A probability problem - finding where I've gone wrongIn how many ways can people get out of the elevator?Elevator stop, other approachThe probability that 7 people taking an elevator will leave it in configuration 3-2-1-1Probability of an elevator rising to a certain floor at most and exactlywhat's the probability that in a building of 10 floors and 5 people in elevator, the elevator would reach exactly until the 5th floor and no higher?Calculate the expected value of the highest floor the elevator may reach.Elevator probability calculationPeople leaving an elevator on different floorsProbability question - the denominator in a specific problemIn how many ways can $7$ people take elevator to $5$th floor such that at the last floor all remaining people exit (at least one remains)?
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A probability problem - finding where I've gone wrong
In how many ways can people get out of the elevator?Elevator stop, other approachThe probability that 7 people taking an elevator will leave it in configuration 3-2-1-1Probability of an elevator rising to a certain floor at most and exactlywhat's the probability that in a building of 10 floors and 5 people in elevator, the elevator would reach exactly until the 5th floor and no higher?Calculate the expected value of the highest floor the elevator may reach.Elevator probability calculationPeople leaving an elevator on different floorsProbability question - the denominator in a specific problemIn how many ways can $7$ people take elevator to $5$th floor such that at the last floor all remaining people exit (at least one remains)?
$begingroup$
There are 7 people in an elevator and 5 floors. What is the probability of at least one person exiting each floor?
I've split this into two cases and my solution was:
$$frac7choose3 5*4! + 7choose 255choose 24*3! 5^7$$
Which gives a result $0.37632$. The solution I'm provided with is $0.215$.
I assume I've gone wrong with:
$$7choose 255choose 24*3!$$
But I can't see why exactly.
I would appreciate if someone could point me to the mistake here. Thank you!
probability combinatorics
edited Mar 16 at 13:27
YuiTo Cheng
2,1192837
asked Mar 15 at 16:52
batbat
82
$endgroup$
add a comment |
$begingroup$
There are 7 people in an elevator and 5 floors. What is the probability of at least one person exiting each floor?
I've split this into two cases and my solution was:
$$frac7choose3 5*4! + 7choose 255choose 24*3! 5^7$$
Which gives a result $0.37632$. The solution I'm provided with is $0.215$.
I assume I've gone wrong with:
$$7choose 255choose 24*3!$$
But I can't see why exactly.
I would appreciate if someone could point me to the mistake here. Thank you!
probability combinatorics
edited Mar 16 at 13:27
YuiTo Cheng
2,1192837
asked Mar 15 at 16:52
batbat
82
$endgroup$
$begingroup$
@MohammadZuhairKhan Your comment does not refer to the question.
$endgroup$
– callculus
Mar 15 at 16:59
add a comment |
$begingroup$
There are 7 people in an elevator and 5 floors. What is the probability of at least one person exiting each floor?
I've split this into two cases and my solution was:
$$frac7choose3 5*4! + 7choose 255choose 24*3! 5^7$$
Which gives a result $0.37632$. The solution I'm provided with is $0.215$.
I assume I've gone wrong with:
$$7choose 255choose 24*3!$$
But I can't see why exactly.
I would appreciate if someone could point me to the mistake here. Thank you!
probability combinatorics
edited Mar 16 at 13:27
YuiTo Cheng
2,1192837
asked Mar 15 at 16:52
batbat
82
$endgroup$
There are 7 people in an elevator and 5 floors. What is the probability of at least one person exiting each floor?
I've split this into two cases and my solution was:
$$frac7choose3 5*4! + 7choose 255choose 24*3! 5^7$$
Which gives a result $0.37632$. The solution I'm provided with is $0.215$.
I assume I've gone wrong with:
$$7choose 255choose 24*3!$$
But I can't see why exactly.
I would appreciate if someone could point me to the mistake here. Thank you!
probability combinatorics
probability combinatorics
edited Mar 16 at 13:27
YuiTo Cheng
2,1192837
asked Mar 15 at 16:52
batbat
82
edited Mar 16 at 13:27
YuiTo Cheng
2,1192837
asked Mar 15 at 16:52
batbat
82
edited Mar 16 at 13:27
YuiTo Cheng
2,1192837
edited Mar 16 at 13:27
YuiTo Cheng
2,1192837
edited Mar 16 at 13:27
YuiTo Cheng
2,1192837
2,1192837
asked Mar 15 at 16:52
batbat
82
asked Mar 15 at 16:52
batbat
82
asked Mar 15 at 16:52
batbat
82
82
$begingroup$
@MohammadZuhairKhan Your comment does not refer to the question.
