Salespersons Adams and Jones call on three and four customers, respectively, on

Question

Salespersons Adams and Jones call on three and four customers, respectively, on a given day. Adams could make 0, 1, 2, or 3 sales, whereas Jones could make 0, 1, 2, or 3 sales. The sample space listing the number of possible sales for person on a given day is given in the following table. Here (0, 1) stands for 0 sales by Adams and 1 sale by Jones. Assume that each sample point is equally likely.

Adams

Jones
0------1------2------3

0

{0, 0} {1, 0} {2, 0} {3, 0}

1

{0, 1} {1, 1} {2, 1} {3, 1}

2

{0, 2} {1, 2} {2, 2} {3, 2}

3

{0, 3} {1, 3} {2, 3} {3, 3}

Let us define the events:
A = each made the same number of sales
B = Adams made exactly one sale

1. List the sample space for Event A and event B.

2. Find the total number of the sample space. S(n) =

3. Find the following probabilities:
(1) P(A) =
(2) P(B) =
(3) P(A?B) =
(4) P(A?B) =
(5) P(A|B) =

Adams

Jones
0------1------2------3

0

{0, 0} {1, 0} {2, 0} {3, 0}

1

{0, 1} {1, 1} {2, 1} {3, 1}

2

{0, 2} {1, 2} {2, 2} {3, 2}

3

{0, 3} {1, 3} {2, 3} {3, 3}

Explanation / Answer

Sample space for A = {(0,0),(1,1),(2,2),(3,3)}

Sample space for B = {(0,1),(1,1),),(2,1),(3,1)}

total number of the sample space = 16

P(A) = 4/16 = 0.25

P(B) = 4/16 = 0.25

P(A?B) = 1/16

P(A?B) = 7/16

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