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You flip a coin $10$ times. How many ways can you get at least $7$ heads?
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You flip a coin $10$ times. How many ways can you get at least $7$ heads?
My answer.
$$binom1010+ binom109cdotbinom101 + binom108cdotbinom102+binom107cdotbinom103$$
You have $10$ Heads and $0$ tails $+$ $9$ Heads $cdot$ $1$ Tail $+$ $8$ Heads $cdot$ $2$ tails $+$ $7$ Heads $cdot$ $3$ tails.
The answer is $176$ though.
combinatorics combinations
edited Aug 10 at 10:00
TheSimpliFire
10.5k62053
asked Aug 9 at 23:01
user9995331
1114
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up vote
3
down vote
favorite
You flip a coin $10$ times. How many ways can you get at least $7$ heads?
My answer.
$$binom1010+ binom109cdotbinom101 + binom108cdotbinom102+binom107cdotbinom103$$
You have $10$ Heads and $0$ tails $+$ $9$ Heads $cdot$ $1$ Tail $+$ $8$ Heads $cdot$ $2$ tails $+$ $7$ Heads $cdot$ $3$ tails.
The answer is $176$ though.
combinatorics combinations
edited Aug 10 at 10:00
TheSimpliFire
10.5k62053
asked Aug 9 at 23:01
user9995331
1114
5
Your method of picking 8 heads and 2 tails does not make use of the fact that a coin is heads or tails, but not both or neither. Once you pick the 8 heads, the remaining coins are tails perforce, there is no picking still to do.
– vadim123
Aug 9 at 23:05
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up vote
3
down vote
favorite
up vote
3
down vote
favorite
You flip a coin $10$ times. How many ways can you get at least $7$ heads?
My answer.
$$binom1010+ binom109cdotbinom101 + binom108cdotbinom102+binom107cdotbinom103$$
You have $10$ Heads and $0$ tails $+$ $9$ Heads $cdot$ $1$ Tail $+$ $8$ Heads $cdot$ $2$ tails $+$ $7$ Heads $cdot$ $3$ tails.
The answer is $176$ though.
combinatorics combinations
edited Aug 10 at 10:00
TheSimpliFire
10.5k62053
asked Aug 9 at 23:01
user9995331
1114
You flip a coin $10$ times. How many ways can you get at least $7$ heads?
My answer.
$$binom1010+ binom109cdotbinom101 + binom108cdotbinom102+binom107cdotbinom103$$
You have $10$ Heads and $0$ tails $+$ $9$ Heads $cdot$ $1$ Tail $+$ $8$ Heads $cdot$ $2$ tails $+$ $7$ Heads $cdot$ $3$ tails.
The answer is $176$ though.
combinatorics combinations
edited Aug 10 at 10:00
TheSimpliFire
10.5k62053
asked Aug 9 at 23:01
user9995331
1114
edited Aug 10 at 10:00
TheSimpliFire
10.5k62053
edited Aug 10 at 10:00
TheSimpliFire
10.5k62053
edited Aug 10 at 10:00
TheSimpliFire
10.5k62053
10.5k62053
asked Aug 9 at 23:01
user9995331
1114
asked Aug 9 at 23:01
user9995331
1114
asked Aug 9 at 23:01
user9995331
1114
1114
5
Your method of picking 8 heads and 2 tails does not make use of the fact that a coin is heads or tails, but not both or neither. Once you pick the 8 heads, the remaining coins are tails perforce, there is no picking still to do.
– vadim123
Aug 9 at 23:05
add a comment |Â
5
Your method of picking 8 heads and 2 tails does not make use of the fact that a coin is heads or tails, but not both or neither. Once you pick the 8 heads, the remaining coins are tails perforce, there is no picking still to do.
– vadim123
Aug 9 at 23:05
5
5
Your method of picking 8 heads and 2 tails does not make use of the fact that a coin is heads or tails, but not both or neither. Once you pick the 8 heads, the remaining coins are tails perforce, there is no picking still to do.
