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1. |
Part (a) is asking about a geometric r.v. X where p = 1/36, q = 35/36, and X = # of throws needed to obtain the first success.
P(X ³ 20) = 1 P(X < 20)
= 1 P(X £ 19)
= 1 .414475 ...
= .5855
Alternate (common sense) method for part (a):
P(X ³ 20) = P(first 19 are failures)
= q19
= .5855.
Part (b) is asking a completely different question, namely concerning a binomial r.v. Y where n = 20, p = 1/36, q = 35/36, and Y = # of successes in 20 throws.
P(Y ³ 2) = 1 P(Y < 2)
= 1 P(Y £ 1)
= 1 .89455 ...
= .1055
Or, if you have forgotten how to use binomcdf, you can use the complicated formulas provided on the AP formula sheet:
P(Y ³ 2) = 1 P(Y < 2)
= 1 P(Y = 0 È Y = 1)
= 1 [P(Y = 0) + P(Y = 1)]
= 1 S nCk pkqn k where k = first 0, then 1
= 1 [20C0 p0q20 + 20C1 p1q19]
= 1 [q20 + 20p1q19]
= 1 [.56926... + .32529...]
= 1 .89455...
= .1055
ANSWER: D |
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2. |
Since the basketball has no computer chip inside that remembers its history, it is as if we are beginning a new contest, starting on the 15th shot. We are asking for the probability of making the first shot in 1, 2, 3, or 4 tries looking forward from this point. Let Z = geometric r.v. for # of shots needed when starting with shot 15, and let p = .12.
P(Z £ 4) = .4003 by calc.
[Note: If this were a free-response problem, you would not be allowed to write geometcdf(.12,4) on your paper. Calculator notation is not permitted. However, in a multiple-choice problem, work is ignored, and you may write whatever you wish.]
ANSWER: B |
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3. |
Let L = # of invitations Larry must extend in order to achieve his first acceptance. Clearly, L is a geometric r.v. with p = .007. [Presumably the 007 in the James Bond films signified something other than Bonds probability of success with the ladies.] At any rate, by the formula we learned,
mL = E(L) = 1/p = 1/.007 = 142.857
Note that it is perfectly acceptable for an expected value to be a noninteger.
ANSWER: D |