Probability and Statistics for Engineering and the Sciences 8th Edition Chapter 2 Probability
Page 71 Problem 1 Answer
Given- 8 bottles of zinfandel, 10 of merlot, and 12 of cabernet all from different wineries.
To find- ways to serve 3bottles of zinfandel and serving order is important.We will use permutation formula P(n,r)=n!
(n−r)! to solve the problem.
If 3 bottles of zinfandel are to be served, so the number of ways for selecting 3 out of 8 are P(8,3).
P(8,3)=8!
(8−3)!=8!
5!=8×7×6×5!
5!=8×7×6=336
There are 336 ways to serve 3 bottles of zinfandel and serving order is important.
Page 71 Problem 2 Answer
Given-8 bottles of zinfandel,10 of merlot, and 12 of cabernet all from different wineries. To find- ways to serve randomly selected 6 bottles of wine from the 30.
Since the order is not important we will use combination formula C(n,r)=n!
r!(n−r)! to solve the problem.
The number of ways of to serve 6 bottles of randomly selected wine out of 30 servings-
C(30,6)=30!
6!(30−6)!=30!
6!×24!=593,775
The number of ways of to serve 6 bottles of randomly selected wine out of 30 servings are 593,775.
Page 71 Problem 3 Answer
Given-8 bottles of zinfandel, 10 of merlot, and 12 of cabernet all from different wineries.
To find- ways to obtain two bottles of each variety when 6bottles are randomly selected.
We will use combination formula C(n,r)=n!
r!(n−r)! to solve the problem because we need two of each variety thus order will matter.
For 6 bottles we need 2 from 8 bottles of zinfandel, 2 from 10 bottles of merlot, 2 from 12 bottles of cabernet, so total number of ways -8C2×10
C2×12
C2=8!
2!(8−2)!×10!
2!(10−2)!×12!
2!(12−2)!=8!
2!×6!×10!
2!×8!×12!
2!×10!=12!
2!×2!×2!×6!
=83,160
There are 83,160 ways to obtain two bottles of each variety when 6 bottles are randomly selected.
Page 71 Problem 4 Answer
Given- 8 bottles of zinfandel, 10 of merlot, and 12 of cabernet all from different wineries.
Number of ways of selecting two bottles of each variety being chosen when 6 bottles are randomly selected obtained in c): 83160.
To find- probability that two bottles of each variety being chosen when 6 bottles are randomly selected.
We will find the probability of the event which is given by: Number of favourable outcomes.
Total number of outcomes
As we know that there are 83,160 ways of selecting two bottles of each variety being chosen when 6 bottles are randomly selected.
So, the probability that two bottles of each variety being chosen when 6 bottles are randomly selected.
P( Two bottles of each variety )=83,160/30
C6=0.14
The probability that two bottles of each variety being chosen when 6 bottles are randomly selected is 0.14.
Page 71 Problem 5 Answer
Given-8 bottles of zinfandel, 10 of merlot, and 12 of cabernet all from different wineries.
To find- the probability that all of the bottles are of same variety when 6 bottles are randomly selected.
We will use combination to solve the problem of all of them being the same variety and then use probability to find the answer.
If 6 bottles are randomly selected from 30 bottles, then the probability that all of them are of the same variety is,
P( all of them are the same variety )=8
C6+10
C6+12
C6/30
C6=28+210+924/593,775
=0.002
The probability that all of the bottles are of same variety when 6 bottles are randomly selected is0.002 .
Page 71 Problem 6 Answer
Given- A production facility employs 20 workers on the day shift, 15 workers on the swing shift, and 6
workers on the graveyard shift. A quality control consultant is to select 6 of the workers in-depth interview.
To find- number of selections that 6 of the workers to be selected for an in-depth interviews will be from the day shift and the probability that all 6 selected workers will be from the day shift.
We will firstly find the combination to select the workers using C(n,r)=n!
r!(n−r)! the find the probability.
The 6 workers are selected from 20 day shift workers. So, the number of ways are(20/6) by using combination formula-
C(n,r)=n!
r!(n−r)!
=(20/6)
=20!
6!×14!