– vadim123
Aug 9 at 23:05
Your method of picking 8 heads and 2 tails does not make use of the fact that a coin is heads or tails, but not both or neither. Once you pick the 8 heads, the remaining coins are tails perforce, there is no picking still to do.
– vadim123
Aug 9 at 23:05
add a comment |Â
6 Answers
6
active
oldest
votes
up vote
6
down vote
accepted
Since we need at least $7$ heads from $10$ trails
First we get $7$ heads and $3$ tails in $dbinom107$
Second we get $8$ heads and $2$ tails in $dbinom108$
Third we get $9$ heads and $1$ tail in $dbinom109$
Fourth we $10$ heads and $0$ tails in $dbinom1010$
Now total number of permutations $=dbinom107+dbinom108+dbinom109+dbinom1010=176$
answered Aug 9 at 23:02
Key Flex
1
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up vote
5
down vote
Let's look at $binom107 cdot binom103$. This counts the number of ways you can
- assign the label "heads" to seven coins, and
- assign the label "tails" to three coins.
This includes, for example, choosing the label the first seven coins as "heads" and the first three coins as "tails".
This choice of labels cannot possibly describe a result of flipping a coin ten times, because three coins have both labels and three coins have neither label.
The simple fix is to recognize that all you need to do is to count the number of ways you can
- assign the label "heads" to seven coins
Once you've done this, you can then proceed to label the three remaining coins as "tails".
In this system, the possible choices for labels are in bijective correspondence with the possible ways to get seven heads in ten coin flips.
answered Aug 9 at 23:06
Hurkyl
109k9113254
I have a fairly basic understanding or combinations and permutations. I memorized the formulas and kind of know when to differentiate the two, but don't really understand how and why you would multiply certain values like this. When you say I you want to pick 7 heads from 10 possible (coins/heads?) and get 3 tails. I thought that meant the data was 2-dimensional? and must be multiplied. Say for example is was a 3 sided die. You wanted to get 1 side 7 times. Would the answer be what I wrote then? I added another possibility to the data.
– user9995331
Aug 9 at 23:15
add a comment |Â
up vote
1
down vote
the combinations notation $10choose 7 = frac 10!7!3!$ accounts for the fact that if you have 7 heads you also have 3 non-heads. $10choose 710choose 3$ is effectively squaring the value that you need for that term. Similar for the other terms.
answered Aug 9 at 23:06
Doug M
39.4k31749
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up vote
1
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The problem is that once you have chosen which $9 $ of the ten flips are going to be heads, there's is no choice remaining as to which one is a tail. It has to be the last one not selected.
Hence the formula starts with $ldots +10choose 9 $, times nothing (or times $1choose 1 $ if you will).
answered Aug 9 at 23:08
Arnaud Mortier
19.6k22159
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up vote
1
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The number of sequences of ten tosses that contain exactly seven heads and three tails is
$$binom107binom33 = binom107$$
since there are $binom107$ ways to select exactly seven of the ten positions for the heads and $binom33$ ways to select all three of the remaining three positions for the tails.
By similar reasoning, the number of sequences of ten tosses that contain exactly $k$ heads and $10 - k$ tails is
$$binom10kbinom10 - k10 - k = binom10k$$
since there are $binom10k$ ways to select exactly $k$ of the ten positions in the sequence for the heads and $binom10 - k10 - k$ to select all $10 - k$ of the remaining $10 - k$ positions for the tails.
Hence, the number of sequences of ten coin tosses in which at least seven heads occur is
beginalign*
sum_k = 7^10 binom10kbinom10 - k10 - k & = sum_k = 7^10 binom10k\
& = binom107 + binom108 + binom109 + binom1010\
& = 120 + 45 + 10 + 1\
& = 176
endalign*
In your attempt, when you calculated the number of ways of selecting seven heads and three tails, you first selected seven of the ten positions for the heads and then selected three of the ten positions for the tails without taking into account that the three tails can only occupy the $10 - 7 = 3$ positions not already occupied by heads. That means you allowed heads and tails to occupy the same positions in the sequence, a physical impossibility.
answered Aug 10 at 10:26
N. F. Taussig
38.8k93153
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up vote
0
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Consider thinking 10 flipping of a coin and getting n<10 heads as putting n black balls(assume tails are white balls) in 10 boxes...
