A fair coin is flipped by Bill $k$ times and by Mary $n-k$ times. Find the probability that they flip the...
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A fair coin is flipped by Bill $k$ times and by Mary $n-k$ times. Find the probability that they flip the same number of heads.
I have two answers but not sure which one is correct.
a. I use hypergeometric distribution.
$$P=sum_{m=0}^{min(k,n-k)}{begin{pmatrix}k\mend{pmatrix}begin{pmatrix}n-k\mend{pmatrix}overbegin{pmatrix}n\2mend{pmatrix}}$$
b.I use product of two binomial random variable and then sum them up.
Let X be the number of heads flipped by Bill and Y for Mary.
Since Bill and Mary are two independent person,
$$P(X=m,Y=m)=P(X=m)P(Y=m)$$
$$P=sum_{m=0}^{min(k,n-k)}P(X=m)P(Y=m)$$
$$=sum_{m=0}^{min(k,n-k)}begin{pmatrix}k\mend{pmatrix}begin{pmatrix}n-k\mend{pmatrix}{1over2^n}$$
If there is a correct answer in either of them, how can I calculate then?
probability
asked Jan 2 at 2:06
Yibei HeYibei He
3139
$endgroup$
add a comment |
$begingroup$
A fair coin is flipped by Bill $k$ times and by Mary $n-k$ times. Find the probability that they flip the same number of heads.
I have two answers but not sure which one is correct.
a. I use hypergeometric distribution.
$$P=sum_{m=0}^{min(k,n-k)}{begin{pmatrix}k\mend{pmatrix}begin{pmatrix}n-k\mend{pmatrix}overbegin{pmatrix}n\2mend{pmatrix}}$$
b.I use product of two binomial random variable and then sum them up.
Let X be the number of heads flipped by Bill and Y for Mary.
Since Bill and Mary are two independent person,
$$P(X=m,Y=m)=P(X=m)P(Y=m)$$
$$P=sum_{m=0}^{min(k,n-k)}P(X=m)P(Y=m)$$
$$=sum_{m=0}^{min(k,n-k)}begin{pmatrix}k\mend{pmatrix}begin{pmatrix}n-k\mend{pmatrix}{1over2^n}$$
If there is a correct answer in either of them, how can I calculate then?
probability
asked Jan 2 at 2:06
Yibei HeYibei He
3139
$endgroup$
$begingroup$
The second one looks correct. I have no idea how to calculate it.
$endgroup$
– herb steinberg
Jan 2 at 2:54
add a comment |
$begingroup$
A fair coin is flipped by Bill $k$ times and by Mary $n-k$ times. Find the probability that they flip the same number of heads.
I have two answers but not sure which one is correct.
a. I use hypergeometric distribution.
$$P=sum_{m=0}^{min(k,n-k)}{begin{pmatrix}k\mend{pmatrix}begin{pmatrix}n-k\mend{pmatrix}overbegin{pmatrix}n\2mend{pmatrix}}$$
b.I use product of two binomial random variable and then sum them up.
Let X be the number of heads flipped by Bill and Y for Mary.
Since Bill and Mary are two independent person,
$$P(X=m,Y=m)=P(X=m)P(Y=m)$$
$$P=sum_{m=0}^{min(k,n-k)}P(X=m)P(Y=m)$$
$$=sum_{m=0}^{min(k,n-k)}begin{pmatrix}k\mend{pmatrix}begin{pmatrix}n-k\mend{pmatrix}{1over2^n}$$
If there is a correct answer in either of them, how can I calculate then?
probability
asked Jan 2 at 2:06
Yibei HeYibei He
3139
$endgroup$
A fair coin is flipped by Bill $k$ times and by Mary $n-k$ times. Find the probability that they flip the same number of heads.
I have two answers but not sure which one is correct.
a. I use hypergeometric distribution.
$$P=sum_{m=0}^{min(k,n-k)}{begin{pmatrix}k\mend{pmatrix}begin{pmatrix}n-k\mend{pmatrix}overbegin{pmatrix}n\2mend{pmatrix}}$$
b.I use product of two binomial random variable and then sum them up.
Let X be the number of heads flipped by Bill and Y for Mary.
