| << Chapter < Page | Chapter >> Page > |
The number N of customers in a shop in a given day is Poisson (120). Customers pay with cash or by MasterCard or Visa charge cards, with respective probabilties 0.25, 0.40, 0.35.Make the usual independence assumptions. Let N 1 , N 2 , N 3 be the numbers of cash sales, MasterCard charges, Visa card charges, respectively. Determine , , , and .
X = 0:120;
PX = ipoisson(120*0.4,X);Y = 0:120;
PY = ipoisson(120*0.35,Y);icalc
Enter row matrix of X values XEnter row matrix of Y values Y
Enter X probabilities PXEnter Y probabilities PY
Use array opertions on matrices X, Y, PX, PY, t, u, and PM = t>u;
PM = total(M.*P)PM = 0.7190 A discount retail store has two outlets in Houston, with a common warehouse. Customer requests are phoned to the warehouse for pickup. Two items, a and b,are featured in a special sale. The number of orders in a day from store A is Poisson (30); from store B, the nember of orders is Poisson (40).
For store A, the probability an order for a is 0.3, and for b is 0.7.
For store B, the probability an order for a is 0.4, and for b is 0.6. What is the probability the total order for item b in a day is 50 or more?
P = cpoisson(30*0.7+40*0.6,50) = 0.2468
The number of bids on a job is a random variable binomial (7, 0.6). Bids (in thousands of dollars) are iid with Y uniform on . What is the probability of at least one bid of $3,500 or less? Note that “no bid” is not a bid of 0.
% First solution --- FY(t) = 1 - gN[P(Y>t)]
P = 1-(0.4 + 0.6*0.75)^7P = 0.6794
% Second solution --- Positive number of satisfactory bids,% i.e. the outcome is indicator for event E, with P(E) = 0.25
pN = ibinom(7,0.6,0:7);gY = [3/4 1/4]; % Generator function for indicator[D,PD] = gendf(pN,gY); % D is number of successesPa = (D>0)*PD' % D>0 means at least one successful bid
Pa = 0.6794 The number of customers during the noon hour at a bank teller's station is a random number N with distribution
The amounts they want to withdraw can be represented by an iid class having the common distribution exponential (0.01). Determine the probabilities that the maximum withdrawal is less than or equal to t for .
Use
gN = 0.01*[0 5 7 10 11 12 13 12 11 10 9];t = 100:100:500;
PY = 1 - exp(-0.01*t);FW = polyval(fliplr(gN),PY) % fliplr puts coeficients in
% descending order of powersFW = 0.1330 0.4598 0.7490 0.8989 0.9615 A job is put out for bids. Experience indicates the number N of bids is a random variable having values 0 through 8, with respective probabilities
| Value | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
| Probability | 0.05 | 0.10 | 0.15 | 0.20 | 0.20 | 0.10 | 0.10 | 0.07 | 0.03 |
The market is such that bids (in thousands of dollars) are iid, uniform [100, 200].Determine the probability of at least one bid of $125,000 or less.
Probability of a successful bid
PY =0.25;
gN = 0.01*[5 10 15 20 20 10 10 7 3];
P = 1 - polyval(fliplr(gN),PY)P = 0.9116 A property is offered for sale. Experience indicates the number N of bids is a random variable having values 0 through 10, with respective probabilities
| Value | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
| Probability | 0.05 | 0.15 | 0.15 | 0.20 | 0.10 | 0.10 | 0.05 | 0.05 | 0.05 | 0.05 | 0.05 |
The market is such that bids (in thousands of dollars) are iid, uniform [150, 200] Determine theprobability of at least one bid of $180,000 or more.
Consider a sequence of N trials with probabiliy .
gN = 0.01*[5 15 15 20 10 10 5 5 5 5 5];gY = [0.4 0.6];[D,PD] = gendf(gN,gY);P = (D>0)*PD'
P = 0.8493 
Notification Switch
Would you like to follow the 'Applied probability' conversation and receive update notifications?
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|