=38760 ways
The probability that all the 6 members are selected from day shift is-
P(all 6 selected workers will be from the day shift )
=(20/6)/(45/6)
=38,760/8,145,060
=0.0048
The number of selections that 6 of the workers to be selected for an in-depth interviews will be from the day shift are 38760 .
The probability that all 6 selected workers will be from the day shift is 0.0048.
Page 71 Problem 7 Answer
Given: Given- A production facility employs 20 workers on the day shift, 15 workers on the swing shift, and 10 workers on the graveyard shift.
A quality control consultant is to select 6of the workers in-depth interview.
To find the probability that all 6selected workers will be from the same shift.
We will first use the formula n/Cr=n!
r!(n−r)! to choose from different shifts and then take the probability.
Calculate the probability of all the chosen people to be in day shift,
D=20
C6/20
C6=20!
6!(20−6)!
20/C6=38760
Calculate the probability of all the chosen people to be in night shift,
N=15/C6
15/C6=15!
6!(15−6)!
15/C6=5005
Calculate of probability of all the chosen people to be in the graveyard shift,
G=10/C6
10/C6=10!
6!(10−6)!
10/C6=210
Calculate the total number of the probability of picking 6 people out of 45,
T=45/C6
45/C6=45!
6!(45−6)!
45/C6=814506
Now substitute the values in the formula,
P=38760+5005+210/814506
P=0.00539
The probability that all 6 selected workers will be from the same shift is:0.00539.
Page 71 Problem 8 Answer
Given- A production facility employs 20 workers on the day shift, 15 workers on the swing shift, and 10 workers on the graveyard shift.
A quality control consultant is to select 6 of the workers in-depth interview.
To find the probability that at least two different shifts will be represented among the selected workers The probability that all 6 selected workers will be from the same shift is obtained as: 0.00539 Hence by using this we will find the required probability.
The probability of all 6 chosen can be from one shift was calculated from the previous question,
P=0.00539
Hence the probability that at least two different shifts will be represented among the selected workers
=1−Probability that all 6 selected workers will be from the same shift.
=1−0.00539
=0.9946
The total probability of at least two different shifts represented among the selected workers is 0.9946.
Page 71 Problem 9 Answer
Given- A production facility 20 employs workers on the day shift, 15 workers on the swing shift, and 10 workers on the graveyard shift.
A quality control consultant is to select 6 of the workers in-depth interview.
To find the probability that at least one of the shifts will be unrepresented in the sample of workers.
We will first find the number of ways of selection for different shifts not represented using nCr=n!
r!(n−r)! and then get the probability.
Finding the number of ways in which day shift is not represented,
A1=15+10
C6/25
C6=25!
6!(25−6)!
25/C6=177100
Finding the number of ways in which swing shift is not represented,
A2=10+20/C6
30/C6=30!
6!(30−6)!
30/C6=593775
Finding the probability of graveyard shift not represented,
A3=20+15
C6/35
C6=35!
6!(35−6)!
35/C6=1623160
Hence we get:P(A)=25
C6/45
C6=0.022
P(A2)=30
C6/45
C6=0.073
P(A3)=35
C6,45
C6=0.199
Similarly we get:
Now add all the possibilities and subtract individual probabilities to find the solution,
P3=A1+A2+A3−(P(A1∩A2)+P(A2∩A3)+P(A1∩A3))+P(A1∩A2∩A3)
Since P(A1∩A2)+P(A2∩A3)+P(A1∩A3)is nothing but the probability of all the swifts together, which was calculated earlier, which is 312.
The value of P(A1∩A2∩A3)=0
Now substitute the value of all the values calculated in the formula,
Hence the required probability will be:
P3=P(A1)+P(A2)+P(A3)−(P(A1∩A2)+P(A2∩A3)+P(A1∩A3))+P(A1∩A2∩A3)
=0.022+0.073+0.199+0.00003+0.0006+0.005+0
=0.289
The probability that at least one of the shifts will be unrepresented in the sample of workers is 0.289.
Page 72 Problem 10 Answer
Given: A box in a certain supply room contains four 40-W lightbulbs, five 60-W bulbs, and six 75-W bulbs.
Given: three bulbs are randomly selected.To find the probability that exactly two of the selected bulbs are rated 75-W.
