Statistics 1040 Summer 2009 Exam III 1. For the following basic probability questions. Give the RULE used in the appropriate blank (BEFORE the question), for each of the following situations, using one of the following letters: a. Simple multiplication rule: P(A) * P(B) b. More complex multiplication rule: P(A) * P(B A) c. Simple addition rule: P(A) + P(B) d. More complex addition rule: P(A) + P(B) - something. e. Complement rule. (a). Use rule -- for cases such as the odds of flipping 5 heads in a row with a fair coin. AND = multiply; INDEPENDENT events (c). Use rule -- for cases such as the probability of finding a student who scored EITHER above 700 OR below 400 on the SAT. OR = add; EXCLUSIVE events (b). Use rule --- for cases such as the chances of drawing a flush (5 cards of the same suit) in a 5-card deal. AND = multiply; DEPENDENT events (d). Use rule -- for cases such as finding the probability of finding books which have EITHER the term "statistics" or the term "regression" as part of their title or description in the card catalog. OR = add; events NOT EXCLUSIVE (e). Use rule -- for cases such as those of rolling at least one double six in 24 rolls of a pair of dice. Complement rule: 1 - Prob (rolling NO double sixes) 2. Explain what the notation P(B A) means in option (b) of question 1, exactly what it would mean in the case or cases you applied it to. Conditional probability of B GIVEN A. Probability of flush = 13/52 * 12/51 * 11/50 * 10/49 * 9/48 3. Explain what the "something" means in option (d) of question 1, and exactly what it means in the case or cases you applied it to. Subtract the probability of BOTH events happening at the same time. P ( statistics OR regression ) = P( statistics ) + P( regression ) - P ( statistics AND regression ) 4. What is the chance of rolling at least one double 6 in 24 rolls of a pair of dice? Set up the problem as carefully as possible, using one of the rules above, but without using your calculator -- but be sure you set it up so that it takes the smallest possible number of calculator punches to solve. The probability of NO double six on a single roll is 35/36; hence the probability of NO double sixes in 24 rolls is 35/36 th to the 24 th power, and the complement rule will give us the probability of AT LEAST one double six in 24 rolls: 1 P(no double sixes) = 1 - (35/36) 24 = 1.0 0.5086 = 0.4914 or 49.14 percent
5. A die (= one of a pair of dice) is rolled 6 times. Leave the answers as appropriate fractions (that is, don't use your calculator here). a. The chance that the first roll is an ace (one-spot) OR that the last roll is an ace is: EXTENDED ADDITION RULE: 1/6 + 1/6-1/36 = 11/36 Events are NOT exclusive b. The chance that the first roll is an ace (one-spot) AND the last roll is an ace is: SIMPLE MULTIPLICATION RULE: 1/6 * 1/6 = 1/36 Events are INDEPENDENT. c. The chance that the first card in the deck is an ace OR the last card is an ace is EXTENDED ADDITION RULE : 4/52 + 4/52-4/52 * 3/51 Events are not exclusive. d. The chance that the first card in the deck is an ace AND the last card is an ace is MULTIPLICATION RULE FOR DEPENDENT EVENTS 4/52 * 3/51 Events are not independent. 6. In question 5, how and why do your answers to (b) and (d) differ in terms of the rules employed? In b, events are independent; in d, they are dependent. 7. The formula N! / k! (N - k)! is part of the binomial formula. True or false (and of course state and explain the correct answer fully if the statement or parts of the statement are false). _F a. The exclamation points mean that the formula is really, really important. Notation for factorial: (for example, 6! = 6 * 5 * 4 * 3 * 2 * 1) _T_b. The formula is used in counting the number of ways to get k heads in N tosses of a coin. _F c. The rest of the binomial formula involves adding the above formula to p * (1 - p), where p is the probability of heads. The rest of the formula is (pow p k) * (pow (1 p) (N k), and we MULTIPLY this by the binomial coefficient, not add. 8. Find the probability of getting exactly 3 heads in 8 tosses of a coin, if the coin is weighted so that the probability of heads is 0.6. Set up the problem AND use your calculator to give the answer to 4 places. [8! / (3! * 5!)] * (pow 0.6 3) * (pow 0.4 5) = 56 * 0.2160 * 0.0102 = 0.1239 or 12.39 percent.
