Probability and Statistics for Engineering and Science, 9th Edition
Authors: Jay L. Devore
ISBN-13: 978-1305251809
See our solution for Question 6E from Chapter 2 from Devore's Probability and Statistics for Engineering and Science.
Problem 6E
Chapter:
Problem:
A college library has five copies of a certain text on reserve. Two copies (1 and 2) are first printings, and the other three (3, 4, and 5) are second printings. A student examines these books in random order, stopping only when a second printing has been selected. One possible outcome is 5, and another is 213...
Step-by-Step Solution
Step 1
The college library has 5 copies of a textbook. The first printings are (1,2), second printings are (3,4,5). The sample space, when students selects these books and stops at second printing is:
Step 2: (a) All outcomes
The all possible cases are listed below\[S = \left\{ \begin{array}{l}\left( 3 \right),\left( 4 \right),\left( 5 \right),\left( {1,3} \right),\left( {1,4} \right),\left( {1,5} \right),\left( {2,3} \right),\left( {2,4} \right),\left( {2,5} \right)\\\left( {1,2,3} \right),\left( {1,2,4} \right),\left( {1,2,5} \right),\left( {2,1,3} \right),\left( {2,1,4} \right),\left( {2,1,5} \right)\end{array} \right\}\]
Step 3: (b) Exactly one book is examined
There are three cases when exactly one book is examined. THe cases correspond to each second print. \[A = \left\{ {\left( 3 \right),\left( 4 \right),\left( 5 \right)} \right\}\]
Step 4: (c) The selected book is number 5
Out the sample space given given above, there are five cases when second print found is number 5. \[B = \left\{ {\left( 5 \right),\left( {1,5} \right),\left( {2,5} \right),\left( {1,2,5} \right),\left( {2,1,5} \right)} \right\}\]
Step 5: (d) Book 1 is not examined
Out the sample space S, there are six cases when book 1 is not examined. \[C = \left\{ {\left( 3 \right),\left( 4 \right),\left( 5 \right),\left( {2,5} \right),\left( {2,4} \right),\left( {2,5} \right)} \right\}\]
The college library has 5 copies of a textbook. The first printings are (1,2), second printings are (3,4,5). The sample space, when students selects these books and stops at second printing is:
Step 2: (a) All outcomes
The all possible cases are listed below\[S = \left\{ \begin{array}{l}\left( 3 \right),\left( 4 \right),\left( 5 \right),\left( {1,3} \right),\left( {1,4} \right),\left( {1,5} \right),\left( {2,3} \right),\left( {2,4} \right),\left( {2,5} \right)\\\left( {1,2,3} \right),\left( {1,2,4} \right),\left( {1,2,5} \right),\left( {2,1,3} \right),\left( {2,1,4} \right),\left( {2,1,5} \right)\end{array} \right\}\]
Step 3: (b) Exactly one book is examined
There are three cases when exactly one book is examined. THe cases correspond to each second print. \[A = \left\{ {\left( 3 \right),\left( 4 \right),\left( 5 \right)} \right\}\]
Step 4: (c) The selected book is number 5
Out the sample space given given above, there are five cases when second print found is number 5. \[B = \left\{ {\left( 5 \right),\left( {1,5} \right),\left( {2,5} \right),\left( {1,2,5} \right),\left( {2,1,5} \right)} \right\}\]
Step 5: (d) Book 1 is not examined
Out the sample space S, there are six cases when book 1 is not examined. \[C = \left\{ {\left( 3 \right),\left( 4 \right),\left( 5 \right),\left( {2,5} \right),\left( {2,4} \right),\left( {2,5} \right)} \right\}\]