$endgroup$
– callculus
Mar 15 at 16:59
add a comment |
$begingroup$
@MohammadZuhairKhan Your comment does not refer to the question.
$endgroup$
– callculus
Mar 15 at 16:59
$begingroup$
@MohammadZuhairKhan Your comment does not refer to the question.
$endgroup$
– callculus
Mar 15 at 16:59
$begingroup$
@MohammadZuhairKhan Your comment does not refer to the question.
$endgroup$
– callculus
Mar 15 at 16:59
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
In your second term, you are overcounting by a factor of two. The $binom72$ chooses one of the pairs, and $binom52$ chooses the other pair. You then assign the first pair's floor in $5$ ways, and the seccond pair's floor in $4$ ways. But it does not matter which pair was chosen first or second, so you need to divide by $2$.
Another way to see it; choose the two floors that have two people get off in $binom52=frac5cdot 42$ ways. Then choose the people who get off on the higher floor in $binom72$ ways, and choose the people who get off on the lower floor in $binom52$ ways. Your mistake was having $5cdot 4$ instead of $binom52=frac5cdot 42$.
answered Mar 15 at 17:05
Mike EarnestMike Earnest
25.6k22151
$endgroup$
add a comment |
$begingroup$
It is useful to check with smaller numbers: assume $5$ people ($A,B,C,D,E$) get off on $3$ floors.
The two cases are: 1) $3,1,1$ and 2) $2,2,1$.
The first case: select three people in $5choose 3$ ways and distribute three units (triple, single, single) in $3!$ waysL
$$5choose 3cdot 3!$$
The second case: select two people in $5choose 2$ ways, select next two people in $3choose 2$ ways, however, here it is double counted:
$$beginarrayc
colorred(AB)(CD)E & (BC)(AD)E & colorred(CD)(AB)E & colorgreen(DE)(AB)C\
colorblue(AB)(CE)D & (BC)(AE)D & (CD)(AE)B & (DE)(AC)B\
colorgreen(AB)(DE)C & (BC)(DE)A & (CD)(BE)A & (DE)(BC)A\
(AC)(BD)E & (BD)(AC)E & (CE)(BD)A & \
(AC)(BE)D & (BD)(AE)C & colorblue(CE)(AB)D & \
(AC)(DE)B & (BD)(CE)A & (CE)(AD)B & \
(AD)(BC)E & (BE)(AC)D & & \
(AD)(BE)C & (BE)(AD)C & & \
(AD)(CE)B & (BE)(CD)A & & \
(AE)(BC)D & & & \
(AE)(BD)C & & & \
(AE)(CD)B & & & \
endarray$$
Note: Only $3$ pairs are highlighted, whereas there are $15$ pairs in total.
Hence, in the second case:
$$frac5choose 23choose 22cdot 3!.$$
answered Mar 16 at 13:23
farruhotafarruhota
21.5k2842
$endgroup$
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
votes
active
oldest
votes
active
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votes
$begingroup$
In your second term, you are overcounting by a factor of two. The $binom72$ chooses one of the pairs, and $binom52$ chooses the other pair. You then assign the first pair's floor in $5$ ways, and the seccond pair's floor in $4$ ways. But it does not matter which pair was chosen first or second, so you need to divide by $2$.
Another way to see it; choose the two floors that have two people get off in $binom52=frac5cdot 42$ ways. Then choose the people who get off on the higher floor in $binom72$ ways, and choose the people who get off on the lower floor in $binom52$ ways. Your mistake was having $5cdot 4$ instead of $binom52=frac5cdot 42$.
answered Mar 15 at 17:05
Mike EarnestMike Earnest
25.6k22151
$endgroup$
add a comment |
$begingroup$
In your second term, you are overcounting by a factor of two. The $binom72$ chooses one of the pairs, and $binom52$ chooses the other pair. You then assign the first pair's floor in $5$ ways, and the seccond pair's floor in $4$ ways. But it does not matter which pair was chosen first or second, so you need to divide by $2$.