Now,
the question transforms to how many ways can you put at least 7 black balls in 10 boxes?
The answer is 10c10+10c9+10c8+10c7=176
. The permutation of white balls doesn't matter as interchanging the box no of white balls counts as a same result.....
answered Aug 9 at 23:15
Sadil Khan
3437
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6 Answers
6
active
oldest
votes
6 Answers
6
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
6
down vote
accepted
Since we need at least $7$ heads from $10$ trails
First we get $7$ heads and $3$ tails in $dbinom107$
Second we get $8$ heads and $2$ tails in $dbinom108$
Third we get $9$ heads and $1$ tail in $dbinom109$
Fourth we $10$ heads and $0$ tails in $dbinom1010$
Now total number of permutations $=dbinom107+dbinom108+dbinom109+dbinom1010=176$
answered Aug 9 at 23:02
Key Flex
1
add a comment |Â
up vote
6
down vote
accepted
Since we need at least $7$ heads from $10$ trails
First we get $7$ heads and $3$ tails in $dbinom107$
Second we get $8$ heads and $2$ tails in $dbinom108$
Third we get $9$ heads and $1$ tail in $dbinom109$
Fourth we $10$ heads and $0$ tails in $dbinom1010$
Now total number of permutations $=dbinom107+dbinom108+dbinom109+dbinom1010=176$
answered Aug 9 at 23:02
Key Flex
1
add a comment |Â
up vote
6
down vote
accepted
up vote
6
down vote
accepted
Since we need at least $7$ heads from $10$ trails
First we get $7$ heads and $3$ tails in $dbinom107$
Second we get $8$ heads and $2$ tails in $dbinom108$
Third we get $9$ heads and $1$ tail in $dbinom109$
Fourth we $10$ heads and $0$ tails in $dbinom1010$
Now total number of permutations $=dbinom107+dbinom108+dbinom109+dbinom1010=176$
answered Aug 9 at 23:02
Key Flex
1
Since we need at least $7$ heads from $10$ trails
First we get $7$ heads and $3$ tails in $dbinom107$
Second we get $8$ heads and $2$ tails in $dbinom108$
Third we get $9$ heads and $1$ tail in $dbinom109$
Fourth we $10$ heads and $0$ tails in $dbinom1010$
Now total number of permutations $=dbinom107+dbinom108+dbinom109+dbinom1010=176$
answered Aug 9 at 23:02
Key Flex
1
edited Aug 9 at 23:18
answered Aug 9 at 23:02
Key Flex
1
answered Aug 9 at 23:02
Key Flex
1
answered Aug 9 at 23:02
Key Flex
1
1
add a comment |Â
add a comment |Â
up vote
5
down vote
Let's look at $binom107 cdot binom103$. This counts the number of ways you can
- assign the label "heads" to seven coins, and
- assign the label "tails" to three coins.
This includes, for example, choosing the label the first seven coins as "heads" and the first three coins as "tails".
This choice of labels cannot possibly describe a result of flipping a coin ten times, because three coins have both labels and three coins have neither label.
The simple fix is to recognize that all you need to do is to count the number of ways you can
- assign the label "heads" to seven coins
Once you've done this, you can then proceed to label the three remaining coins as "tails".
In this system, the possible choices for labels are in bijective correspondence with the possible ways to get seven heads in ten coin flips.
answered Aug 9 at 23:06
Hurkyl
109k9113254
I have a fairly basic understanding or combinations and permutations. I memorized the formulas and kind of know when to differentiate the two, but don't really understand how and why you would multiply certain values like this. When you say I you want to pick 7 heads from 10 possible (coins/heads?) and get 3 tails. I thought that meant the data was 2-dimensional? and must be multiplied. Say for example is was a 3 sided die. You wanted to get 1 side 7 times. Would the answer be what I wrote then? I added another possibility to the data.
– user9995331
Aug 9 at 23:15
add a comment |Â
up vote
5
down vote
Let's look at $binom107 cdot binom103$. This counts the number of ways you can
- assign the label "heads" to seven coins, and
- assign the label "tails" to three coins.
