when a dime is tossed 4 times. c. (C or D) For C, first toss is heads = {HHHH, HTHH, HHHT, HTHT, HHTH, HTTH, HHTT, HTTT} For D, all four tosses come up the same = {HHHH, TTTT} P (C or D) = P(C) + P(D) – P (C & D) = 1/2 + 1/8 – (1/2 * 1/8) = 9/16 or 56.25%

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So we can see that there is 56.25% chance of getting either C & D when a dime is tossed 4 times.

3. For high school students graduating in 2010, college admissions to the nation’s most selective schools were the most competitive in memory. Harvard accepted only about 9% of the applicants, Stanford 10% and Penn 16%. John has applied to all three schools. Assuming that he is a typical applicant, and that admissions to the three schools are independent events, compute the following probabilities. a) What is the probability that John will be accepted at both Harvard and Stanford, but not at Penn? P (getting accepted at Harvard) = 9% or 9/100 P (getting accepted at Stanford) = 10% or 1/10 P (getting accepted at Penn) = 16% or 4/25 Since the admission to three schools are independent events, P (getting accepted at Harvard & Stanford) = P (getting accepted at Harvard) * P (getting accepted at Stanford) = 9/100 * 1/10 = 9/1000 or 0.009% P (not getting acceptance at Penn) = 1 – P (getting accepted at Penn) = 1 – 4/25 = 21/25 Therefore, P (getting accepted to Harvard & Stanford but not in Penn) = 9/1000 * 21/25 = 0.00756 or 0.756% So we can see that there is 0.756% chance of getting accepted to Harvard & Stanford but not in Penn which is a rare possibility. b) What is the probability that John will not be accepted by any of the three schools? P (not getting acceptance at Harvard) = 1 – P (getting accepted at Harvard) = 1 – 9/100 = 91/100 or 0.91 P (not getting acceptance at Stanford) = 1 – P (getting accepted at Stanford) = 1 – 1/10 = 9/10 or 0.9 P (not getting acceptance at Penn) = 1 – P (getting accepted at Penn) = 1 – 4/25 = 21/25 or 0.84 P (not getting accepted by any of three schools) = 91/100 * 9/10 * 21/25 = 0.68796 or 68.796% So we can see that there is 68.796% chance of not getting accepted to any of the three schools which is quite a possibility.

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c) What is the probability that at least one school will accept him? P (getting accepted by at least one school) = 1 – P (not getting accepted by any of three schools) = 1 – 0.68796 = 0.31204 or 31.204% So we can see that there is 31.204% chance of getting accepted to at least one school which is a possibility.

4. A usability engineer claims that in usability experiments there is a 59% chance (or a probability value of 0.59) that a user tester discovers a minor usability problem of a certain product. Suppose the usability engineer wants to conduct a usability study to evaluate the quality of a new computer interface, and recruits 5 user testers. Compute the following probabilities: a) What is the probability that all 5 user testers will find a minor problem with the interface? P (chance of getting minor usability problem) = 0.59 or 59% P (getting all 5 user testers to find a minor problem with interface) = P (User 1 finding it) * P (User 2 finding it) * P (User 3 finding it) * P (User 4 finding it) * P (User 5 finding it) =0.59*0.59*0.59*0.59*0.59 = 0.0714 or 7.14% So we can see that there is 7.14% chance that all 5 user testers will find a minor problem with the interface. b) What is the probability that at least one user out of the 5 user testers will find a minor problem? P (getting at least one user out of 5 user testers) = (1 – 0.59) ^ 5 = 0.41 * 0.41 * 0.41 * 0.41 * 0.41 = 0.0115 or 1.15% So we can see that there is 1.15% chance that at least one user out of all 5 user testers will find a minor problem with the interface which is quite a rare possibility. 5. Social media tools such as Facebook or Twitter are dominated by younger population, but adult Internet users are increasingly using online social networks. Here is the distribution of social network users by age group: Age

Probability

Teen (12-18 years old) 0.45

Young Adult (19-29 years old) 0.35

Adult (30-49 years old) 0.15

Elderly population (50-70 years old) 0.05

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a) Is this is a valid probability table? Explain your answer. For checking the validity, we add all the probabilities of all age groups. 0.45 + 0.35 + 0.15 + 0.05 = 1 or 100% This shows that the The probability of an event, expressed as P(event), is always between 0 and 1 (inclusive) which in this case, all over probabilities were inclusive between 0 & 1. Also the probability of the entire table is 1 which shows that the probability is certain (at 1) and hence it is a valid probability table. b) What’s the probability that a randomly chosen social network user is between the age of 19 and 49 years? P (randomly choosing social network user between age of 19 and 49 years) = P (user having age 19-29 years old) * P (user having age 30-49 years old) = 0.35 * 0.15 = 0.0525 or 5.25% So we can see that there is 5.25% chance that a randomly chosen social network user is between the age of 19 and 49 years which is possibility.

c) What’s the probability that a social network user is not a teenager (that is older than 18)? P (social network user not a teenager) = P (user having age 19-29 years old) * P (user having age 30-49 years old) * P (user having age 50-70 years old) = 0.35 * 0.15 * 0.05= 0.002625 or 0.2625% So we can see that there is 0.2625% chance that a social network user is not a teenager which is quite a rare possibility as social networking is quite a rage especially amongst teenagers. 6. A 65-year-old woman takes out a $100,000 term life insurance policy. The company charges a premium of $520. Mortality tables indicate that only 2.6% of women age 65 die within a year. a) Create a probability table that displays the probability associated to insurance gain/loss on the individual policy (similar to the examples in the notes and in the book) X insurance profit/loss ($) Probability

$520

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$ -99,480 (payout in case of death) 0.026

b. Estimate the company’s expected profit on such policies for 65-year-old women.

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520*1 + 0.026(-99480) = 520 - 2586.48 = -2066.48 The company made loss of approximately $2066 on such policies for 65-year-old women.

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