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Thirty members of a class each flip a coin ten times. What is the probability that at least five of them get seven or more heads?
p = cbinom(10,0.5,7) = 0.1719
P = cbinom(30,p,5) = 0.6052 For the system in [link] , call a day in which one or more failures occur among the 350 lamps a “service day.” Since a Bernoulli sequence “starts over”at any time, the sequence of service/nonservice days may be considered a Bernoulli sequence with probability p 1 , the probability of one or more lamp failures in a day.
p1 = 1 - (1 - 0.0017)^350 = 0.4487 k = 1:5; (prob given day is a service day)
P = p1*(1 - p1).^(k-1) = 0.4487 0.2474 0.1364 0.0752 0.0414 P0 = (1 - p1)^7 = 0.0155 For the system in [link] and [link] assume the plant works seven days a week. What is the probability the third service day occursby the end of 10 days? Solve using the negative binomial distribution; repeat using the binomial distribution.
p1 = 1 - (1 - 0.0017)^350 = 0.4487
P = sum(nbinom(3,p1,3:10)) = 0.8990 Pa = cbinom(10,p1,3) = 0.8990 A residential College plans to raise money by selling “chances” on a board. Fifty chances are sold. A player pays $10 to play; he or she wins $30with probability . The profit to the College is
Determine the distribution for X and calculate , , and
.
N = 0:50;
PN = ibinom(50,0.2,0:50);X = 500 - 30*N;
Ppos = (X>0)*PN'
Ppos = 0.9856P200 = (X>=200)*PN'
P200 = 0.5836P300 = (X>=300)*PN'
P300 = 0.1034 A single six-sided die is rolled repeatedly until either a one or a six turns up. What is the probability that the first appearance of either of these numbers is achievedby the fifth trial or sooner?
P = 1 - (2/3)^5 = 0.8683
Consider a Bernoulli sequence with probability of success on any component trial.
P = sum(nbinom(4,0.53,4:10)) = 0.8729 Pa = cbinom(10,0.53,4) = 0.8729 Fifty percent of the components coming off an assembly line fail to meet specifications for a special job. It is desired to select three units which meet thestringent specifications. Items are selected and tested in succession. Under the usual assumptions for Bernoulli trials, what is the probability the third satisfactoryunit will be found on six or fewer trials?
P = cbinom(6,0.5,3) = 0.6562
The number of cars passing a certain traffic count position in an hour has Poisson (53) distribution. What is the probability the number of carspassing in an hour lies between 45 and 55 (inclusive)? What is the probability of more than 55?
P1 = cpoisson(53,45) - cpoisson(53,56) = 0.5224
P2 = cpoisson(53,56) = 0.3581 Compare and for binomial(5000, 0.001) and Poisson (5), for . Do this directly with ibinom and ipoisson. Then use the m-procedure bincomp to obtain graphical results (including acomparison with the normal distribution).
k = 0:10;
Pb = 1 - cbinom(5000,0.001,k+1);Pp = 1 - cpoisson(5,k+1);
disp([k;Pb;Pp]')
0 0.0067 0.00671.0000 0.0404 0.0404
2.0000 0.1245 0.12473.0000 0.2649 0.2650
4.0000 0.4404 0.44055.0000 0.6160 0.6160
6.0000 0.7623 0.76227.0000 0.8667 0.8666
8.0000 0.9320 0.93199.0000 0.9682 0.9682
10.0000 0.9864 0.9863
bincomp
Enter the parameter n 5000Enter the parameter p 0.001
Binomial-- stairsPoisson-- -.-.
Adjusted Gaussian-- o o ogtext('Exercise 17') 
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