Mixing
Combinatorics Issues
$begingroup$
I have three homework questions, all of which have the mutual problem of combinatorics
The first one:
A library subscribes to two different weekly news magazines, each of which is supposed to arrive in Wednesday’s mail. In actuality, each one may arrive on Wednesday, Thursday, Friday, or Saturday. Suppose the two arrive independently of one another, and for each one $P(Wed.)=3$, $P(Thurs.)=.4$, $P(Fri.)=.2$, and $P(Sat.)=.1$. Let Y= the number of days beyond Wednesday that it takes for both magazines to arrive (so possible Y values are 0, 1, 2, or 3). Compute the pmf of $Y$. [Hint: There are 16 possible outcomes; $Y(W,W) = 0$, $Y(F, TH) = 2$, and so on.]
Now, I know that the explicitly tell you the number of outcomes in the sample space, and it is small enough that you could generate the sample space $S$; but I was wondering, are there any fancy combinatoric techniques that you could use to count them? I have rather poor skills when it comes to combinatorics.
The second question is:
A new battery’s voltage may be acceptable (A) or unacceptable (U). A certain flashlight requires two batteries, so batteries will be independently selected and tested until two
acceptable ones have been found. Suppose that 90% of all batteries have acceptable voltages. Let Y denote the number of batteries that must be tested.
a.What is p(2), that is,$P(Y=2)$?
b.What is p(3)? [Hint:There are two different outcomes that result in $Y=3$]
c. To have $Y=5$ , what must be true of the fifth battery
selected? List the four outcomes for which $Y=5$ and then determine $p(5)$.
d.Use the pattern in your answers for parts (a)–(c) to obtain a general formula for p(y).
For this particular question, I am working on part c). I know for part c), that the 5th battery test in the sequence must be an acceptable battery; if it occurred any sooner in the sequence, then would wouldn't have to test 5 batteries. So, the last battery tested is fixed; that is _ _ _ _ A, where the first four slots can take on one other A (acceptable and 3 different U (unacceptable).
From this reasoning, I thought that $P(5) would be:
$P(5)=P(Y=5) = (.9){{4}choose{1}}(.9){{3}choose{1}}(.1)$
However, this isn't the correct answer. Could someone help me with part c), and thoroughly explain what you are doing, and why you are doing? As I've alluded to earlier, I am not very good with combinatorics, I just don't have an intuition for them.
probability combinatorics
asked Feb 17 '13 at 14:34
MackMack
2,0511766119
$endgroup$
add a comment |
$begingroup$
I have three homework questions, all of which have the mutual problem of combinatorics
The first one:
A library subscribes to two different weekly news magazines, each of which is supposed to arrive in Wednesday’s mail. In actuality, each one may arrive on Wednesday, Thursday, Friday, or Saturday. Suppose the two arrive independently of one another, and for each one $P(Wed.)=3$, $P(Thurs.)=.4$, $P(Fri.)=.2$, and $P(Sat.)=.1$. Let Y= the number of days beyond Wednesday that it takes for both magazines to arrive (so possible Y values are 0, 1, 2, or 3). Compute the pmf of $Y$. [Hint: There are 16 possible outcomes; $Y(W,W) = 0$, $Y(F, TH) = 2$, and so on.]
Now, I know that the explicitly tell you the number of outcomes in the sample space, and it is small enough that you could generate the sample space $S$; but I was wondering, are there any fancy combinatoric techniques that you could use to count them? I have rather poor skills when it comes to combinatorics.
The second question is:
A new battery’s voltage may be acceptable (A) or unacceptable (U). A certain flashlight requires two batteries, so batteries will be independently selected and tested until two
acceptable ones have been found. Suppose that 90% of all batteries have acceptable voltages. Let Y denote the number of batteries that must be tested.
a.What is p(2), that is,$P(Y=2)$?
b.What is p(3)? [Hint:There are two different outcomes that result in $Y=3$]
c. To have $Y=5$ , what must be true of the fifth battery
selected? List the four outcomes for which $Y=5$ and then determine $p(5)$.
d.Use the pattern in your answers for parts (a)–(c) to obtain a general formula for p(y).
For this particular question, I am working on part c). I know for part c), that the 5th battery test in the sequence must be an acceptable battery; if it occurred any sooner in the sequence, then would wouldn't have to test 5 batteries. So, the last battery tested is fixed; that is _ _ _ _ A, where the first four slots can take on one other A (acceptable and 3 different U (unacceptable).
From this reasoning, I thought that $P(5) would be:
$P(5)=P(Y=5) = (.9){{4}choose{1}}(.9){{3}choose{1}}(.1)$
However, this isn't the correct answer. Could someone help me with part c), and thoroughly explain what you are doing, and why you are doing? As I've alluded to earlier, I am not very good with combinatorics, I just don't have an intuition for them.
probability combinatorics
asked Feb 17 '13 at 14:34
MackMack
2,0511766119
$endgroup$
1
$begingroup$
For the second question: It's almost right. Suppose the sequence was $AUUUA$. The probability of this is $(.9)(.1)(.1)(.1)(.9)=(.1)^3(.9)^2$. Of course, this is only one way in which the fifth battery is the second acceptable one chosen. (Note, there are $4choose 3$ ways to select the "slots" for the three defective batteries.)