First, we will find the number of outcomes that is in how many ways two bulbs of 75−W can be removed by using combination Cr,n=(n/r) and then find the probability.
Number of 40 W bulbs-4
Number of 60 W bulbs-5
Number of 75 W bulbs-6
Hence 2 bulbs shall be taken from a group of 6 items and remaining one can be taken from 4+5=9
Hence the number of ways to do so is:
C2,6×C1,9=(6/2)×(9/1)
⇒C2,6×C1,9
=135
The total outcomes are:
C3,15=(15/3)
C3,15=455
Hence if we consider A to be the event where exactly two bulbs of 75−W are selected then:
P(A)= number of favorable outcomes in A number of outcomes in the sample space
P(A)=135/455
P(A)=0.297
The probability that exactly two of the selected bulbs are rated 75-W is 0.297.
Page 72 Problem 11 Answer
Given: A box in a certain supply room contains four 40-W lightbulbs, five 60-W bulbs, and six 75-W bulbs.
Given: Three bulbs are randomly selected.to find the probability that all three bulbs selected have the same rating.
We will use the formula for disjoint events: P(A∪B∪C)=P(A)+P(B)+P(C)
It is possible that the 3 selected light bulbs are all from either of the groups 40 W, 60 W and 75 W.
That is, denote events B1,B2,B3 as all 3 light bulbs are from 40W,60W , and 75W, respectively.
The event B is the initial one – all 3 randomly selected are from the same group.
Hence, as the 3 events B1,B2,B3 are disjoint we can apply:
P(A∪B∪C)=P(A)+P(B)+P(C)
⇒P(B)=P(B1∪B2∪B3)
P(B1∪B2∪B3)=P(B1)+P(B2)+P(B3)=C3,4/C3,15+C3,5/C3,15+C3,6/C3,15
=4/455+10/455+20/455
=34/455
=0.075.
The probability that all three of the selected bulbs have the same rating is 0.075.
Page 72 Problem 12 Answer
Given: A box in a certain supply room contains four 40-W lightbulbs, five 60-W bulbs, and six 75
-W bulbs Given: Three bulbs are randomly selected.
To find the probability that one bulb of each type is selected.
We will solve this by getting the favorable outcomes and using (n/1)=n.
We denote with C event that the 3 randomly selected light bulbs have representative in each of the three groups.
Then the required probability is given by:
P(C)= number of favorable outcomes in C number of outcomes in the sample space
P(C)=(4/1)×(5/1)×(6/1)(15/3)
P(C)=4×5×6/455
P(C)=0.2637
The probability that one bulb of each type is selected is 0.2637.
Page 72 Problem 13 Answer
Given: A box in a certain supply room contains four 40-W lightbulbs, five 60-W bulbs, and six 75-W bulbs Given: Bulbs are to be selected one by one until a 75-W bulb is found.
To find the probability that it is necessary to examine at least six bulbs.
We will use complement rule P(A)=1−P(Ac) to find the required probability.
Let D be the event that is is necessary to examine at least six bulbs
The probability of the required event may be calculated using the facts that the light bulbs are selected one-by-one, and that the selected light bulb before sixth is from group of 75 W (complement of D , in 1./2./3./4 or 5. turn 75W is selected) Therefore we have:
P{ At least six bulbs selected to get 75W Bulb }=1−P{75W bulb selected within first five bulbs }
P(D)=1−P(Dc)
=1−6/15−9/15×6/14−9⋅8/15⋅14×6/13−9⋅8⋅7
15⋅14⋅13×6
12−9⋅8⋅7⋅6
15⋅14⋅13⋅12×6/11
=1−0.958
=0.042
The probability that it is necessary to examine at least six bulbs is 0.042.
Page 72 Problem 14 Answer
Given: An ATM personal identification number (PIN) consists of four digits, each a0,1,2,…8, or 9, in succession.To find the number of different possible PINs if there are no restrictions on the choice of digits.
We will use the product rule for k−tuples.
Finding the number of ways the PIN can be arranged,When there is no restrictions each digit can be selected in 10 ways.
