Conditional probability -Basketball
$begingroup$
How does the number of shots made by a player in a game get distribute if we are given that the number of shots missed equal to the number of shots scored?
data:
P(score)=$0.83$, P(miss)=$0.17$
no. of shots in a game $Zsim Poi(6)$
game duration: 48 min
$underline{textrm{My work so far:}}$
from what is given Y=no.of misses=no. of hits=X
so $Xsim Poi(6cdot 0.83=4.98)$ and $Ysim Poi(1.02)$
$$P(X|X=Y)=P(X=k,Y=k)/P(X=Y)= frac{P(X=k)P(Y=k)}{sumlimits_{n=0}^{infty} P(X=n)cdot P(Y=n)}$$
The problem is that i dont sure if $P(X=Y)=sumlimits_{n=0}^{infty} P(X=n)cdot P(Y=n)$.
probability conditional-probability
edited Jan 7 at 17:09
callculus
18.7k31428
asked Jan 7 at 12:34
tt600tt600
63
$endgroup$
|
show 4 more comments
$begingroup$
How does the number of shots made by a player in a game get distribute if we are given that the number of shots missed equal to the number of shots scored?
data:
P(score)=$0.83$, P(miss)=$0.17$
no. of shots in a game $Zsim Poi(6)$
game duration: 48 min
$underline{textrm{My work so far:}}$
from what is given Y=no.of misses=no. of hits=X
so $Xsim Poi(6cdot 0.83=4.98)$ and $Ysim Poi(1.02)$
$$P(X|X=Y)=P(X=k,Y=k)/P(X=Y)= frac{P(X=k)P(Y=k)}{sumlimits_{n=0}^{infty} P(X=n)cdot P(Y=n)}$$
The problem is that i dont sure if $P(X=Y)=sumlimits_{n=0}^{infty} P(X=n)cdot P(Y=n)$.
probability conditional-probability
edited Jan 7 at 17:09
callculus
18.7k31428
asked Jan 7 at 12:34
tt600tt600
63
$endgroup$
1
$begingroup$
Your question isn't clear. "How does the number of shots of a player in a game get divided by if we are given that the number of misses equal to the number of scores?" doesn't make sense. What data are you given? What do you wish to compute?
$endgroup$
– lulu
Jan 7 at 12:36
$begingroup$
i am sorry English isn't my first language. i want to find How does the number of shots made by a player in a game get distribute if we are given that the number of shots missed equal to the number of shots scored. the ive been given is p(score)=0.83 p(miss)=0.17 no. of shots in a game Z~poi(6) game duration: 48 min
$endgroup$
– tt600
Jan 7 at 12:49
$begingroup$
Ok...and I gather that the number of shots is assumed to be poisson with $lambda=6$? You don't actually say that anywhere, and the fact that you bring up the length of a regulation game suggests that maybe you meant something else? But assuming I have guessed correctly, then knowing that scores= misses tells us that an even number of shots were taken. For each $N$ compute the probability that exactly $2N$ shots were taken and that of those exactly $N$ were scores and $N$ were misses, then use Bayes' Theorem. The probability drops off very fast as $N$ grows.
$endgroup$
– lulu
Jan 7 at 12:54
$begingroup$
you assumed correctly- poisson with λ=6. i didnt mean any thing with game duration it is just data given. How do i do this "For each N compute the probability that exactly 2N shots were taken and that of those exactly N were scores and N were misses" that is my problam
$endgroup$
– tt600
Jan 7 at 13:07
$begingroup$
Where is the difficulty? Let's say $N=1$. Then $2N=2$. The probability that he takes exactly $2$ shots is $0.0446$. Conditioned on that, the probability that he gets exactly one score and one miss is $binom 21times .83^1times .17^1=0.2822$. Thus, for $N=1$ the value you want is $.0446times 0.2822=0.01259$. Now do this for more values of $N$. You don't need many values of $N$...the probabilities rapidly become insignificant as $N$ grows.