$endgroup$
– David Mitra
Feb 17 '13 at 14:53
$begingroup$
@DavidMitra Oh, I see. There are four ways to arrange the 3 U's, and 4 ways to arrange that single A. Sometimes it's so frustrating how a simple counting problem can be so elusive for me. Thank you!
$endgroup$
– Mack
Feb 17 '13 at 15:22
$begingroup$
Note the total number of ways to have the fifth choice being the second $A$ is just $4$, though. (If you chose where the "A" is in the first four slots, you've also chosen the three slots for the $U$'s.)
$endgroup$
– David Mitra
Feb 17 '13 at 15:29
add a comment |
$begingroup$
I have three homework questions, all of which have the mutual problem of combinatorics
The first one:
A library subscribes to two different weekly news magazines, each of which is supposed to arrive in Wednesday’s mail. In actuality, each one may arrive on Wednesday, Thursday, Friday, or Saturday. Suppose the two arrive independently of one another, and for each one $P(Wed.)=3$, $P(Thurs.)=.4$, $P(Fri.)=.2$, and $P(Sat.)=.1$. Let Y= the number of days beyond Wednesday that it takes for both magazines to arrive (so possible Y values are 0, 1, 2, or 3). Compute the pmf of $Y$. [Hint: There are 16 possible outcomes; $Y(W,W) = 0$, $Y(F, TH) = 2$, and so on.]
Now, I know that the explicitly tell you the number of outcomes in the sample space, and it is small enough that you could generate the sample space $S$; but I was wondering, are there any fancy combinatoric techniques that you could use to count them? I have rather poor skills when it comes to combinatorics.
The second question is:
A new battery’s voltage may be acceptable (A) or unacceptable (U). A certain flashlight requires two batteries, so batteries will be independently selected and tested until two
acceptable ones have been found. Suppose that 90% of all batteries have acceptable voltages. Let Y denote the number of batteries that must be tested.
a.What is p(2), that is,$P(Y=2)$?
b.What is p(3)? [Hint:There are two different outcomes that result in $Y=3$]
c. To have $Y=5$ , what must be true of the fifth battery
selected? List the four outcomes for which $Y=5$ and then determine $p(5)$.
d.Use the pattern in your answers for parts (a)–(c) to obtain a general formula for p(y).
For this particular question, I am working on part c). I know for part c), that the 5th battery test in the sequence must be an acceptable battery; if it occurred any sooner in the sequence, then would wouldn't have to test 5 batteries. So, the last battery tested is fixed; that is _ _ _ _ A, where the first four slots can take on one other A (acceptable and 3 different U (unacceptable).
From this reasoning, I thought that $P(5) would be:
$P(5)=P(Y=5) = (.9){{4}choose{1}}(.9){{3}choose{1}}(.1)$
However, this isn't the correct answer. Could someone help me with part c), and thoroughly explain what you are doing, and why you are doing? As I've alluded to earlier, I am not very good with combinatorics, I just don't have an intuition for them.
probability combinatorics
asked Feb 17 '13 at 14:34
MackMack
2,0511766119
$endgroup$
I have three homework questions, all of which have the mutual problem of combinatorics
The first one:
A library subscribes to two different weekly news magazines, each of which is supposed to arrive in Wednesday’s mail. In actuality, each one may arrive on Wednesday, Thursday, Friday, or Saturday. Suppose the two arrive independently of one another, and for each one $P(Wed.)=3$, $P(Thurs.)=.4$, $P(Fri.)=.2$, and $P(Sat.)=.1$. Let Y= the number of days beyond Wednesday that it takes for both magazines to arrive (so possible Y values are 0, 1, 2, or 3). Compute the pmf of $Y$. [Hint: There are 16 possible outcomes; $Y(W,W) = 0$, $Y(F, TH) = 2$, and so on.]
Now, I know that the explicitly tell you the number of outcomes in the sample space, and it is small enough that you could generate the sample space $S$; but I was wondering, are there any fancy combinatoric techniques that you could use to count them? I have rather poor skills when it comes to combinatorics.
The second question is:
A new battery’s voltage may be acceptable (A) or unacceptable (U). A certain flashlight requires two batteries, so batteries will be independently selected and tested until two
acceptable ones have been found. Suppose that 90% of all batteries have acceptable voltages. Let Y denote the number of batteries that must be tested.
a.What is p(2), that is,$P(Y=2)$?
b.What is p(3)? [Hint:There are two different outcomes that result in $Y=3$]
c. To have $Y=5$ , what must be true of the fifth battery
selected? List the four outcomes for which $Y=5$ and then determine $p(5)$.
d.Use the pattern in your answers for parts (a)–(c) to obtain a general formula for p(y).