This includes, for example, choosing the label the first seven coins as "heads" and the first three coins as "tails".
This choice of labels cannot possibly describe a result of flipping a coin ten times, because three coins have both labels and three coins have neither label.
The simple fix is to recognize that all you need to do is to count the number of ways you can
- assign the label "heads" to seven coins
Once you've done this, you can then proceed to label the three remaining coins as "tails".
In this system, the possible choices for labels are in bijective correspondence with the possible ways to get seven heads in ten coin flips.
answered Aug 9 at 23:06
Hurkyl
109k9113254
I have a fairly basic understanding or combinations and permutations. I memorized the formulas and kind of know when to differentiate the two, but don't really understand how and why you would multiply certain values like this. When you say I you want to pick 7 heads from 10 possible (coins/heads?) and get 3 tails. I thought that meant the data was 2-dimensional? and must be multiplied. Say for example is was a 3 sided die. You wanted to get 1 side 7 times. Would the answer be what I wrote then? I added another possibility to the data.
– user9995331
Aug 9 at 23:15
add a comment |Â
up vote
5
down vote
up vote
5
down vote
Let's look at $binom107 cdot binom103$. This counts the number of ways you can
- assign the label "heads" to seven coins, and
- assign the label "tails" to three coins.
This includes, for example, choosing the label the first seven coins as "heads" and the first three coins as "tails".
This choice of labels cannot possibly describe a result of flipping a coin ten times, because three coins have both labels and three coins have neither label.
The simple fix is to recognize that all you need to do is to count the number of ways you can
- assign the label "heads" to seven coins
Once you've done this, you can then proceed to label the three remaining coins as "tails".
In this system, the possible choices for labels are in bijective correspondence with the possible ways to get seven heads in ten coin flips.
answered Aug 9 at 23:06
Hurkyl
109k9113254
Let's look at $binom107 cdot binom103$. This counts the number of ways you can
- assign the label "heads" to seven coins, and
- assign the label "tails" to three coins.
This includes, for example, choosing the label the first seven coins as "heads" and the first three coins as "tails".
This choice of labels cannot possibly describe a result of flipping a coin ten times, because three coins have both labels and three coins have neither label.
The simple fix is to recognize that all you need to do is to count the number of ways you can
- assign the label "heads" to seven coins
Once you've done this, you can then proceed to label the three remaining coins as "tails".
In this system, the possible choices for labels are in bijective correspondence with the possible ways to get seven heads in ten coin flips.
answered Aug 9 at 23:06
Hurkyl
109k9113254
answered Aug 9 at 23:06
Hurkyl
109k9113254
answered Aug 9 at 23:06
Hurkyl
109k9113254
answered Aug 9 at 23:06
Hurkyl
109k9113254
109k9113254
I have a fairly basic understanding or combinations and permutations. I memorized the formulas and kind of know when to differentiate the two, but don't really understand how and why you would multiply certain values like this. When you say I you want to pick 7 heads from 10 possible (coins/heads?) and get 3 tails. I thought that meant the data was 2-dimensional? and must be multiplied. Say for example is was a 3 sided die. You wanted to get 1 side 7 times. Would the answer be what I wrote then? I added another possibility to the data.
– user9995331
Aug 9 at 23:15
add a comment |Â
I have a fairly basic understanding or combinations and permutations. I memorized the formulas and kind of know when to differentiate the two, but don't really understand how and why you would multiply certain values like this. When you say I you want to pick 7 heads from 10 possible (coins/heads?) and get 3 tails. I thought that meant the data was 2-dimensional? and must be multiplied. Say for example is was a 3 sided die. You wanted to get 1 side 7 times. Would the answer be what I wrote then? I added another possibility to the data.
– user9995331
Aug 9 at 23:15
I have a fairly basic understanding or combinations and permutations. I memorized the formulas and kind of know when to differentiate the two, but don't really understand how and why you would multiply certain values like this. When you say I you want to pick 7 heads from 10 possible (coins/heads?) and get 3 tails. I thought that meant the data was 2-dimensional? and must be multiplied. Say for example is was a 3 sided die. You wanted to get 1 side 7 times. Would the answer be what I wrote then? I added another possibility to the data.