$endgroup$
– lulu
Jan 7 at 13:26
|
show 4 more comments
$begingroup$
How does the number of shots made by a player in a game get distribute if we are given that the number of shots missed equal to the number of shots scored?
data:
P(score)=$0.83$, P(miss)=$0.17$
no. of shots in a game $Zsim Poi(6)$
game duration: 48 min
$underline{textrm{My work so far:}}$
from what is given Y=no.of misses=no. of hits=X
so $Xsim Poi(6cdot 0.83=4.98)$ and $Ysim Poi(1.02)$
$$P(X|X=Y)=P(X=k,Y=k)/P(X=Y)= frac{P(X=k)P(Y=k)}{sumlimits_{n=0}^{infty} P(X=n)cdot P(Y=n)}$$
The problem is that i dont sure if $P(X=Y)=sumlimits_{n=0}^{infty} P(X=n)cdot P(Y=n)$.
probability conditional-probability
edited Jan 7 at 17:09
callculus
18.7k31428
asked Jan 7 at 12:34
tt600tt600
63
$endgroup$
How does the number of shots made by a player in a game get distribute if we are given that the number of shots missed equal to the number of shots scored?
data:
P(score)=$0.83$, P(miss)=$0.17$
no. of shots in a game $Zsim Poi(6)$
game duration: 48 min
$underline{textrm{My work so far:}}$
from what is given Y=no.of misses=no. of hits=X
so $Xsim Poi(6cdot 0.83=4.98)$ and $Ysim Poi(1.02)$
$$P(X|X=Y)=P(X=k,Y=k)/P(X=Y)= frac{P(X=k)P(Y=k)}{sumlimits_{n=0}^{infty} P(X=n)cdot P(Y=n)}$$
The problem is that i dont sure if $P(X=Y)=sumlimits_{n=0}^{infty} P(X=n)cdot P(Y=n)$.
probability conditional-probability
probability conditional-probability
edited Jan 7 at 17:09
callculus
18.7k31428
asked Jan 7 at 12:34
tt600tt600
63
edited Jan 7 at 17:09
callculus
18.7k31428
asked Jan 7 at 12:34
tt600tt600
63
edited Jan 7 at 17:09
callculus
18.7k31428
edited Jan 7 at 17:09
callculus
18.7k31428
edited Jan 7 at 17:09
callculus
18.7k31428
18.7k31428
asked Jan 7 at 12:34
tt600tt600
63
asked Jan 7 at 12:34
tt600tt600
63
asked Jan 7 at 12:34
tt600tt600
63
63
1
$begingroup$
Your question isn't clear. "How does the number of shots of a player in a game get divided by if we are given that the number of misses equal to the number of scores?" doesn't make sense. What data are you given? What do you wish to compute?
$endgroup$
– lulu
Jan 7 at 12:36
$begingroup$
i am sorry English isn't my first language. i want to find How does the number of shots made by a player in a game get distribute if we are given that the number of shots missed equal to the number of shots scored. the ive been given is p(score)=0.83 p(miss)=0.17 no. of shots in a game Z~poi(6) game duration: 48 min
$endgroup$
– tt600
Jan 7 at 12:49
$begingroup$
Ok...and I gather that the number of shots is assumed to be poisson with $lambda=6$? You don't actually say that anywhere, and the fact that you bring up the length of a regulation game suggests that maybe you meant something else? But assuming I have guessed correctly, then knowing that scores= misses tells us that an even number of shots were taken. For each $N$ compute the probability that exactly $2N$ shots were taken and that of those exactly $N$ were scores and $N$ were misses, then use Bayes' Theorem. The probability drops off very fast as $N$ grows.
$endgroup$
– lulu
Jan 7 at 12:54
$begingroup$
you assumed correctly- poisson with λ=6. i didnt mean any thing with game duration it is just data given. How do i do this "For each N compute the probability that exactly 2N shots were taken and that of those exactly N were scores and N were misses" that is my problam
$endgroup$
– tt600
Jan 7 at 13:07
$begingroup$
Where is the difficulty? Let's say $N=1$. Then $2N=2$. The probability that he takes exactly $2$ shots is $0.0446$. Conditioned on that, the probability that he gets exactly one score and one miss is $binom 21times .83^1times .17^1=0.2822$. Thus, for $N=1$ the value you want is $.0446times 0.2822=0.01259$. Now do this for more values of $N$. You don't need many values of $N$...the probabilities rapidly become insignificant as $N$ grows.