For this particular question, I am working on part c). I know for part c), that the 5th battery test in the sequence must be an acceptable battery; if it occurred any sooner in the sequence, then would wouldn't have to test 5 batteries. So, the last battery tested is fixed; that is _ _ _ _ A, where the first four slots can take on one other A (acceptable and 3 different U (unacceptable).
From this reasoning, I thought that $P(5) would be:
$P(5)=P(Y=5) = (.9){{4}choose{1}}(.9){{3}choose{1}}(.1)$
However, this isn't the correct answer. Could someone help me with part c), and thoroughly explain what you are doing, and why you are doing? As I've alluded to earlier, I am not very good with combinatorics, I just don't have an intuition for them.
probability combinatorics
probability combinatorics
asked Feb 17 '13 at 14:34
MackMack
2,0511766119
asked Feb 17 '13 at 14:34
MackMack
2,0511766119
asked Feb 17 '13 at 14:34
MackMack
2,0511766119
asked Feb 17 '13 at 14:34
MackMack
2,0511766119
asked Feb 17 '13 at 14:34
MackMack
2,0511766119
2,0511766119
1
$begingroup$
For the second question: It's almost right. Suppose the sequence was $AUUUA$. The probability of this is $(.9)(.1)(.1)(.1)(.9)=(.1)^3(.9)^2$. Of course, this is only one way in which the fifth battery is the second acceptable one chosen. (Note, there are $4choose 3$ ways to select the "slots" for the three defective batteries.)
$endgroup$
– David Mitra
Feb 17 '13 at 14:53
$begingroup$
@DavidMitra Oh, I see. There are four ways to arrange the 3 U's, and 4 ways to arrange that single A. Sometimes it's so frustrating how a simple counting problem can be so elusive for me. Thank you!
$endgroup$
– Mack
Feb 17 '13 at 15:22
$begingroup$
Note the total number of ways to have the fifth choice being the second $A$ is just $4$, though. (If you chose where the "A" is in the first four slots, you've also chosen the three slots for the $U$'s.)
$endgroup$
– David Mitra
Feb 17 '13 at 15:29
add a comment |
1
$begingroup$
For the second question: It's almost right. Suppose the sequence was $AUUUA$. The probability of this is $(.9)(.1)(.1)(.1)(.9)=(.1)^3(.9)^2$. Of course, this is only one way in which the fifth battery is the second acceptable one chosen. (Note, there are $4choose 3$ ways to select the "slots" for the three defective batteries.)
$endgroup$
– David Mitra
Feb 17 '13 at 14:53
$begingroup$
@DavidMitra Oh, I see. There are four ways to arrange the 3 U's, and 4 ways to arrange that single A. Sometimes it's so frustrating how a simple counting problem can be so elusive for me. Thank you!
$endgroup$
– Mack
Feb 17 '13 at 15:22
$begingroup$
Note the total number of ways to have the fifth choice being the second $A$ is just $4$, though. (If you chose where the "A" is in the first four slots, you've also chosen the three slots for the $U$'s.)
$endgroup$
– David Mitra
Feb 17 '13 at 15:29
1
1
$begingroup$
For the second question: It's almost right. Suppose the sequence was $AUUUA$. The probability of this is $(.9)(.1)(.1)(.1)(.9)=(.1)^3(.9)^2$. Of course, this is only one way in which the fifth battery is the second acceptable one chosen. (Note, there are $4choose 3$ ways to select the "slots" for the three defective batteries.)
$endgroup$
– David Mitra
Feb 17 '13 at 14:53
$begingroup$
For the second question: It's almost right. Suppose the sequence was $AUUUA$. The probability of this is $(.9)(.1)(.1)(.1)(.9)=(.1)^3(.9)^2$. Of course, this is only one way in which the fifth battery is the second acceptable one chosen. (Note, there are $4choose 3$ ways to select the "slots" for the three defective batteries.)
$endgroup$
– David Mitra
Feb 17 '13 at 14:53
$begingroup$
@DavidMitra Oh, I see. There are four ways to arrange the 3 U's, and 4 ways to arrange that single A. Sometimes it's so frustrating how a simple counting problem can be so elusive for me. Thank you!
$endgroup$
– Mack
Feb 17 '13 at 15:22
$begingroup$
@DavidMitra Oh, I see. There are four ways to arrange the 3 U's, and 4 ways to arrange that single A. Sometimes it's so frustrating how a simple counting problem can be so elusive for me. Thank you!
$endgroup$
– Mack
Feb 17 '13 at 15:22
$begingroup$
Note the total number of ways to have the fifth choice being the second $A$ is just $4$, though. (If you chose where the "A" is in the first four slots, you've also chosen the three slots for the $U$'s.)
$endgroup$
– David Mitra
Feb 17 '13 at 15:29
$begingroup$
Note the total number of ways to have the fifth choice being the second $A$ is just $4$, though. (If you chose where the "A" is in the first four slots, you've also chosen the three slots for the $U$'s.)
