When a fair die is thrown, the probability of obtaining a "1" is 1/6. Alexander throws such a die repeatedly. Calculate the probability that:
(a) he throws at least three "1"'s in his first twenty throws
(b) he throws his first "1" on his third throw;
(c) he throws his second "1" on his twelfth throw
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answer:
(a)0.6713
(b)25/216
(c)0.0493
(math expert please urgent!) Discrete Distribution Functions?
2012-02-16 11:14 am
回答 (1)
2012-02-16 11:32 am
✔ 最佳答案
qa5^20 ways of getting zero 1's,
20c1*1*5^19 ways of getting one 1,
20c2*1*5^18 ways of getting two 1's,
total 5^18(5^2 + 20*5 + 190) ways for < three 1's
now total # of ways = 6^20, so
P[≥3] = 1- P[<3] = 1 - 5^18*315/6^20 = .6713 <-----
qb
Pr = 5/6 *5/6 *1/6 = 25/216 <-------
qc
1st 1 anywhere in 1st 11 throws, 2nd on 12th
11*(5/6)^10*(1/6)^2 = .0493 <-------
(a) was binomdist, (b) wa geometric & (c) was negative binomial
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