$endgroup$
– lulu
Jan 7 at 13:26
|
show 4 more comments
1
$begingroup$
Your question isn't clear. "How does the number of shots of a player in a game get divided by if we are given that the number of misses equal to the number of scores?" doesn't make sense. What data are you given? What do you wish to compute?
$endgroup$
– lulu
Jan 7 at 12:36
$begingroup$
i am sorry English isn't my first language. i want to find How does the number of shots made by a player in a game get distribute if we are given that the number of shots missed equal to the number of shots scored. the ive been given is p(score)=0.83 p(miss)=0.17 no. of shots in a game Z~poi(6) game duration: 48 min
$endgroup$
– tt600
Jan 7 at 12:49
$begingroup$
Ok...and I gather that the number of shots is assumed to be poisson with $lambda=6$? You don't actually say that anywhere, and the fact that you bring up the length of a regulation game suggests that maybe you meant something else? But assuming I have guessed correctly, then knowing that scores= misses tells us that an even number of shots were taken. For each $N$ compute the probability that exactly $2N$ shots were taken and that of those exactly $N$ were scores and $N$ were misses, then use Bayes' Theorem. The probability drops off very fast as $N$ grows.
$endgroup$
– lulu
Jan 7 at 12:54
$begingroup$
you assumed correctly- poisson with λ=6. i didnt mean any thing with game duration it is just data given. How do i do this "For each N compute the probability that exactly 2N shots were taken and that of those exactly N were scores and N were misses" that is my problam
$endgroup$
– tt600
Jan 7 at 13:07
$begingroup$
Where is the difficulty? Let's say $N=1$. Then $2N=2$. The probability that he takes exactly $2$ shots is $0.0446$. Conditioned on that, the probability that he gets exactly one score and one miss is $binom 21times .83^1times .17^1=0.2822$. Thus, for $N=1$ the value you want is $.0446times 0.2822=0.01259$. Now do this for more values of $N$. You don't need many values of $N$...the probabilities rapidly become insignificant as $N$ grows.
$endgroup$
– lulu
Jan 7 at 13:26
1
1
$begingroup$
Your question isn't clear. "How does the number of shots of a player in a game get divided by if we are given that the number of misses equal to the number of scores?" doesn't make sense. What data are you given? What do you wish to compute?
$endgroup$
– lulu
Jan 7 at 12:36
$begingroup$
Your question isn't clear. "How does the number of shots of a player in a game get divided by if we are given that the number of misses equal to the number of scores?" doesn't make sense. What data are you given? What do you wish to compute?
$endgroup$
– lulu
Jan 7 at 12:36
$begingroup$
i am sorry English isn't my first language. i want to find How does the number of shots made by a player in a game get distribute if we are given that the number of shots missed equal to the number of shots scored. the ive been given is p(score)=0.83 p(miss)=0.17 no. of shots in a game Z~poi(6) game duration: 48 min
$endgroup$
– tt600
Jan 7 at 12:49
$begingroup$
i am sorry English isn't my first language. i want to find How does the number of shots made by a player in a game get distribute if we are given that the number of shots missed equal to the number of shots scored. the ive been given is p(score)=0.83 p(miss)=0.17 no. of shots in a game Z~poi(6) game duration: 48 min
$endgroup$
– tt600
Jan 7 at 12:49
$begingroup$
Ok...and I gather that the number of shots is assumed to be poisson with $lambda=6$? You don't actually say that anywhere, and the fact that you bring up the length of a regulation game suggests that maybe you meant something else? But assuming I have guessed correctly, then knowing that scores= misses tells us that an even number of shots were taken. For each $N$ compute the probability that exactly $2N$ shots were taken and that of those exactly $N$ were scores and $N$ were misses, then use Bayes' Theorem. The probability drops off very fast as $N$ grows.
$endgroup$
– lulu
Jan 7 at 12:54
$begingroup$
Ok...and I gather that the number of shots is assumed to be poisson with $lambda=6$? You don't actually say that anywhere, and the fact that you bring up the length of a regulation game suggests that maybe you meant something else? But assuming I have guessed correctly, then knowing that scores= misses tells us that an even number of shots were taken. For each $N$ compute the probability that exactly $2N$ shots were taken and that of those exactly $N$ were scores and $N$ were misses, then use Bayes' Theorem. The probability drops off very fast as $N$ grows.