$endgroup$
– David Mitra
Feb 17 '13 at 15:29
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
1)
The question said, for each magazine,
$$
begin{align*}
P(text{Wed}) &= 0.3\P(text{Thu}) &= 0.4\P(text{Fri}) &= 0.2\P(text{Sat}) &= 0.1\
end{align*}
$$
Consider both magazines to be $A$ and $B$ and let's just call Wed = $3$, Thu = $4$ and so on. You mentioned you have completed the sample space which looks something like...
$$
begin{bmatrix}
* & B_3&B_4 & B_5 & B_6 & TOTAL\
A_3 &3& 4& 5 spadesuit& 6& diamondsuit\
A_4 & 4&4 & 5&6 & diamondsuit \
A_5 & 5& 5&5 &6 & diamondsuit \
A_6 &6 & 6&6 &6 & diamondsuit \
TOTAL &clubsuit & clubsuit & clubsuit & clubsuit & 1 \
end{bmatrix}
$$
One way to check is to ensure that the "total" row & column, when summed, adds up to 1. ie the clubs add to 1 & the diamonds add to 1. To the best of my knowledge, I would do it this way too. The number in the matrix represents the value of $Y$, number of days taken for the LAST paper to arrive. You should know how to do the multiplication here eg. as denoted by a spade, $A_3B_5 = (0.3)(0.2) $ and so on.
2c)
I shall only focus on part (c) as per your request. Yes, the 5th battery must be acceptable. Consider the first 4 slots. There are 4 different sequences to draw to end up having 3 $U$ and 1 $A$. They are:
(assume a club denotes not working & a heart denotes working)
$$
begin{pmatrix}
heartsuit-clubsuit-clubsuit-clubsuit\
clubsuit-heartsuit-clubsuit-clubsuit\
clubsuit-clubsuit-heartsuit-clubsuit\
clubsuit-clubsuit-clubsuit-heartsuit
end{pmatrix}
$$
So, the probability of drawing 3 $U$ and 1 $A$ is
$$
(0.1)^3(0.9)times 4
$$
Having this combination, you need to draw a working one on the 5th draw. Hence the answer is
$$
(0.1)^3(0.9)times 4 times 0.9
$$
where the last $0.9$ represents a $A$ on the last draw.
If you want to find out more, you can read about the Negative Binomial Probability Distribution which addresses how many times do you need to take in order to get $r$ successes.
answered Feb 17 '13 at 15:09
bryanblackbeebryanblackbee
2,71621944
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add a comment |
$begingroup$
1)
Let the first letter be magazine1 and the second letter be magazine2 eg. WW = wednesday-wednesday.
y = the number of days beyond Wednesday that it takes for both magazines to arrive.
E(y=0) = {WW(0-0) }
P(y=0) = .3*.3 = .0900
E(y=1) = { WT(0-1) , TW (1-0) , TT (1-1) } i.g. y=maximum of 2 values that represents day
P(y=1) = .4*.3 + .4*.3 + .4*.4 = 0.4000
E(y=2) = {WF(0-2) , FW(2-0) , FF(2-2) , TF(1-2) , FT(2-1) } i.g. y= maximum of 2 values that represents day
P(y=2) = .3*.2+.2*.3 + .2*.2 + .4*.2 + .2*.4 = 0.3200
E(y=3) = {WS(0-3) , SW(3-0) , SS(3-3) , FS(2-3) , SF(3-2) , TS(1-3) or ST(3-1) } i.g. y = maximum of 2 values that represents day
P(y=3) = .3*.1+.1*.3+.1*.1 +.2*.1 + .1*.2 + .4*.1+.1*.4 = 0.1900
Control: sum([ .0900 0.4000 0.3200 0.1900 ]) = 1
answered Feb 27 '18 at 12:47
MaSaMaSa
1
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add a comment |
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Another approach to the Library magazine question.
Let P(Wed) = 0.3, P(Thur) = 0.4, P(Fri) = 0.2, P(Sat) = 0.1 . The stated probabilities values from above.
For the sake of an equation, let P(0) = P(Wed), P(1) = P(Thur), P(2) = P(Fri), P(3) = P(Sat) where 'y' below represents the appropriate day.
For P(0) its: $$P(0)^2 = 0.3*0.3=0.09$$
For P(1) its: $$ P(1) = (P(0)+P(1))^2 - P(0)^2$$
For any P value > 2$$ P(y) = (P(0)+ dots +P(y))^2 - (P(0)+ dots + P(y-1))^2$$
An example with P(3) or Sat. with probability values:
$$P(3)=(0.3+0.4+0.2+0.1)^2-(0.3+0.4+0.2)^2=0.19$$
answered Jan 26 at 18:33
CS StudentCS Student
34
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add a comment |
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3 Answers
3
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3 Answers
3
active
oldest
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oldest
votes
$begingroup$
1)
The question said, for each magazine,
$$
begin{align*}
P(text{Wed}) &= 0.3\P(text{Thu}) &= 0.4\P(text{Fri}) &= 0.2\P(text{Sat}) &= 0.1\
end{align*}
$$
Consider both magazines to be $A$ and $B$ and let's just call Wed = $3$, Thu = $4$ and so on. You mentioned you have completed the sample space which looks something like...