9. An oil company thinks that there is a 40 percent chance (on the basis of the geological formations) that a well in a certain spot will hit substantial amounts of oil. They hire a geologist who drills test holes to be more certain; in the past, he has been right 70 percent of the time whether or not substantial amounts of oil are in a spot. If the geologist's tests indicate a substantial amount of oil, the oil company should: (a) Disregard the geologist, since he is right only a bit more often than not, and assume the real chance is still forty percent. (b) Conclude that there is a seventy percent chance of hitting substantial amounts of oil when they drill. (c) Revise their prior probability to somewhere between 40 and 70 percent. (d) Revise their prior probability to another value. The correct option is: (c) Explain your answer, and give the exact revised probability HERE: 60.87 percent Show your work below, setting up a table with OIL / NO OIL at the top, and the geologists indicator of substantial oil. (TESTS YES / TESTS NO) along the side. Note that the past record of the geologist shows that he gives the right information 70 percent of the time, but the question is about the probability of oil given his information, and our prior probability for oil is only 40 percent. While his information should raise the probability, it will not raise it to 70 percent. With some notation: ( plus = geologist says there is oil; minus = geologist says there is no oil) P (+ oil) = 0.70 P (+ no oil) = 0.30 P (- oil) = 0.30 P(- no oil) = 0.70 We fill in the table below by noting that P (+ AND oil) = P(+ oil) * P(oil) = 0.7 * 0.4 = 0.28 P (+ AND no oil) = P(+ no oil) * P(no oil) = 0.3 * 0.6 = 0.18 and so on: OIL NO OIL TESTS YES 0.28 0.18 TESTS NO 0.12 0.42 We know that the geologist said OIL, so we know we are in the first row of the table. Given our prior probability, he will say oil 0.28 + 0.18 = 0.46 or 46 percent of the time; he will be right 28 / 46 = 60.87 percent of the time, and wrong 18 / 46 = 39.13 percent of the time
10. A gambler is going to play roulette 25 times, putting a dollar on a SECTION BET covering numbers 13 through 24 each time (a section bet will cover 12 numbers --of the 38 numbers on the wheel -- and will pay 2 to 1, which means he wins 2 dollars if one of the 12 numbers comes up, and loses a dollar otherwise. Remember that the wheel also has ). We want to set up the simplest accurate box model of this, and are considering the following possibilities: Box A: [25 tickets labeled (2) and 25 tickets labeled (-1) ] Box B: [ 12 tickets labeled 1through 12, 12 tickets labeled 13 through 24, 12 tickets labeled 25 through 36, 2 tickets labeled 0 and 00] Box C: [ 12 tickets labeled (2) and 26 tickets labeled (-1)] Box D: [12 tickets labeled 1 and 26 tickets labeled 0] (c)_ is the best choice. Because it corresponds to the odds and to the winnings and losings he actually faces. Option (a) has the odds wrong; option (b) has the numbers on the wheel but not the winnings, option (d) would work for COUNTING the wins and the losses, but does not reflect his winnings, so could not answer the questions below. 11. The expected value of the gambler's winnings (or losings) after 25 plays is Show work below: Mean of the box model: 2 * 12/38 + (-1) * 26/38 = - 2/38 EV of winnings after 25 plays = 25 * (- 2 /38) = -1.3158 The standard deviation of the correct box model is 1.3945 Explain below how we found this value: (Big Little) * (sqrt 12/38 * 26/38) = 3 / 38 (sqrt 12 * 26) = 1.3945 So the standard error of the winnings or losings after 25 plays is = 6.9725 Explain below: (sqrt 25) * 1.3945 = 5 * 1.3945 = 6.9725 27 percent 12. There is a --- percent chance for the gambler to WIN more than $ 3 after 25 plays of the game. Explain carefully how you arrived at your answer; the attached tables will help. (a) $ 3 = $ 114/38 is 114/38 - (- 50/38) = 164/38 = 4.3158 dollars higher than expected (b) so the Z-score is 4.3158 / 6.9725 = 0.6190 standard units The central area for a z-score of 0.60 is 45.15 percent, which implies that the two tail area is 54.85 percent, and the one tail area (winning $ 3 or more) is half of this or 27.425 percent.