So possible numbers of pin are: P=10×10×10×10
P=10000
The total number of different possible PINs can be generated, if there are no restrictions on the choice of digits is 10000.
Page 72 Problem 15 Answer
Given: An ATM personal identification number (PIN) consists of four digits, each a0,1,2,…8 , or 9 , in succession.
Given: According to a representative at the author’s local branch of Chase Bank, there are in fact restrictions on the choice of digits. The following choices are prohibited:
(1) all four digits identical
(2) sequences of consecutive ascending or descending digits, such as 6543
(3) any sequence starting with 19
To find the probability that the sequence of pin will be legitimate.
Calculating the probability of all the conditions,. The identical numbers are totally 10.
There are 7 combinations of ascending and 7 descending combinations. A total of 14 .
The first two numbers are 1and 9, then the probability is 10=100.
Now remove all the invalid probabilities from the total,
P=10000−10−14−100/10000
P=0.9876
The probability that the random pick will be a legitimate PIN is 0.9876
Page 72 Problem 16 Answer
Given: An ATM personal identification number (PIN) consists of four digits, each0,1,2,…8, or 9, in succession.
Given: Someone has stolen an ATM card and knows that the first and last digits of the PIN are 8 and 1, respectively.
He has three tries before the card is retained by the ATM (but does not realize that).
So he randomly selects the 2nd and 3rd digits for the first try, then randomly selects a different pair of digits for the second try, and yet another randomly selected pair of digits for the third try .
To find the probability that the individual gains access to the account.
First, we will find the favorable outcomes by using the product rule and then find the probability.
Since there is only two slots left and 10 digits, the number of favorable outcomes will be
S=10⋅10
S=100
There was three attempt for trying a combination,
P=3/100
P=0.03 .
The probability that the individual gains access to the account is 0.03
Page 72 Problem 17 Answer
Given: An ATM personal identification number (PIN) consists of four digits, each0,1,2,…8, or 9, in succession.
Obtained probability in part c) 0.03
To find the probability that the individual gains access to the account if the first and last digits are 1 and 1, respectively.
We will get the favorable outcomes and then find the required probability.
Since the numbers are identical, there is a possibility of violation in the password.
It can not be1111 and can not contain continues numbers (1901,1902,1991).
Therefore, the total number of violation is 10+1=11
The total probability with three possible combination is,
P=3/100−11
P=3/89
P=0.0337
The probability for the individual to gain access to the account is 0.0337.
Page 72 Problem 18 Answer
Given: A starting lineup in basketball consists of two guards, two forwards, and a center.
Given: A certain college team has on its roster three centers, four guards, four forwards, and one individual (X) who can play either guard or forward.
To find the number of different lineups that can be created.
We shall use the combination formula nCr=n!
r!(n−r)! for the possibilities for which X can be picked or not.
There are three possibilities of player(X),
When (X) is not picked,
P=(2 out of 4 guards )+(2 out of 4 frowards )+(1 out of 3 center )
P=(4/C2)⋅(4/C2)⋅(3/C1)
P=6⋅6⋅3
P=108
When (X) is picked as guard,
P=(1 out of 4 guards )+(2 out of 4 frowards )+(1 out of 3 center )
P=(4/C1)⋅(4/C2)⋅(3/C1)
P=4⋅6⋅3
P=72
The same number is when the player is picked as forward.
P=(2 out of 4 guards )+(1 out of 4 forwards )+(1 out of 3 center )
P=(4/C2)⋅(4/C1)⋅(3/C1)
P=4⋅6⋅3
P=72
Now add all the found variables,
T=108+72+72
T=252
The number of starting lineups that can be created is 252.
Page 72 Problem 19 Answer
Given: A starting lineup in basketball consists of two guards, two forwards, and a center.
Given: the roster has 5 guards, 5 forwards, 3 centers, and 2 “swing players” (X and Y) who can play either guard or forward.
To find the probability that the team consists of a legitimate starting lineup.
We will find the combination with different position of the individual player X (old team member) or (given team member) and them sum up all.
When the starting lineup is made without X or Y: C2,5⋅C2,5⋅C1,3
=(5/2)⋅(5/2)⋅(3/1)
=5!/2(5−2)!⋅5!/2!(5−2)!⋅3!/1!(3−1)!