$$
begin{bmatrix}
* & B_3&B_4 & B_5 & B_6 & TOTAL\
A_3 &3& 4& 5 spadesuit& 6& diamondsuit\
A_4 & 4&4 & 5&6 & diamondsuit \
A_5 & 5& 5&5 &6 & diamondsuit \
A_6 &6 & 6&6 &6 & diamondsuit \
TOTAL &clubsuit & clubsuit & clubsuit & clubsuit & 1 \
end{bmatrix}
$$
One way to check is to ensure that the "total" row & column, when summed, adds up to 1. ie the clubs add to 1 & the diamonds add to 1. To the best of my knowledge, I would do it this way too. The number in the matrix represents the value of $Y$, number of days taken for the LAST paper to arrive. You should know how to do the multiplication here eg. as denoted by a spade, $A_3B_5 = (0.3)(0.2) $ and so on.
2c)
I shall only focus on part (c) as per your request. Yes, the 5th battery must be acceptable. Consider the first 4 slots. There are 4 different sequences to draw to end up having 3 $U$ and 1 $A$. They are:
(assume a club denotes not working & a heart denotes working)
$$
begin{pmatrix}
heartsuit-clubsuit-clubsuit-clubsuit\
clubsuit-heartsuit-clubsuit-clubsuit\
clubsuit-clubsuit-heartsuit-clubsuit\
clubsuit-clubsuit-clubsuit-heartsuit
end{pmatrix}
$$
So, the probability of drawing 3 $U$ and 1 $A$ is
$$
(0.1)^3(0.9)times 4
$$
Having this combination, you need to draw a working one on the 5th draw. Hence the answer is
$$
(0.1)^3(0.9)times 4 times 0.9
$$
where the last $0.9$ represents a $A$ on the last draw.
If you want to find out more, you can read about the Negative Binomial Probability Distribution which addresses how many times do you need to take in order to get $r$ successes.
answered Feb 17 '13 at 15:09
bryanblackbeebryanblackbee
2,71621944
$endgroup$
add a comment |
$begingroup$
1)
The question said, for each magazine,
$$
begin{align*}
P(text{Wed}) &= 0.3\P(text{Thu}) &= 0.4\P(text{Fri}) &= 0.2\P(text{Sat}) &= 0.1\
end{align*}
$$
Consider both magazines to be $A$ and $B$ and let's just call Wed = $3$, Thu = $4$ and so on. You mentioned you have completed the sample space which looks something like...
$$
begin{bmatrix}
* & B_3&B_4 & B_5 & B_6 & TOTAL\
A_3 &3& 4& 5 spadesuit& 6& diamondsuit\
A_4 & 4&4 & 5&6 & diamondsuit \
A_5 & 5& 5&5 &6 & diamondsuit \
A_6 &6 & 6&6 &6 & diamondsuit \
TOTAL &clubsuit & clubsuit & clubsuit & clubsuit & 1 \
end{bmatrix}
$$
One way to check is to ensure that the "total" row & column, when summed, adds up to 1. ie the clubs add to 1 & the diamonds add to 1. To the best of my knowledge, I would do it this way too. The number in the matrix represents the value of $Y$, number of days taken for the LAST paper to arrive. You should know how to do the multiplication here eg. as denoted by a spade, $A_3B_5 = (0.3)(0.2) $ and so on.
2c)
I shall only focus on part (c) as per your request. Yes, the 5th battery must be acceptable. Consider the first 4 slots. There are 4 different sequences to draw to end up having 3 $U$ and 1 $A$. They are:
(assume a club denotes not working & a heart denotes working)
$$
begin{pmatrix}
heartsuit-clubsuit-clubsuit-clubsuit\
clubsuit-heartsuit-clubsuit-clubsuit\
clubsuit-clubsuit-heartsuit-clubsuit\
clubsuit-clubsuit-clubsuit-heartsuit
end{pmatrix}
$$
So, the probability of drawing 3 $U$ and 1 $A$ is
$$
(0.1)^3(0.9)times 4
$$
Having this combination, you need to draw a working one on the 5th draw. Hence the answer is
$$
(0.1)^3(0.9)times 4 times 0.9
$$
where the last $0.9$ represents a $A$ on the last draw.
If you want to find out more, you can read about the Negative Binomial Probability Distribution which addresses how many times do you need to take in order to get $r$ successes.
answered Feb 17 '13 at 15:09
bryanblackbeebryanblackbee
2,71621944
$endgroup$
add a comment |
$begingroup$
1)
The question said, for each magazine,
$$
begin{align*}
P(text{Wed}) &= 0.3\P(text{Thu}) &= 0.4\P(text{Fri}) &= 0.2\P(text{Sat}) &= 0.1\
end{align*}
$$
Consider both magazines to be $A$ and $B$ and let's just call Wed = $3$, Thu = $4$ and so on. You mentioned you have completed the sample space which looks something like...
$$
begin{bmatrix}
* & B_3&B_4 & B_5 & B_6 & TOTAL\
A_3 &3& 4& 5 spadesuit& 6& diamondsuit\
A_4 & 4&4 & 5&6 & diamondsuit \
A_5 & 5& 5&5 &6 & diamondsuit \
A_6 &6 & 6&6 &6 & diamondsuit \
TOTAL &clubsuit & clubsuit & clubsuit & clubsuit & 1 \
end{bmatrix}
$$
One way to check is to ensure that the "total" row & column, when summed, adds up to 1. ie the clubs add to 1 & the diamonds add to 1. To the best of my knowledge, I would do it this way too. The number in the matrix represents the value of $Y$, number of days taken for the LAST paper to arrive. You should know how to do the multiplication here eg. as denoted by a spade, $A_3B_5 = (0.3)(0.2) $ and so on.