$endgroup$
– lulu
Jan 7 at 12:54
$begingroup$
you assumed correctly- poisson with λ=6. i didnt mean any thing with game duration it is just data given. How do i do this "For each N compute the probability that exactly 2N shots were taken and that of those exactly N were scores and N were misses" that is my problam
$endgroup$
– tt600
Jan 7 at 13:07
$begingroup$
you assumed correctly- poisson with λ=6. i didnt mean any thing with game duration it is just data given. How do i do this "For each N compute the probability that exactly 2N shots were taken and that of those exactly N were scores and N were misses" that is my problam
$endgroup$
– tt600
Jan 7 at 13:07
$begingroup$
Where is the difficulty? Let's say $N=1$. Then $2N=2$. The probability that he takes exactly $2$ shots is $0.0446$. Conditioned on that, the probability that he gets exactly one score and one miss is $binom 21times .83^1times .17^1=0.2822$. Thus, for $N=1$ the value you want is $.0446times 0.2822=0.01259$. Now do this for more values of $N$. You don't need many values of $N$...the probabilities rapidly become insignificant as $N$ grows.
$endgroup$
– lulu
Jan 7 at 13:26
$begingroup$
Where is the difficulty? Let's say $N=1$. Then $2N=2$. The probability that he takes exactly $2$ shots is $0.0446$. Conditioned on that, the probability that he gets exactly one score and one miss is $binom 21times .83^1times .17^1=0.2822$. Thus, for $N=1$ the value you want is $.0446times 0.2822=0.01259$. Now do this for more values of $N$. You don't need many values of $N$...the probabilities rapidly become insignificant as $N$ grows.
$endgroup$
– lulu
Jan 7 at 13:26
|
show 4 more comments
1 Answer
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$begingroup$
I assume that "the number of shots made by a player " refers to the total number of shots, scored and missed: $Z = X + Y$
Then we want
$$
begin{align}
P(Z =z| X = Y)&=frac{P(X = Y|Z=z) P(Z=z)}{P(X=Y)}\
tag{1}end{align}
$$
Let's compute $P(X = Y|Z=z)$ . First, we note that this is zero if $z$ is odd.
Now, assuming $z$ is even, $P(X = Y|Z=z)$ corresponds to the event of having, in $z=2t$ tries of a Bernoulli experiment with hit probability $p$, $t$ misses and $t$ scores. The probability of this event is given by a Binomial: $binom{z}{t} p^t (1-p)^t $
Then the numerator of $(1)$ is
$$ binom{z}{z/2} (p q)^{z/2} e^{-lambda} frac{lambda^z}{z!} tag{2}$$
where $p=0.83$, $q=1-p$, $lambda = 6$
(actually, it's not clear what you mean by $Z∼Poi(6)$ and "game duration=48 minutes" - it would seem that the latter is not necessary - unless the poisson parameter is actually some rate (scores per time unit) , but then the statement is confusing)
The denominator is just a normalization constant:
$$
begin{align}
P(X=Y)&=
sum_{z ,{rm even}} binom{z}{z/2} (p q)^{z/2} e^{-lambda} frac{lambda^z}{z!} \
&=e^{-lambda} sum_{t=0}^infty frac{a^t }{(t!)^2}
end{align}
tag3$$
with $a=p q sqrt{lambda}$. I'm not sure if that has a simple closed form expression.
answered Jan 7 at 21:05
leonbloyleonbloy
42.4k647108
$endgroup$
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$begingroup$
I assume that "the number of shots made by a player " refers to the total number of shots, scored and missed: $Z = X + Y$
Then we want
$$
begin{align}
P(Z =z| X = Y)&=frac{P(X = Y|Z=z) P(Z=z)}{P(X=Y)}\
tag{1}end{align}
$$
Let's compute $P(X = Y|Z=z)$ . First, we note that this is zero if $z$ is odd.