– user9995331
Aug 9 at 23:15
I have a fairly basic understanding or combinations and permutations. I memorized the formulas and kind of know when to differentiate the two, but don't really understand how and why you would multiply certain values like this. When you say I you want to pick 7 heads from 10 possible (coins/heads?) and get 3 tails. I thought that meant the data was 2-dimensional? and must be multiplied. Say for example is was a 3 sided die. You wanted to get 1 side 7 times. Would the answer be what I wrote then? I added another possibility to the data.
– user9995331
Aug 9 at 23:15
add a comment |Â
up vote
1
down vote
the combinations notation $10choose 7 = frac 10!7!3!$ accounts for the fact that if you have 7 heads you also have 3 non-heads. $10choose 710choose 3$ is effectively squaring the value that you need for that term. Similar for the other terms.
answered Aug 9 at 23:06
Doug M
39.4k31749
add a comment |Â
up vote
1
down vote
the combinations notation $10choose 7 = frac 10!7!3!$ accounts for the fact that if you have 7 heads you also have 3 non-heads. $10choose 710choose 3$ is effectively squaring the value that you need for that term. Similar for the other terms.
answered Aug 9 at 23:06
Doug M
39.4k31749
add a comment |Â
up vote
1
down vote
up vote
1
down vote
the combinations notation $10choose 7 = frac 10!7!3!$ accounts for the fact that if you have 7 heads you also have 3 non-heads. $10choose 710choose 3$ is effectively squaring the value that you need for that term. Similar for the other terms.
answered Aug 9 at 23:06
Doug M
39.4k31749
the combinations notation $10choose 7 = frac 10!7!3!$ accounts for the fact that if you have 7 heads you also have 3 non-heads. $10choose 710choose 3$ is effectively squaring the value that you need for that term. Similar for the other terms.
answered Aug 9 at 23:06
Doug M
39.4k31749
answered Aug 9 at 23:06
Doug M
39.4k31749
answered Aug 9 at 23:06
Doug M
39.4k31749
answered Aug 9 at 23:06
Doug M
39.4k31749
39.4k31749
add a comment |Â
add a comment |Â
up vote
1
down vote
The problem is that once you have chosen which $9 $ of the ten flips are going to be heads, there's is no choice remaining as to which one is a tail. It has to be the last one not selected.
Hence the formula starts with $ldots +10choose 9 $, times nothing (or times $1choose 1 $ if you will).
answered Aug 9 at 23:08
Arnaud Mortier
19.6k22159
add a comment |Â
up vote
1
down vote
The problem is that once you have chosen which $9 $ of the ten flips are going to be heads, there's is no choice remaining as to which one is a tail. It has to be the last one not selected.
Hence the formula starts with $ldots +10choose 9 $, times nothing (or times $1choose 1 $ if you will).
answered Aug 9 at 23:08
Arnaud Mortier
19.6k22159
add a comment |Â
up vote
1
down vote
up vote
1
down vote
The problem is that once you have chosen which $9 $ of the ten flips are going to be heads, there's is no choice remaining as to which one is a tail. It has to be the last one not selected.
Hence the formula starts with $ldots +10choose 9 $, times nothing (or times $1choose 1 $ if you will).
answered Aug 9 at 23:08
Arnaud Mortier
19.6k22159
The problem is that once you have chosen which $9 $ of the ten flips are going to be heads, there's is no choice remaining as to which one is a tail. It has to be the last one not selected.
Hence the formula starts with $ldots +10choose 9 $, times nothing (or times $1choose 1 $ if you will).
answered Aug 9 at 23:08
Arnaud Mortier
19.6k22159
answered Aug 9 at 23:08
Arnaud Mortier
19.6k22159
answered Aug 9 at 23:08
Arnaud Mortier
19.6k22159
answered Aug 9 at 23:08
Arnaud Mortier
19.6k22159
19.6k22159
add a comment |Â
add a comment |Â
up vote
1
down vote
The number of sequences of ten tosses that contain exactly seven heads and three tails is
$$binom107binom33 = binom107$$
since there are $binom107$ ways to select exactly seven of the ten positions for the heads and $binom33$ ways to select all three of the remaining three positions for the tails.