Another way to see it; choose the two floors that have two people get off in $binom52=frac5cdot 42$ ways. Then choose the people who get off on the higher floor in $binom72$ ways, and choose the people who get off on the lower floor in $binom52$ ways. Your mistake was having $5cdot 4$ instead of $binom52=frac5cdot 42$.
answered Mar 15 at 17:05
Mike EarnestMike Earnest
25.6k22151
$endgroup$
add a comment |
$begingroup$
In your second term, you are overcounting by a factor of two. The $binom72$ chooses one of the pairs, and $binom52$ chooses the other pair. You then assign the first pair's floor in $5$ ways, and the seccond pair's floor in $4$ ways. But it does not matter which pair was chosen first or second, so you need to divide by $2$.
Another way to see it; choose the two floors that have two people get off in $binom52=frac5cdot 42$ ways. Then choose the people who get off on the higher floor in $binom72$ ways, and choose the people who get off on the lower floor in $binom52$ ways. Your mistake was having $5cdot 4$ instead of $binom52=frac5cdot 42$.
answered Mar 15 at 17:05
Mike EarnestMike Earnest
25.6k22151
$endgroup$
In your second term, you are overcounting by a factor of two. The $binom72$ chooses one of the pairs, and $binom52$ chooses the other pair. You then assign the first pair's floor in $5$ ways, and the seccond pair's floor in $4$ ways. But it does not matter which pair was chosen first or second, so you need to divide by $2$.
Another way to see it; choose the two floors that have two people get off in $binom52=frac5cdot 42$ ways. Then choose the people who get off on the higher floor in $binom72$ ways, and choose the people who get off on the lower floor in $binom52$ ways. Your mistake was having $5cdot 4$ instead of $binom52=frac5cdot 42$.
answered Mar 15 at 17:05
Mike EarnestMike Earnest
25.6k22151
answered Mar 15 at 17:05
Mike EarnestMike Earnest
25.6k22151
answered Mar 15 at 17:05
Mike EarnestMike Earnest
25.6k22151
answered Mar 15 at 17:05
Mike EarnestMike Earnest
25.6k22151
25.6k22151
add a comment |
add a comment |
$begingroup$
It is useful to check with smaller numbers: assume $5$ people ($A,B,C,D,E$) get off on $3$ floors.
The two cases are: 1) $3,1,1$ and 2) $2,2,1$.
The first case: select three people in $5choose 3$ ways and distribute three units (triple, single, single) in $3!$ waysL
$$5choose 3cdot 3!$$
The second case: select two people in $5choose 2$ ways, select next two people in $3choose 2$ ways, however, here it is double counted:
$$beginarrayc
colorred(AB)(CD)E & (BC)(AD)E & colorred(CD)(AB)E & colorgreen(DE)(AB)C\
colorblue(AB)(CE)D & (BC)(AE)D & (CD)(AE)B & (DE)(AC)B\
colorgreen(AB)(DE)C & (BC)(DE)A & (CD)(BE)A & (DE)(BC)A\
(AC)(BD)E & (BD)(AC)E & (CE)(BD)A & \
(AC)(BE)D & (BD)(AE)C & colorblue(CE)(AB)D & \
(AC)(DE)B & (BD)(CE)A & (CE)(AD)B & \
(AD)(BC)E & (BE)(AC)D & & \
(AD)(BE)C & (BE)(AD)C & & \
(AD)(CE)B & (BE)(CD)A & & \
(AE)(BC)D & & & \
(AE)(BD)C & & & \
(AE)(CD)B & & & \
endarray$$
Note: Only $3$ pairs are highlighted, whereas there are $15$ pairs in total.
Hence, in the second case:
$$frac5choose 23choose 22cdot 3!.$$
answered Mar 16 at 13:23
farruhotafarruhota
21.5k2842
$endgroup$
add a comment |
$begingroup$
It is useful to check with smaller numbers: assume $5$ people ($A,B,C,D,E$) get off on $3$ floors.
The two cases are: 1) $3,1,1$ and 2) $2,2,1$.
The first case: select three people in $5choose 3$ ways and distribute three units (triple, single, single) in $3!$ waysL
$$5choose 3cdot 3!$$
The second case: select two people in $5choose 2$ ways, select next two people in $3choose 2$ ways, however, here it is double counted:
$$beginarrayc
colorred(AB)(CD)E & (BC)(AD)E & colorred(CD)(AB)E & colorgreen(DE)(AB)C\
colorblue(AB)(CE)D & (BC)(AE)D & (CD)(AE)B & (DE)(AC)B\
colorgreen(AB)(DE)C & (BC)(DE)A & (CD)(BE)A & (DE)(BC)A\
(AC)(BD)E & (BD)(AC)E & (CE)(BD)A & \
(AC)(BE)D & (BD)(AE)C & colorblue(CE)(AB)D & \
(AC)(DE)B & (BD)(CE)A & (CE)(AD)B & \
(AD)(BC)E & (BE)(AC)D & & \
(AD)(BE)C & (BE)(AD)C & & \
(AD)(CE)B & (BE)(CD)A & & \
(AE)(BC)D & & & \
(AE)(BD)C & & & \
(AE)(CD)B & & & \
endarray$$
Note: Only $3$ pairs are highlighted, whereas there are $15$ pairs in total.