2c)
I shall only focus on part (c) as per your request. Yes, the 5th battery must be acceptable. Consider the first 4 slots. There are 4 different sequences to draw to end up having 3 $U$ and 1 $A$. They are:
(assume a club denotes not working & a heart denotes working)
$$
begin{pmatrix}
heartsuit-clubsuit-clubsuit-clubsuit\
clubsuit-heartsuit-clubsuit-clubsuit\
clubsuit-clubsuit-heartsuit-clubsuit\
clubsuit-clubsuit-clubsuit-heartsuit
end{pmatrix}
$$
So, the probability of drawing 3 $U$ and 1 $A$ is
$$
(0.1)^3(0.9)times 4
$$
Having this combination, you need to draw a working one on the 5th draw. Hence the answer is
$$
(0.1)^3(0.9)times 4 times 0.9
$$
where the last $0.9$ represents a $A$ on the last draw.
If you want to find out more, you can read about the Negative Binomial Probability Distribution which addresses how many times do you need to take in order to get $r$ successes.
answered Feb 17 '13 at 15:09
bryanblackbeebryanblackbee
2,71621944
$endgroup$
1)
The question said, for each magazine,
$$
begin{align*}
P(text{Wed}) &= 0.3\P(text{Thu}) &= 0.4\P(text{Fri}) &= 0.2\P(text{Sat}) &= 0.1\
end{align*}
$$
Consider both magazines to be $A$ and $B$ and let's just call Wed = $3$, Thu = $4$ and so on. You mentioned you have completed the sample space which looks something like...
$$
begin{bmatrix}
* & B_3&B_4 & B_5 & B_6 & TOTAL\
A_3 &3& 4& 5 spadesuit& 6& diamondsuit\
A_4 & 4&4 & 5&6 & diamondsuit \
A_5 & 5& 5&5 &6 & diamondsuit \
A_6 &6 & 6&6 &6 & diamondsuit \
TOTAL &clubsuit & clubsuit & clubsuit & clubsuit & 1 \
end{bmatrix}
$$
One way to check is to ensure that the "total" row & column, when summed, adds up to 1. ie the clubs add to 1 & the diamonds add to 1. To the best of my knowledge, I would do it this way too. The number in the matrix represents the value of $Y$, number of days taken for the LAST paper to arrive. You should know how to do the multiplication here eg. as denoted by a spade, $A_3B_5 = (0.3)(0.2) $ and so on.
2c)
I shall only focus on part (c) as per your request. Yes, the 5th battery must be acceptable. Consider the first 4 slots. There are 4 different sequences to draw to end up having 3 $U$ and 1 $A$. They are:
(assume a club denotes not working & a heart denotes working)
$$
begin{pmatrix}
heartsuit-clubsuit-clubsuit-clubsuit\
clubsuit-heartsuit-clubsuit-clubsuit\
clubsuit-clubsuit-heartsuit-clubsuit\
clubsuit-clubsuit-clubsuit-heartsuit
end{pmatrix}
$$
So, the probability of drawing 3 $U$ and 1 $A$ is
$$
(0.1)^3(0.9)times 4
$$
Having this combination, you need to draw a working one on the 5th draw. Hence the answer is
$$
(0.1)^3(0.9)times 4 times 0.9
$$
where the last $0.9$ represents a $A$ on the last draw.
If you want to find out more, you can read about the Negative Binomial Probability Distribution which addresses how many times do you need to take in order to get $r$ successes.
answered Feb 17 '13 at 15:09
bryanblackbeebryanblackbee
2,71621944
edited Feb 17 '13 at 15:24
answered Feb 17 '13 at 15:09
bryanblackbeebryanblackbee
2,71621944
answered Feb 17 '13 at 15:09
bryanblackbeebryanblackbee
2,71621944
answered Feb 17 '13 at 15:09
bryanblackbeebryanblackbee
2,71621944
2,71621944
add a comment |
add a comment |
$begingroup$
1)
Let the first letter be magazine1 and the second letter be magazine2 eg. WW = wednesday-wednesday.
y = the number of days beyond Wednesday that it takes for both magazines to arrive.