By similar reasoning, the number of sequences of ten tosses that contain exactly $k$ heads and $10 - k$ tails is
$$binom10kbinom10 - k10 - k = binom10k$$
since there are $binom10k$ ways to select exactly $k$ of the ten positions in the sequence for the heads and $binom10 - k10 - k$ to select all $10 - k$ of the remaining $10 - k$ positions for the tails.
Hence, the number of sequences of ten coin tosses in which at least seven heads occur is
beginalign*
sum_k = 7^10 binom10kbinom10 - k10 - k & = sum_k = 7^10 binom10k\
& = binom107 + binom108 + binom109 + binom1010\
& = 120 + 45 + 10 + 1\
& = 176
endalign*
In your attempt, when you calculated the number of ways of selecting seven heads and three tails, you first selected seven of the ten positions for the heads and then selected three of the ten positions for the tails without taking into account that the three tails can only occupy the $10 - 7 = 3$ positions not already occupied by heads. That means you allowed heads and tails to occupy the same positions in the sequence, a physical impossibility.
answered Aug 10 at 10:26
N. F. Taussig
38.8k93153
add a comment |Â
up vote
1
down vote
The number of sequences of ten tosses that contain exactly seven heads and three tails is
$$binom107binom33 = binom107$$
since there are $binom107$ ways to select exactly seven of the ten positions for the heads and $binom33$ ways to select all three of the remaining three positions for the tails.
By similar reasoning, the number of sequences of ten tosses that contain exactly $k$ heads and $10 - k$ tails is
$$binom10kbinom10 - k10 - k = binom10k$$
since there are $binom10k$ ways to select exactly $k$ of the ten positions in the sequence for the heads and $binom10 - k10 - k$ to select all $10 - k$ of the remaining $10 - k$ positions for the tails.
Hence, the number of sequences of ten coin tosses in which at least seven heads occur is
beginalign*
sum_k = 7^10 binom10kbinom10 - k10 - k & = sum_k = 7^10 binom10k\
& = binom107 + binom108 + binom109 + binom1010\
& = 120 + 45 + 10 + 1\
& = 176
endalign*
In your attempt, when you calculated the number of ways of selecting seven heads and three tails, you first selected seven of the ten positions for the heads and then selected three of the ten positions for the tails without taking into account that the three tails can only occupy the $10 - 7 = 3$ positions not already occupied by heads. That means you allowed heads and tails to occupy the same positions in the sequence, a physical impossibility.
answered Aug 10 at 10:26
N. F. Taussig
38.8k93153
add a comment |Â
up vote
1
down vote
up vote
1
down vote
The number of sequences of ten tosses that contain exactly seven heads and three tails is
$$binom107binom33 = binom107$$
since there are $binom107$ ways to select exactly seven of the ten positions for the heads and $binom33$ ways to select all three of the remaining three positions for the tails.
By similar reasoning, the number of sequences of ten tosses that contain exactly $k$ heads and $10 - k$ tails is
$$binom10kbinom10 - k10 - k = binom10k$$
since there are $binom10k$ ways to select exactly $k$ of the ten positions in the sequence for the heads and $binom10 - k10 - k$ to select all $10 - k$ of the remaining $10 - k$ positions for the tails.
Hence, the number of sequences of ten coin tosses in which at least seven heads occur is
beginalign*
sum_k = 7^10 binom10kbinom10 - k10 - k & = sum_k = 7^10 binom10k\
& = binom107 + binom108 + binom109 + binom1010\
& = 120 + 45 + 10 + 1\
& = 176
endalign*
In your attempt, when you calculated the number of ways of selecting seven heads and three tails, you first selected seven of the ten positions for the heads and then selected three of the ten positions for the tails without taking into account that the three tails can only occupy the $10 - 7 = 3$ positions not already occupied by heads. That means you allowed heads and tails to occupy the same positions in the sequence, a physical impossibility.
answered Aug 10 at 10:26
N. F. Taussig
38.8k93153
The number of sequences of ten tosses that contain exactly seven heads and three tails is
$$binom107binom33 = binom107$$
since there are $binom107$ ways to select exactly seven of the ten positions for the heads and $binom33$ ways to select all three of the remaining three positions for the tails.