=10⋅10⋅3
=300
X is guard and there is no Y:C1,5⋅C2,5⋅C1,3
=(5/1)⋅(5/2)⋅(3/1)
=5!/1((5−1)!⋅5!/2(5−2)!⋅3!/1(3−1)!
=5⋅10⋅3
=150
Xis guard and there is no Y:C2,5⋅C1,5⋅C1,3
=(5/2)⋅(5/1)⋅(3/1)
=5!/2(5−2)!⋅5!/1!(5−1)!⋅3!/1!(3−1)!
=10⋅5⋅3
=150
Y is guard and there is no X:
C1,5⋅C2,5⋅C1,3
=(5/1)⋅(5/2)⋅(3/1)
=5!/1((5−1)!⋅5!/2!(5−2)!⋅3!/1(3−1)!
=5⋅10⋅3
=150
Y is forward and there is no X:C2,5⋅C1,5⋅C1,3
=(5/2)⋅(5/1)⋅(3/1)
=5!/2!(5−2)!⋅5!/1!(5−1)!⋅3!/1(3−1)!
=10⋅5⋅3
=150
X is guard and Y is also guard:
1⋅C2,5⋅C1,3
=1⋅(5/2)⋅(3/1)
=1⋅5!/2!(5−2)!⋅3!/1!(3−1)!
=1⋅10⋅3
=30
X is guard and Y is forward:
C1,5⋅C1,5⋅C1,3
=(5/1)⋅(5/1)⋅(3/1)
=5!/1(5−1)!⋅5!/1(5−1)!⋅3!/1(3−1)!
=5⋅5⋅3
=75
Xis forward and Y is guard: C1,5⋅C1,5⋅C1,3
=(5/1)⋅(5/1)⋅(3/1)
=5!/1!(5−1)!⋅5!/1!(5−1)!⋅3!/1(3−1)!
=5⋅5⋅3
=75
X is forward and Y is forward:
C2,5⋅1⋅C1,3
=(5/2)⋅1⋅(3/1)
=5!/2!(5−2)!⋅1⋅3!/1!(3−1)!
=10⋅1⋅3
=30
We will now sum up all the favorable possibilities.
300+4⋅150+2⋅30+2⋅75=1110
Total number of ways to choose 5 players from 15 members:
(15/5)
=15!
5!(15−5)!
=3003
The required probability of five constituting the legitimate starting lineup is:
P(A)= Number of favorable outcomes in A
Number of outcomes in the sample space
=1110/3003
=0.3696
The probability of five constituting the legitimate starting lineup is 0.3696.
Page 72 Problem 20 Answer
Given: In five-card poker, a straight consists of five cards with adjacent denominations. You are dealt with a five-card hand.
To find the probability that it will be a straight with high card 10.
To find the probability that it will be straight.
To find the probability that it will be a straight flush.
We will use the combination formula Ck,n=n!
k!(n−k)! to find the favorable outcomes for each case and then find the probability.
Total ways to select 5 cards out of 52are
C5,52 =2,598,960
(52/5)/52!
= 5!(52−5)!
We have a 10 high card, straight consists of six, seven, eight, nine, ten.
We will have 4 different types of cards of each type.
So, favorable outcomes will be 4⋅4⋅4⋅4⋅4=45
=1024
So, probability that out of 5 dealt cards we get a straight with high card10:
Number of favorable outcomes in A
Number of outcomes in the sample space =1024/2,598,960 = 0.000394
We have the straights with high card,5 high card,6 high card,7 high card,8 high card,9 high card,10 high card,J high card,Q high card K and high card A.So, there are10 cards.
We have calculated the desired outcomes to be 1024
Hence, we get P({ straight })=10⋅1024/2,598,960=0.00394.
The number of ways to select straight flush is 10.4=40 as there are 10 different straights and we have 4 different straight flush for each straight.
We know that total ways of selecting 5 cards out of 52 is 2,598,960.
Therefore, the probability that the card is a straight flush is P({ straight flush })=40/2,598,960=0.00001539
The probability that out of 5 dealt cards, it is straight with high card 10 is 0.000394.
The probability that the selected c.