Probability and Statistics for Engineering and Science, 9th Edition
Authors: Jay L. Devore
ISBN-13: 978-1305251809
See our solution for Question 34E from Chapter 2 from Devore's Probability and Statistics for Engineering and Science.
Problem 34E
Chapter:
Problem:
Computer keyboard failures can be attributed to electrical defects or mechanical defects. A repair facility currently has 25 failed keyboards, 6 of which have electrical defects and 19 of which have mechanical defects. a. How many ways are there to randomly select 5 of these keyboards for a thorough inspection (without regard to order)? b. In how many ways can a sample of 5 keyboards be selected so that exactly two have an electrical defect? c. If a sample of 5 keyboards is randomly selected, what is the probability that at least 4 of these will have a mechanical defect?
Step-by-Step Solution
Step 1
We are given that total defective keyboards are 25. Out of these, six have electrical defects and five have mechanical.
Step 2: (a)
Here, we have to select 5 defective keyboards out of 25 defective keyboards.\[n = 25,\,\,\,r = 5\]Total combinations are:\[\begin{array}{l}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}\\\\^n{C_r} = \dfrac{{25!}}{{5!\left( {25 - 5} \right)!}}\\\\^n{C_r} = \dfrac{{25!}}{{5!\left( {20} \right)!}}\\\\^n{C_r} = 53,130\end{array}\]Therefore, \[^n{C_r} = 53,130\]
Step 3: (b)
Here, we have to select 5 defective keyboards out of 25 defective keyboards sch that two keyboards have an electrical defect.\[\begin{array}{l}{n_1} = 6,\,\,\,{r_1} = 2\\{n_2} = 19,\,\,\,{r_1} = 3\end{array}\]\[\begin{array}{l}B{ = ^{{n_1}}}{C_{{r_1}}}{ \times ^{{n_2}}}{C_{{r_2}}}\\\\B = \dfrac{{{n_1}!}}{{{r_1}!\left( {{n_1} - {r_1}} \right)!}} \times \dfrac{{{n_2}!}}{{{r_2}!\left( {{n_2} - {r_2}} \right)!}}\\\\B = \dfrac{{6!}}{{2!\left( {6 - 2} \right)!}} \times \dfrac{{19!}}{{3!\left( {19 - 3} \right)!}}\\\\B = \dfrac{{6!}}{{2!\left( 4 \right)!}} \times \dfrac{{19!}}{{3!\left( {16} \right)!}}\\\\B = 14535\end{array}\]Therefore, B = 14535
Step 4: (c)
We have to select 5 keyboards such that at least 4 have mechanical defect. This is sum of two cases, first is that there are exactly 4 mechanical defects other could be when all 5 have mechanical defects.
Total Number of Cases: \[\begin{array}{l}N{ = ^{19}}{C_4}{ \times ^6}{C_1}{ + ^{19}}{C_5}{ \times ^6}{C_0}\\\\ = \dfrac{{19!}}{{4!\left( {19 - 4} \right)!}} + \dfrac{{19!}}{{5!\left( {19 - 5} \right)!}}\\\\= 23256 + 11628\\\\ = 34884\end{array}\]Total possible cases of choosing 5 keyboards: \[\begin{array}{l}{N_a}{ = ^{25}}{C_5}\\\\ = \dfrac{{25!}}{{5!\left( {25 - 5} \right)!}}\\\\ = 53130\end{array}\]Required probability\[\begin{array}{l}P\left( A \right) = \dfrac{N}{{{N_a}}}\\\\ = \dfrac{{34884}}{{53130}}\\\\ = 0.6566\end{array}\]Therefore, \[P = 0.6566\]
We are given that total defective keyboards are 25. Out of these, six have electrical defects and five have mechanical.
Step 2: (a)
Here, we have to select 5 defective keyboards out of 25 defective keyboards.\[n = 25,\,\,\,r = 5\]Total combinations are:\[\begin{array}{l}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}\\\\^n{C_r} = \dfrac{{25!}}{{5!\left( {25 - 5} \right)!}}\\\\^n{C_r} = \dfrac{{25!}}{{5!\left( {20} \right)!}}\\\\^n{C_r} = 53,130\end{array}\]Therefore, \[^n{C_r} = 53,130\]
Step 3: (b)
Here, we have to select 5 defective keyboards out of 25 defective keyboards sch that two keyboards have an electrical defect.\[\begin{array}{l}{n_1} = 6,\,\,\,{r_1} = 2\\{n_2} = 19,\,\,\,{r_1} = 3\end{array}\]\[\begin{array}{l}B{ = ^{{n_1}}}{C_{{r_1}}}{ \times ^{{n_2}}}{C_{{r_2}}}\\\\B = \dfrac{{{n_1}!}}{{{r_1}!\left( {{n_1} - {r_1}} \right)!}} \times \dfrac{{{n_2}!}}{{{r_2}!\left( {{n_2} - {r_2}} \right)!}}\\\\B = \dfrac{{6!}}{{2!\left( {6 - 2} \right)!}} \times \dfrac{{19!}}{{3!\left( {19 - 3} \right)!}}\\\\B = \dfrac{{6!}}{{2!\left( 4 \right)!}} \times \dfrac{{19!}}{{3!\left( {16} \right)!}}\\\\B = 14535\end{array}\]Therefore, B = 14535
Step 4: (c)
We have to select 5 keyboards such that at least 4 have mechanical defect. This is sum of two cases, first is that there are exactly 4 mechanical defects other could be when all 5 have mechanical defects.
Total Number of Cases: \[\begin{array}{l}N{ = ^{19}}{C_4}{ \times ^6}{C_1}{ + ^{19}}{C_5}{ \times ^6}{C_0}\\\\ = \dfrac{{19!}}{{4!\left( {19 - 4} \right)!}} + \dfrac{{19!}}{{5!\left( {19 - 5} \right)!}}\\\\= 23256 + 11628\\\\ = 34884\end{array}\]Total possible cases of choosing 5 keyboards: \[\begin{array}{l}{N_a}{ = ^{25}}{C_5}\\\\ = \dfrac{{25!}}{{5!\left( {25 - 5} \right)!}}\\\\ = 53130\end{array}\]Required probability\[\begin{array}{l}P\left( A \right) = \dfrac{N}{{{N_a}}}\\\\ = \dfrac{{34884}}{{53130}}\\\\ = 0.6566\end{array}\]Therefore, \[P = 0.6566\]