By similar reasoning, the number of sequences of ten tosses that contain exactly $k$ heads and $10 - k$ tails is
$$binom10kbinom10 - k10 - k = binom10k$$
since there are $binom10k$ ways to select exactly $k$ of the ten positions in the sequence for the heads and $binom10 - k10 - k$ to select all $10 - k$ of the remaining $10 - k$ positions for the tails.
Hence, the number of sequences of ten coin tosses in which at least seven heads occur is
beginalign*
sum_k = 7^10 binom10kbinom10 - k10 - k & = sum_k = 7^10 binom10k\
& = binom107 + binom108 + binom109 + binom1010\
& = 120 + 45 + 10 + 1\
& = 176
endalign*
In your attempt, when you calculated the number of ways of selecting seven heads and three tails, you first selected seven of the ten positions for the heads and then selected three of the ten positions for the tails without taking into account that the three tails can only occupy the $10 - 7 = 3$ positions not already occupied by heads. That means you allowed heads and tails to occupy the same positions in the sequence, a physical impossibility.
answered Aug 10 at 10:26
N. F. Taussig
38.8k93153
answered Aug 10 at 10:26
N. F. Taussig
38.8k93153
answered Aug 10 at 10:26
N. F. Taussig
38.8k93153
answered Aug 10 at 10:26
N. F. Taussig
38.8k93153
38.8k93153
add a comment |Â
add a comment |Â
up vote
0
down vote
Consider thinking 10 flipping of a coin and getting n<10 heads as putting n black balls(assume tails are white balls) in 10 boxes...
Now,
the question transforms to how many ways can you put at least 7 black balls in 10 boxes?
The answer is 10c10+10c9+10c8+10c7=176
. The permutation of white balls doesn't matter as interchanging the box no of white balls counts as a same result.....
answered Aug 9 at 23:15
Sadil Khan
3437
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Consider thinking 10 flipping of a coin and getting n<10 heads as putting n black balls(assume tails are white balls) in 10 boxes...
Now,
the question transforms to how many ways can you put at least 7 black balls in 10 boxes?
The answer is 10c10+10c9+10c8+10c7=176
. The permutation of white balls doesn't matter as interchanging the box no of white balls counts as a same result.....
answered Aug 9 at 23:15
Sadil Khan
3437
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Consider thinking 10 flipping of a coin and getting n<10 heads as putting n black balls(assume tails are white balls) in 10 boxes...
Now,
the question transforms to how many ways can you put at least 7 black balls in 10 boxes?
The answer is 10c10+10c9+10c8+10c7=176
. The permutation of white balls doesn't matter as interchanging the box no of white balls counts as a same result.....
answered Aug 9 at 23:15
Sadil Khan
3437
Consider thinking 10 flipping of a coin and getting n<10 heads as putting n black balls(assume tails are white balls) in 10 boxes...
Now,
the question transforms to how many ways can you put at least 7 black balls in 10 boxes?
The answer is 10c10+10c9+10c8+10c7=176
. The permutation of white balls doesn't matter as interchanging the box no of white balls counts as a same result.....
answered Aug 9 at 23:15
Sadil Khan
3437
edited Aug 10 at 0:07
answered Aug 9 at 23:15
Sadil Khan
3437
answered Aug 9 at 23:15
Sadil Khan
3437
answered Aug 9 at 23:15
Sadil Khan
3437
3437
add a comment |Â
add a comment |Â
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5
Your method of picking 8 heads and 2 tails does not make use of the fact that a coin is heads or tails, but not both or neither. Once you pick the 8 heads, the remaining coins are tails perforce, there is no picking still to do.
– vadim123
Aug 9 at 23:05