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A pair of dice is rolled five times. What is the probability the first two results are “sevens” and the others are not?

P = ( 1 / 6 ) 2 ( 5 / 6 ) 3 = 0 . 0161 .

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David, Mary, Joan, Hal, Sharon, and Wayne take an exam in their probability course. Their probabilities of making 90 percent or more are

0 . 72 0 . 83 0 . 75 0 . 92 0 . 65 0 . 79

respectively. Assume these are independent events. What is the probability three or more, four or more,five or more make grades of at least 90 percent?

P = 0.01*[72 83 75 92 65 79];y = ckn(P,[3 4 5])y = 0.9780 0.8756 0.5967
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Two independent random numbers between 0 and 1 are selected (say by a random number generator on a calculator). What is the probability the first is no greaterthan 0.33 and the other is at least 57?

P = 0 . 33 ( 1 - 0 . 57 ) = 0 . 1419

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Helen is wondering how to plan for the weekend. She will get a letter from home (with money) with probability 0.05. There is a probability of 0.85 that she willget a call from Jim at SMU in Dallas. There is also a probability of 0.5 that William will ask for a date. What is the probability she will get money and Jim will not call orthat both Jim will call and William will ask for a date?

A letter with money, B call from Jim, C William ask for date

P = 0.01*[5 85 50];minvec3 Variables are A, B, C, Ac, Bc, CcThey may be renamed, if desired. pm = minprob(P);p = ((A&Bc)|(B&C))*pm' p = 0.4325
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A basketball player takes ten free throws in a contest. On her first shot she is nervous and has probability 0.3 of making the shot. She begins to settle down and probabilitieson the next seven shots are 0.5, 0.6 0.7 0.8 0.8, 0.8 and 0.85, respectively. Then she realizes her opponent is doing well, and becomes tense as she takes the last two shots, withprobabilities reduced to 0.75, 0.65. Assuming independence between the shots, what is the probability she will make k or more for k = 2 , 3 , 10 ?

P = 0.01*[30 50 60 70 80 80 80 85 75 65];k = 2:10; p = ckn(P,k)p = Columns 1 through 70.9999 0.9984 0.9882 0.9441 0.8192 0.5859 0.3043 Columns 8 through 90.0966 0.0134
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In a group there are M men and W women; m of the men and w of the women are college graduates. An individual is picked at random. Let A be the event the individual is a woman and B be the event he or she is a college graduate. Under what condition is the pair { A , B } independent?

P ( A | B ) = w / ( m + w ) = W / ( W + M ) = P ( A )

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Consider the pair { A , B } of events. Let P ( A ) = p , P ( A c ) = q = 1 - p , P ( B | A ) = p 1 , and P ( B | A c ) = p 2 . Under what condition is the pair { A , B } independent?

p 1 = P ( B | A ) = P ( B | A c ) = p 2 (see table of equivalent conditions).

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Show that if event A is independent of itself, then P ( A ) = 0 or P ( A ) = 1 . (This fact is key to an important “zero-one law.”)

P ( A ) = P ( A A ) = P ( A ) P ( A ) . x 2 = x iff x = 0 or x = 1 .

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Does { A , B } independent and { B , C } independent imply { A , C } is independent? Justify your answer.

% No. Consider for example the following minterm probabilities: pm = [0.2 0.05 0.125 0.125 0.05 0.2 0.125 0.125]; minvec3Variables are A, B, C, Ac, Bc, Cc They may be renamed, if desired.PA = A*pm' PA = 0.5000PB = B*pm' PB = 0.5000PC = C*pm' PC = 0.5000PAB = (A&B)*pm' % Product rule holds PAB = 0.2500PBC = (B&C)*pm' % Product rule holds PBC = 0.2500PAC = (A&C)*pm' % Product rule fails PAC = 0.3250
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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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