SECTION ONE: Multiple Choice (5 points for each answer: 50 points total)
1. On the planet Dangerboy, Captain Bananahands has stumbled across what appears to be a 3-
sided coin. Let’s call the sides heads, tails, and feet. Bananahands flips the coin 12 times and
gets 6 feet. Assuming the coin is fair, what is the probability of this happening?
A. .500 B. .111 C. .667 D. .392
BThe experiment is a Binomial process. Hence, the answer is !12
6"(1/3)6(2/3)6≈.111.
2. In a magical, far-away land, I found a 5-sided die. On Earth, a 5-sided die is necessarily not fair.
I decide to test this newly found die for fairness by rolling it 400 times and counting the number
of !
•’s. I counted 52.
i) Assuming the die is fair, the number of !
•’s rolled is , give or take , or so.
A. 80; 8 B. 80; 16 C. 200; 8 D. 200; 16
AOur distribution here is binary: 1,0,0,0,0. We expect (400)(1/5)=80 !
•’s (this is
EVsum ). The second blank is SEsum =√400#(1/5)(4/5) = 8.
ii) Assuming the die is fair, the chance of rolling 52 or fewer !
•’s is
A) ≈0% B) .03% C) .5% D) 99.997%
BWe expecet 400(1/5) = 80 with SEsum =√400#(1/5)(4/5) = 8. Hence, using the
Central Limit Theorem, our approximate probability is the area to the left of 52.5 (using
the .5correction for 0-1 boxes) on a Normal curve with mean 80 and standard deviation
8. We convert: 52.5−80
8≈ −3.44. The corresponding area is .03%.
3. A strange lottery at Britneyland Casinos is played as follows. The numbers 2, 3, 5, 7, and
13 are placed in a hat. The chairman of the lottery then picks at random 120 times (with
replacement) from the hat. There are two types of winning “hands”:
(A) The number 5 is picked more than 35 times;
(B) The sum of the picks is less than 500.
Here are the questions:
i) The number of fives picked is , give or take , or so.
A. 24;.037 B. 24;3.65 C. 24; 4.38 D. 24; 8.76
CThe distribution is binary: 0,0,1,0,0. Hence, the expected number of ones is 120(1/5) =
24 with SEsum =√120#(1/5)(4/5) ≈4.38.
ii) The sum of the picks is , give or take , or so.
A. 720; 42.7 B. 600; 36.8 C. 720; 36.8 D. 600; 42.7
AWe use the distribution: 2,3,5,7,13 which has an average of 6 and a standard deviation
of 3.9. Hence, our expected sum is 120(6) = 720 with SEsum =√120(3.9) ≈42.7.
4. Someone rolls 100 four-sided dice. The sides are numbered 1 through 4 and are all equally
likely to be rolled.
i) The sum of the rolls should be , give or take , or so.