Hence, in the second case:
$$frac5choose 23choose 22cdot 3!.$$
answered Mar 16 at 13:23
farruhotafarruhota
21.5k2842
$endgroup$
add a comment |
$begingroup$
It is useful to check with smaller numbers: assume $5$ people ($A,B,C,D,E$) get off on $3$ floors.
The two cases are: 1) $3,1,1$ and 2) $2,2,1$.
The first case: select three people in $5choose 3$ ways and distribute three units (triple, single, single) in $3!$ waysL
$$5choose 3cdot 3!$$
The second case: select two people in $5choose 2$ ways, select next two people in $3choose 2$ ways, however, here it is double counted:
$$beginarrayc
colorred(AB)(CD)E & (BC)(AD)E & colorred(CD)(AB)E & colorgreen(DE)(AB)C\
colorblue(AB)(CE)D & (BC)(AE)D & (CD)(AE)B & (DE)(AC)B\
colorgreen(AB)(DE)C & (BC)(DE)A & (CD)(BE)A & (DE)(BC)A\
(AC)(BD)E & (BD)(AC)E & (CE)(BD)A & \
(AC)(BE)D & (BD)(AE)C & colorblue(CE)(AB)D & \
(AC)(DE)B & (BD)(CE)A & (CE)(AD)B & \
(AD)(BC)E & (BE)(AC)D & & \
(AD)(BE)C & (BE)(AD)C & & \
(AD)(CE)B & (BE)(CD)A & & \
(AE)(BC)D & & & \
(AE)(BD)C & & & \
(AE)(CD)B & & & \
endarray$$
Note: Only $3$ pairs are highlighted, whereas there are $15$ pairs in total.
Hence, in the second case:
$$frac5choose 23choose 22cdot 3!.$$
answered Mar 16 at 13:23
farruhotafarruhota
21.5k2842
$endgroup$
It is useful to check with smaller numbers: assume $5$ people ($A,B,C,D,E$) get off on $3$ floors.
The two cases are: 1) $3,1,1$ and 2) $2,2,1$.
The first case: select three people in $5choose 3$ ways and distribute three units (triple, single, single) in $3!$ waysL
$$5choose 3cdot 3!$$
The second case: select two people in $5choose 2$ ways, select next two people in $3choose 2$ ways, however, here it is double counted:
$$beginarrayc
colorred(AB)(CD)E & (BC)(AD)E & colorred(CD)(AB)E & colorgreen(DE)(AB)C\
colorblue(AB)(CE)D & (BC)(AE)D & (CD)(AE)B & (DE)(AC)B\
colorgreen(AB)(DE)C & (BC)(DE)A & (CD)(BE)A & (DE)(BC)A\
(AC)(BD)E & (BD)(AC)E & (CE)(BD)A & \
(AC)(BE)D & (BD)(AE)C & colorblue(CE)(AB)D & \
(AC)(DE)B & (BD)(CE)A & (CE)(AD)B & \
(AD)(BC)E & (BE)(AC)D & & \
(AD)(BE)C & (BE)(AD)C & & \
(AD)(CE)B & (BE)(CD)A & & \
(AE)(BC)D & & & \
(AE)(BD)C & & & \
(AE)(CD)B & & & \
endarray$$
Note: Only $3$ pairs are highlighted, whereas there are $15$ pairs in total.
Hence, in the second case:
$$frac5choose 23choose 22cdot 3!.$$
answered Mar 16 at 13:23
farruhotafarruhota
21.5k2842
answered Mar 16 at 13:23
farruhotafarruhota
21.5k2842
answered Mar 16 at 13:23
farruhotafarruhota
21.5k2842
answered Mar 16 at 13:23
farruhotafarruhota
21.5k2842
21.5k2842
add a comment |
add a comment |
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$begingroup$
@MohammadZuhairKhan Your comment does not refer to the question.
$endgroup$
– callculus
Mar 15 at 16:59