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A committee of five is chosen from a group of 20 people. What is the probability that a specified member of the group will be on the committee?

C ( 20 , 5 ) committees; C ( 19 , 4 ) have a designated member.

P = 19 ! 4 ! 15 ! 5 ! 15 ! 20 ! = 5 / 20 = 1 / 4
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Ten employees of a company drive their cars to the city each day and park randomly in ten spots. What is the (classical) probability that on a given day Jim willbe in place three? There are n ! equally likely ways to arrange n items (order important).

10! permutations. 1 × 9 ! permutations with Jim in place 3. P = 9 ! / 10 ! = 1 / 10 .

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An extension of the classical model involves the use of areas. A certain region L (say of land) is taken as a reference. For any subregion A , define P ( A ) = a r e a ( A ) / a r e a ( L ) . Show that P ( ) is a probability measure on the subregions of L .

Additivity follows from additivity of areas of disjoint regions.

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John thinks the probability the Houston Texans will win next Sunday is 0.3 and the probability the Dallas Cowboys will win is 0.7 (they are not playing each other).He thinks the probability both will win is somewhere between—say, 0.5. Is that a reasonable assumption? Justify your answer.

P ( A B ) = 0 . 5 is not reasonable. It must no greater than the minimum of P ( A ) = 0 . 3 and P ( B ) = 0 . 7 .

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Suppose P ( A ) = 0 . 5 and P ( B ) = 0 . 3 . What is the largest possible value of P ( A B ) ? Using the maximum value of P ( A B ) , determine P ( A B c ) , P ( A c B ) , P ( A c B c ) and P ( A B ) . Are these values determined uniquely?

Draw a Venn diagram, or use algebraic expressions P ( A B c ) = P ( A ) - P ( A B ) = 0 . 2

P ( A c B ) = P ( B ) - P ( A B ) = 0 P ( A c B c ) = P ( A c ) - P ( A c B ) = 0 . 5 P ( A B ) = 0 . 5
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For each of the following probability “assignments”, fill out the table. Which assignments are not permissible? Explain why, in each case.

P ( A ) P ( B ) P ( A B ) P ( A B ) P ( A B c ) P ( A c B ) P ( A ) + P ( B )
0.3 0.7 0.4
0.2 0.1 0.4
0.3 0.7 0.2
0.3 0.5 0
0.3 0.8 0
P ( A ) P ( B ) P ( A B ) P ( A B ) P ( A B c ) P ( A c B ) P ( A ) + P ( B )
0.3 0.7 0.4 0.6 -0.1 0.3 1.0
0.2 0.1 0.4 -0.1 -0.2 -0.3 0.3
0.3 0.7 0.2 0.8 0.1 0.5 1.0
0.3 0.5 0 0.8 0.3 0.5 0.8
0.3 0.8 0 1.1 0.3 0.8 1.1

Only the third and fourth assignments are permissible.

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The class { A , B , C } of events is a partition. Event A is twice as likely as C and event B is as likely as the combination A or C . Determine the probabilities P ( A ) , P ( B ) , P ( C ) .

P ( A ) + P ( B ) + P ( C ) = 1 , P ( A ) = 2 P ( C ) , and P ( B ) = P ( A ) + P ( C ) = 3 P ( C ) , which implies

P ( C ) = 1 / 6 , P ( A ) = 1 / 3 , and P ( B ) = 1 / 2
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Determine the probability P ( A B C ) in terms of the probabilities of the events A , B , C and their intersections.

P ( A B C ) = P ( A B ) + P ( C ) - P ( A C B C )

= P ( A ) + P ( B ) - P ( A B ) + P ( C ) - P ( A C ) - P ( B C ) + P ( A B C )
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If occurrence of event A implies occurrence of B , show that P ( A c B ) = P ( B ) - P ( A ) .

P ( A B ) = P ( A ) and P ( A B ) + P ( A c B ) = P ( B ) implies P ( A c B ) = P ( B ) - P ( A ) .

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Show that P ( A B ) P ( A ) + P ( B ) - 1 .

Follows from P ( A ) + P ( B ) - P ( A B ) = P ( A B ) 1 .

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The set combination A B = A B c A c B is known as the disjunctive union or the symetric difference of A and B . This is the event that only one of the events A or B occurs on a trial. Determine P ( A B ) in terms of P ( A ) , P ( B ) , and P ( A B ) .

A Venn diagram shows P ( A B ) = P ( A B c ) + P ( A B c ) = P ( A ) + P ( B ) - 2 P ( A B ) .

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Use fundamental properties of probability to show

  1. P ( A B ) P ( A ) P ( A B ) P ( A ) + P ( B )
  2. P j = 1 E j P ( E i ) P j = 1 E j j = 1 P ( E j )

A B A A B implies P ( A B ) P ( A ) P ( A B ) = P ( A ) + P ( B ) - P ( A B ) P ( A ) + P ( B ) . The general case follows similarly, with the last inequality determined by subadditivity.

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Suppose P 1 , P 2 are probability measures and c 1 , c 2 are positive numbers such that c 1 + c 2 = 1 . Show that the assignment P ( E ) = c 1 P 1 ( E ) + c 2 P 2 ( E ) to the class of events is a probability measure. Such a combination of probability measures is known as a mixture . Extend this to

P ( E ) = i = 1 n c i P i ( E ) , where the P i are probabilities measures, c i > 0 , and i = 1 n c i = 1

Clearly P ( E ) 0 . P ( Ω ) = c 1 P 1 ( Ω ) + c 2 P 2 ( Ω ) = 1 .

E = i = 1 E i implies P ( E ) = c 1 i = 1 P 1 ( E i ) + c 2 i = 1 P 2 ( E i ) = i = 1 P ( E i )

The pattern is the same for the general case, except that the sum of two terms is replaced by the sum of n terms c i P i ( E ) .

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Suppose { A 1 , A 2 , , A n } is a partition and { c 1 , c 2 , , c n } is a class of positive constants. For each event E , let

Q ( E ) = i = 1 n c i P ( E A i ) i = 1 n c i P ( A i )

Show that Q ( ) us a probability measure.

Clearly Q ( E ) 0 and since A i Ω = A i we have Q ( Ω ) = 1 . If

E = k = 1 E k , then P ( E A i ) = k = 1 P ( E k A i ) i

Interchanging the order of summation shows that Q is countably additive.

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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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