Since Bill and Mary are two independent person,
$$P(X=m,Y=m)=P(X=m)P(Y=m)$$
$$P=sum_{m=0}^{min(k,n-k)}P(X=m)P(Y=m)$$
$$=sum_{m=0}^{min(k,n-k)}begin{pmatrix}k\mend{pmatrix}begin{pmatrix}n-k\mend{pmatrix}{1over2^n}$$
If there is a correct answer in either of them, how can I calculate then?
probability
probability
asked Jan 2 at 2:06
Yibei HeYibei He
3139
asked Jan 2 at 2:06
Yibei HeYibei He
3139
asked Jan 2 at 2:06
Yibei HeYibei He
3139
asked Jan 2 at 2:06
Yibei HeYibei He
3139
asked Jan 2 at 2:06
Yibei HeYibei He
3139
3139
$begingroup$
The second one looks correct. I have no idea how to calculate it.
$endgroup$
– herb steinberg
Jan 2 at 2:54
add a comment |
$begingroup$
The second one looks correct. I have no idea how to calculate it.
$endgroup$
– herb steinberg
Jan 2 at 2:54
$begingroup$
The second one looks correct. I have no idea how to calculate it.
$endgroup$
– herb steinberg
Jan 2 at 2:54
$begingroup$
The second one looks correct. I have no idea how to calculate it.
$endgroup$
– herb steinberg
Jan 2 at 2:54
add a comment |
1 Answer
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The second answer you gave is correct, and the logic you give behind it is valid. They are independent events so you just have to sum up the probability of them both getting m heads for each possible m.
As far as how to calculate it, there is no shortcut, you just have to sum it up. Here is some python code that will do it for you quickly though!
from scipy.special import comb
n = 10
k = 7
np.sum([comb(k,m,exact=True)*comb(n-k,m,exact=True)/2**n for m in range(min(k,n-k)+1)])
answered Jan 2 at 3:22
Erik ParkinsonErik Parkinson
1,17519
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add a comment |
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1 Answer
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1 Answer
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$begingroup$
The second answer you gave is correct, and the logic you give behind it is valid. They are independent events so you just have to sum up the probability of them both getting m heads for each possible m.
As far as how to calculate it, there is no shortcut, you just have to sum it up. Here is some python code that will do it for you quickly though!
from scipy.special import comb
n = 10
k = 7
np.sum([comb(k,m,exact=True)*comb(n-k,m,exact=True)/2**n for m in range(min(k,n-k)+1)])
answered Jan 2 at 3:22
Erik ParkinsonErik Parkinson
1,17519
$endgroup$
add a comment |
$begingroup$
The second answer you gave is correct, and the logic you give behind it is valid. They are independent events so you just have to sum up the probability of them both getting m heads for each possible m.
As far as how to calculate it, there is no shortcut, you just have to sum it up. Here is some python code that will do it for you quickly though!
from scipy.special import comb
n = 10
k = 7
np.sum([comb(k,m,exact=True)*comb(n-k,m,exact=True)/2**n for m in range(min(k,n-k)+1)])
answered Jan 2 at 3:22
Erik ParkinsonErik Parkinson
1,17519
$endgroup$
add a comment |
$begingroup$
The second answer you gave is correct, and the logic you give behind it is valid. They are independent events so you just have to sum up the probability of them both getting m heads for each possible m.
As far as how to calculate it, there is no shortcut, you just have to sum it up. Here is some python code that will do it for you quickly though!
from scipy.special import comb
n = 10
k = 7
np.sum([comb(k,m,exact=True)*comb(n-k,m,exact=True)/2**n for m in range(min(k,n-k)+1)])
answered Jan 2 at 3:22
Erik ParkinsonErik Parkinson
1,17519
$endgroup$
The second answer you gave is correct, and the logic you give behind it is valid. They are independent events so you just have to sum up the probability of them both getting m heads for each possible m.
As far as how to calculate it, there is no shortcut, you just have to sum it up. Here is some python code that will do it for you quickly though!
from scipy.special import comb
n = 10
k = 7
np.sum([comb(k,m,exact=True)*comb(n-k,m,exact=True)/2**n for m in range(min(k,n-k)+1)])
answered Jan 2 at 3:22
Erik ParkinsonErik Parkinson
1,17519
answered Jan 2 at 3:22
Erik ParkinsonErik Parkinson
1,17519
answered Jan 2 at 3:22
Erik ParkinsonErik Parkinson
1,17519
answered Jan 2 at 3:22
Erik ParkinsonErik Parkinson
1,17519
1,17519
add a comment |
add a comment |
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The second one looks correct. I have no idea how to calculate it.
$endgroup$
– herb steinberg
Jan 2 at 2:54