E(y=0) = {WW(0-0) }
P(y=0) = .3*.3 = .0900
E(y=1) = { WT(0-1) , TW (1-0) , TT (1-1) } i.g. y=maximum of 2 values that represents day
P(y=1) = .4*.3 + .4*.3 + .4*.4 = 0.4000
E(y=2) = {WF(0-2) , FW(2-0) , FF(2-2) , TF(1-2) , FT(2-1) } i.g. y= maximum of 2 values that represents day
P(y=2) = .3*.2+.2*.3 + .2*.2 + .4*.2 + .2*.4 = 0.3200
E(y=3) = {WS(0-3) , SW(3-0) , SS(3-3) , FS(2-3) , SF(3-2) , TS(1-3) or ST(3-1) } i.g. y = maximum of 2 values that represents day
P(y=3) = .3*.1+.1*.3+.1*.1 +.2*.1 + .1*.2 + .4*.1+.1*.4 = 0.1900
Control: sum([ .0900 0.4000 0.3200 0.1900 ]) = 1
answered Feb 27 '18 at 12:47
MaSaMaSa
1
$endgroup$
add a comment |
$begingroup$
1)
Let the first letter be magazine1 and the second letter be magazine2 eg. WW = wednesday-wednesday.
y = the number of days beyond Wednesday that it takes for both magazines to arrive.
E(y=0) = {WW(0-0) }
P(y=0) = .3*.3 = .0900
E(y=1) = { WT(0-1) , TW (1-0) , TT (1-1) } i.g. y=maximum of 2 values that represents day
P(y=1) = .4*.3 + .4*.3 + .4*.4 = 0.4000
E(y=2) = {WF(0-2) , FW(2-0) , FF(2-2) , TF(1-2) , FT(2-1) } i.g. y= maximum of 2 values that represents day
P(y=2) = .3*.2+.2*.3 + .2*.2 + .4*.2 + .2*.4 = 0.3200
E(y=3) = {WS(0-3) , SW(3-0) , SS(3-3) , FS(2-3) , SF(3-2) , TS(1-3) or ST(3-1) } i.g. y = maximum of 2 values that represents day
P(y=3) = .3*.1+.1*.3+.1*.1 +.2*.1 + .1*.2 + .4*.1+.1*.4 = 0.1900
Control: sum([ .0900 0.4000 0.3200 0.1900 ]) = 1
answered Feb 27 '18 at 12:47
MaSaMaSa
1
$endgroup$
add a comment |
$begingroup$
1)
Let the first letter be magazine1 and the second letter be magazine2 eg. WW = wednesday-wednesday.
y = the number of days beyond Wednesday that it takes for both magazines to arrive.
E(y=0) = {WW(0-0) }
P(y=0) = .3*.3 = .0900
E(y=1) = { WT(0-1) , TW (1-0) , TT (1-1) } i.g. y=maximum of 2 values that represents day
P(y=1) = .4*.3 + .4*.3 + .4*.4 = 0.4000
E(y=2) = {WF(0-2) , FW(2-0) , FF(2-2) , TF(1-2) , FT(2-1) } i.g. y= maximum of 2 values that represents day
P(y=2) = .3*.2+.2*.3 + .2*.2 + .4*.2 + .2*.4 = 0.3200
E(y=3) = {WS(0-3) , SW(3-0) , SS(3-3) , FS(2-3) , SF(3-2) , TS(1-3) or ST(3-1) } i.g. y = maximum of 2 values that represents day
P(y=3) = .3*.1+.1*.3+.1*.1 +.2*.1 + .1*.2 + .4*.1+.1*.4 = 0.1900
Control: sum([ .0900 0.4000 0.3200 0.1900 ]) = 1
answered Feb 27 '18 at 12:47
MaSaMaSa
1
$endgroup$
1)
Let the first letter be magazine1 and the second letter be magazine2 eg. WW = wednesday-wednesday.
y = the number of days beyond Wednesday that it takes for both magazines to arrive.
E(y=0) = {WW(0-0) }
P(y=0) = .3*.3 = .0900
E(y=1) = { WT(0-1) , TW (1-0) , TT (1-1) } i.g. y=maximum of 2 values that represents day
P(y=1) = .4*.3 + .4*.3 + .4*.4 = 0.4000
E(y=2) = {WF(0-2) , FW(2-0) , FF(2-2) , TF(1-2) , FT(2-1) } i.g. y= maximum of 2 values that represents day
P(y=2) = .3*.2+.2*.3 + .2*.2 + .4*.2 + .2*.4 = 0.3200
E(y=3) = {WS(0-3) , SW(3-0) , SS(3-3) , FS(2-3) , SF(3-2) , TS(1-3) or ST(3-1) } i.g. y = maximum of 2 values that represents day
P(y=3) = .3*.1+.1*.3+.1*.1 +.2*.1 + .1*.2 + .4*.1+.1*.4 = 0.1900
Control: sum([ .0900 0.4000 0.3200 0.1900 ]) = 1
answered Feb 27 '18 at 12:47
MaSaMaSa
1
answered Feb 27 '18 at 12:47
MaSaMaSa
1
answered Feb 27 '18 at 12:47
MaSaMaSa
1
answered Feb 27 '18 at 12:47
MaSaMaSa
1
1
add a comment |
add a comment |
$begingroup$
Another approach to the Library magazine question.
