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A random number N of students take a qualifying exam. A grade of 70 or more earns a pass. Suppose N binomial (20, 0.3). If each student has probability p = 0 . 7 of making 70 or more, what is the probability all will pass? Ten or more will pass?

gN = ibinom(20,0.3,0:20); gY = [0.3 0.7]; gendDo not forget zero coefficients for missing powers Enter gen fn COEFFICIENTS for gN gNEnter gen fn COEFFICIENTS for gY gY Results are in N, PN, Y, PY, D, PD, PMay use jcalc or jcalcf on N, D, P To view the distribution, call for gD.Pall = (D==20)*PD' Pall = 2.7822e-14pall = (0.3*0.7)^20 % Alternate: use D binomial (pp0) pall = 2.7822e-14P10 = (D>= 10)*PD' P10 = 0.0038
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Five hundred questionnaires are sent out. The probability of a reply is 0.6. The probability that a reply will be favorable is 0.75. What is the probability of at least200, 225, 250 favorable replies?

n = 500; p = 0.6;p0 = 0.75; D = 0:500;PD = ibinom(500,p*p0,D); k = [200 225 250]; P = zeros(1,3);for i = 1:3 P(i) = (D>=k(i))*PD'; enddisp(P) 0.9893 0.5173 0.0140
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Suppose the number of Japanese visitors to Florida in a week is N 1 Poisson (500) and the number of German visitors is N 2 Poisson (300). If 25 percent of the Japanese and 20 percent of the Germans visit Disney World,what is the distribution for the total number D of German and Japanese visitors to the park? Determine P ( D k ) for k = 150 , 155 , , 245 , 250 .

J D Poisson (500*0.25 = 125); G D Poisson (300*0.20 = 60); D Poisson (185).

k = 150:5:250; PD = cpoisson(185,k); disp([k;PD]') 150.0000 0.9964155.0000 0.9892 160.0000 0.9718165.0000 0.9362 170.0000 0.8736175.0000 0.7785 180.0000 0.6532185.0000 0.5098 190.0000 0.3663195.0000 0.2405 200.0000 0.1435205.0000 0.0776 210.0000 0.0379215.0000 0.0167 220.0000 0.0067225.0000 0.0024 230.0000 0.0008235.0000 0.0002 240.0000 0.0001245.0000 0.0000 250.0000 0.0000
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A junction point in a network has two incoming lines and two outgoing lines. The number of incoming messages N 1 on line one in one hour is Poisson (50); on line 2 the number is N 2 Poisson (45). On incoming line 1 the messages have probability p 1 a = 0 . 33 of leaving on outgoing line a and 1 - p 1 a of leaving on line b. The messages coming in on line 2 have probability p 2 a = 0 . 47 of leaving on line a. Under the usual independence assumptions, what is the distribution of outgoing messages on line a?What are the probabilities of at least 30, 35, 40 outgoing messages on line a?

m1a = 50*0.33; m2a = 45*0.47; ma = m1a + m2a; PNa = cpoisson(ma,[30 35 40]) PNa = 0.9119 0.6890 0.3722
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A computer store sells Macintosh, HP, and various other IBM compatible personal computers. It has two major sources of customers:

  1. Students and faculty from a nearby university
  2. General customers for home and business computing. Suppose the following assumptions are reasonable for monthly purchases.
  • The number of university buyers N 1 Poisson (30). The probabilities for Mac, HP, others are 0.4, 0.2, 0.4, respectively.
  • The number of non-university buyers N 2 Poisson (65). The respective probabilities for Mac, HP, others are 0.2, 0.3, 0.5.
  • For each group, the composite demand assumptions are reasonable, and the two groups buy independently.

What is the distribution for the number of Mac sales? What is the distribution for the total number of Mac and Dell sales?

Mac sales Poisson (30*0.4 + 65*0.2 = 25); HP sales Poisson (30*0.2 + 65*0.3 = 25.5); total Mac plus HP sales Poisson(50.5).

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Source:  OpenStax, Applied probability. OpenStax CNX. Aug 31, 2009 Download for free at http://cnx.org/content/col10708/1.6
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