Now, assuming $z$ is even, $P(X = Y|Z=z)$ corresponds to the event of having, in $z=2t$ tries of a Bernoulli experiment with hit probability $p$, $t$ misses and $t$ scores. The probability of this event is given by a Binomial: $binom{z}{t} p^t (1-p)^t $
Then the numerator of $(1)$ is
$$ binom{z}{z/2} (p q)^{z/2} e^{-lambda} frac{lambda^z}{z!} tag{2}$$
where $p=0.83$, $q=1-p$, $lambda = 6$
(actually, it's not clear what you mean by $Z∼Poi(6)$ and "game duration=48 minutes" - it would seem that the latter is not necessary - unless the poisson parameter is actually some rate (scores per time unit) , but then the statement is confusing)
The denominator is just a normalization constant:
$$
begin{align}
P(X=Y)&=
sum_{z ,{rm even}} binom{z}{z/2} (p q)^{z/2} e^{-lambda} frac{lambda^z}{z!} \
&=e^{-lambda} sum_{t=0}^infty frac{a^t }{(t!)^2}
end{align}
tag3$$
with $a=p q sqrt{lambda}$. I'm not sure if that has a simple closed form expression.
answered Jan 7 at 21:05
leonbloyleonbloy
42.4k647108
$endgroup$
add a comment |
$begingroup$
I assume that "the number of shots made by a player " refers to the total number of shots, scored and missed: $Z = X + Y$
Then we want
$$
begin{align}
P(Z =z| X = Y)&=frac{P(X = Y|Z=z) P(Z=z)}{P(X=Y)}\
tag{1}end{align}
$$
Let's compute $P(X = Y|Z=z)$ . First, we note that this is zero if $z$ is odd.
Now, assuming $z$ is even, $P(X = Y|Z=z)$ corresponds to the event of having, in $z=2t$ tries of a Bernoulli experiment with hit probability $p$, $t$ misses and $t$ scores. The probability of this event is given by a Binomial: $binom{z}{t} p^t (1-p)^t $
Then the numerator of $(1)$ is
$$ binom{z}{z/2} (p q)^{z/2} e^{-lambda} frac{lambda^z}{z!} tag{2}$$
where $p=0.83$, $q=1-p$, $lambda = 6$
(actually, it's not clear what you mean by $Z∼Poi(6)$ and "game duration=48 minutes" - it would seem that the latter is not necessary - unless the poisson parameter is actually some rate (scores per time unit) , but then the statement is confusing)
The denominator is just a normalization constant:
$$
begin{align}
P(X=Y)&=
sum_{z ,{rm even}} binom{z}{z/2} (p q)^{z/2} e^{-lambda} frac{lambda^z}{z!} \
&=e^{-lambda} sum_{t=0}^infty frac{a^t }{(t!)^2}
end{align}
tag3$$
with $a=p q sqrt{lambda}$. I'm not sure if that has a simple closed form expression.
answered Jan 7 at 21:05
leonbloyleonbloy
42.4k647108
$endgroup$
add a comment |
$begingroup$
I assume that "the number of shots made by a player " refers to the total number of shots, scored and missed: $Z = X + Y$
Then we want
$$
begin{align}
P(Z =z| X = Y)&=frac{P(X = Y|Z=z) P(Z=z)}{P(X=Y)}\
tag{1}end{align}
$$
Let's compute $P(X = Y|Z=z)$ . First, we note that this is zero if $z$ is odd.
Now, assuming $z$ is even, $P(X = Y|Z=z)$ corresponds to the event of having, in $z=2t$ tries of a Bernoulli experiment with hit probability $p$, $t$ misses and $t$ scores. The probability of this event is given by a Binomial: $binom{z}{t} p^t (1-p)^t $
Then the numerator of $(1)$ is
$$ binom{z}{z/2} (p q)^{z/2} e^{-lambda} frac{lambda^z}{z!} tag{2}$$
where $p=0.83$, $q=1-p$, $lambda = 6$
(actually, it's not clear what you mean by $Z∼Poi(6)$ and "game duration=48 minutes" - it would seem that the latter is not necessary - unless the poisson parameter is actually some rate (scores per time unit) , but then the statement is confusing)
The denominator is just a normalization constant:
$$
begin{align}
P(X=Y)&=
sum_{z ,{rm even}} binom{z}{z/2} (p q)^{z/2} e^{-lambda} frac{lambda^z}{z!} \
&=e^{-lambda} sum_{t=0}^infty frac{a^t }{(t!)^2}
end{align}
tag3$$
with $a=p q sqrt{lambda}$. I'm not sure if that has a simple closed form expression.