Let P(Wed) = 0.3, P(Thur) = 0.4, P(Fri) = 0.2, P(Sat) = 0.1 . The stated probabilities values from above.
For the sake of an equation, let P(0) = P(Wed), P(1) = P(Thur), P(2) = P(Fri), P(3) = P(Sat) where 'y' below represents the appropriate day.
For P(0) its: $$P(0)^2 = 0.3*0.3=0.09$$
For P(1) its: $$ P(1) = (P(0)+P(1))^2 - P(0)^2$$
For any P value > 2$$ P(y) = (P(0)+ dots +P(y))^2 - (P(0)+ dots + P(y-1))^2$$
An example with P(3) or Sat. with probability values:
$$P(3)=(0.3+0.4+0.2+0.1)^2-(0.3+0.4+0.2)^2=0.19$$
answered Jan 26 at 18:33
CS StudentCS Student
34
$endgroup$
add a comment |
$begingroup$
Another approach to the Library magazine question.
Let P(Wed) = 0.3, P(Thur) = 0.4, P(Fri) = 0.2, P(Sat) = 0.1 . The stated probabilities values from above.
For the sake of an equation, let P(0) = P(Wed), P(1) = P(Thur), P(2) = P(Fri), P(3) = P(Sat) where 'y' below represents the appropriate day.
For P(0) its: $$P(0)^2 = 0.3*0.3=0.09$$
For P(1) its: $$ P(1) = (P(0)+P(1))^2 - P(0)^2$$
For any P value > 2$$ P(y) = (P(0)+ dots +P(y))^2 - (P(0)+ dots + P(y-1))^2$$
An example with P(3) or Sat. with probability values:
$$P(3)=(0.3+0.4+0.2+0.1)^2-(0.3+0.4+0.2)^2=0.19$$
answered Jan 26 at 18:33
CS StudentCS Student
34
$endgroup$
add a comment |
$begingroup$
Another approach to the Library magazine question.
Let P(Wed) = 0.3, P(Thur) = 0.4, P(Fri) = 0.2, P(Sat) = 0.1 . The stated probabilities values from above.
For the sake of an equation, let P(0) = P(Wed), P(1) = P(Thur), P(2) = P(Fri), P(3) = P(Sat) where 'y' below represents the appropriate day.
For P(0) its: $$P(0)^2 = 0.3*0.3=0.09$$
For P(1) its: $$ P(1) = (P(0)+P(1))^2 - P(0)^2$$
For any P value > 2$$ P(y) = (P(0)+ dots +P(y))^2 - (P(0)+ dots + P(y-1))^2$$
An example with P(3) or Sat. with probability values:
$$P(3)=(0.3+0.4+0.2+0.1)^2-(0.3+0.4+0.2)^2=0.19$$
answered Jan 26 at 18:33
CS StudentCS Student
34
$endgroup$
Another approach to the Library magazine question.
Let P(Wed) = 0.3, P(Thur) = 0.4, P(Fri) = 0.2, P(Sat) = 0.1 . The stated probabilities values from above.
For the sake of an equation, let P(0) = P(Wed), P(1) = P(Thur), P(2) = P(Fri), P(3) = P(Sat) where 'y' below represents the appropriate day.
For P(0) its: $$P(0)^2 = 0.3*0.3=0.09$$
For P(1) its: $$ P(1) = (P(0)+P(1))^2 - P(0)^2$$
For any P value > 2$$ P(y) = (P(0)+ dots +P(y))^2 - (P(0)+ dots + P(y-1))^2$$
An example with P(3) or Sat. with probability values:
$$P(3)=(0.3+0.4+0.2+0.1)^2-(0.3+0.4+0.2)^2=0.19$$
answered Jan 26 at 18:33
CS StudentCS Student
34
edited Feb 18 at 5:15
answered Jan 26 at 18:33
CS StudentCS Student
34
answered Jan 26 at 18:33
CS StudentCS Student
34
answered Jan 26 at 18:33
CS StudentCS Student
34
34
add a comment |
add a comment |
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1
$begingroup$
For the second question: It's almost right. Suppose the sequence was $AUUUA$. The probability of this is $(.9)(.1)(.1)(.1)(.9)=(.1)^3(.9)^2$. Of course, this is only one way in which the fifth battery is the second acceptable one chosen. (Note, there are $4choose 3$ ways to select the "slots" for the three defective batteries.)
$endgroup$
– David Mitra
Feb 17 '13 at 14:53
$begingroup$
@DavidMitra Oh, I see. There are four ways to arrange the 3 U's, and 4 ways to arrange that single A. Sometimes it's so frustrating how a simple counting problem can be so elusive for me. Thank you!
$endgroup$
– Mack
Feb 17 '13 at 15:22
$begingroup$
Note the total number of ways to have the fifth choice being the second $A$ is just $4$, though. (If you chose where the "A" is in the first four slots, you've also chosen the three slots for the $U$'s.)
$endgroup$
– David Mitra
Feb 17 '13 at 15:29