answered Jan 7 at 21:05
leonbloyleonbloy
42.4k647108
$endgroup$
I assume that "the number of shots made by a player " refers to the total number of shots, scored and missed: $Z = X + Y$
Then we want
$$
begin{align}
P(Z =z| X = Y)&=frac{P(X = Y|Z=z) P(Z=z)}{P(X=Y)}\
tag{1}end{align}
$$
Let's compute $P(X = Y|Z=z)$ . First, we note that this is zero if $z$ is odd.
Now, assuming $z$ is even, $P(X = Y|Z=z)$ corresponds to the event of having, in $z=2t$ tries of a Bernoulli experiment with hit probability $p$, $t$ misses and $t$ scores. The probability of this event is given by a Binomial: $binom{z}{t} p^t (1-p)^t $
Then the numerator of $(1)$ is
$$ binom{z}{z/2} (p q)^{z/2} e^{-lambda} frac{lambda^z}{z!} tag{2}$$
where $p=0.83$, $q=1-p$, $lambda = 6$
(actually, it's not clear what you mean by $Z∼Poi(6)$ and "game duration=48 minutes" - it would seem that the latter is not necessary - unless the poisson parameter is actually some rate (scores per time unit) , but then the statement is confusing)
The denominator is just a normalization constant:
$$
begin{align}
P(X=Y)&=
sum_{z ,{rm even}} binom{z}{z/2} (p q)^{z/2} e^{-lambda} frac{lambda^z}{z!} \
&=e^{-lambda} sum_{t=0}^infty frac{a^t }{(t!)^2}
end{align}
tag3$$
with $a=p q sqrt{lambda}$. I'm not sure if that has a simple closed form expression.
answered Jan 7 at 21:05
leonbloyleonbloy
42.4k647108
edited Jan 7 at 21:49
answered Jan 7 at 21:05
leonbloyleonbloy
42.4k647108
answered Jan 7 at 21:05
leonbloyleonbloy
42.4k647108
answered Jan 7 at 21:05
leonbloyleonbloy
42.4k647108
42.4k647108
add a comment |
add a comment |
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Your question isn't clear. "How does the number of shots of a player in a game get divided by if we are given that the number of misses equal to the number of scores?" doesn't make sense. What data are you given? What do you wish to compute?
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– lulu
Jan 7 at 12:36
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i am sorry English isn't my first language. i want to find How does the number of shots made by a player in a game get distribute if we are given that the number of shots missed equal to the number of shots scored. the ive been given is p(score)=0.83 p(miss)=0.17 no. of shots in a game Z~poi(6) game duration: 48 min
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– tt600
Jan 7 at 12:49
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Ok...and I gather that the number of shots is assumed to be poisson with $lambda=6$? You don't actually say that anywhere, and the fact that you bring up the length of a regulation game suggests that maybe you meant something else? But assuming I have guessed correctly, then knowing that scores= misses tells us that an even number of shots were taken. For each $N$ compute the probability that exactly $2N$ shots were taken and that of those exactly $N$ were scores and $N$ were misses, then use Bayes' Theorem. The probability drops off very fast as $N$ grows.
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– lulu
Jan 7 at 12:54
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you assumed correctly- poisson with λ=6. i didnt mean any thing with game duration it is just data given. How do i do this "For each N compute the probability that exactly 2N shots were taken and that of those exactly N were scores and N were misses" that is my problam
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– tt600
Jan 7 at 13:07
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Where is the difficulty? Let's say $N=1$. Then $2N=2$. The probability that he takes exactly $2$ shots is $0.0446$. Conditioned on that, the probability that he gets exactly one score and one miss is $binom 21times .83^1times .17^1=0.2822$. Thus, for $N=1$ the value you want is $.0446times 0.2822=0.01259$. Now do this for more values of $N$. You don't need many values of $N$...the probabilities rapidly become insignificant as $N$ grows.
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– lulu
Jan 7 at 13:26