Mixing
Penalty score probability
0
$begingroup$
A footballer has a $0.8$ chance of scoring a penalty. How many
penalties does he need to take to score at least two times with a
$0.9$ probability?
I tried the negative binomial method ($r=2$), so that becomes $binom{m-1}{1}cdot 0.8^2cdot 0.2^{m-2}$. Or should I use hypergeometrical?
probability
edited Jan 27 at 17:32
callculus
18.6k31428
asked Jan 27 at 16:57
NiccoNicco
12
$endgroup$
add a comment |
0
$begingroup$
A footballer has a $0.8$ chance of scoring a penalty. How many
penalties does he need to take to score at least two times with a
$0.9$ probability?
I tried the negative binomial method ($r=2$), so that becomes $binom{m-1}{1}cdot 0.8^2cdot 0.2^{m-2}$. Or should I use hypergeometrical?
probability
edited Jan 27 at 17:32
callculus
18.6k31428
asked Jan 27 at 16:57
NiccoNicco
12
$endgroup$
$begingroup$
What are your thoughts? I don´t think that you get an answer if you don´t show a minimum of effort. See also here: How to ask a good question.
$endgroup$
– callculus
Jan 27 at 17:01
$begingroup$
The answer is $4$. However, if you want a more detailed answer you should put a little effort into your question.
$endgroup$
– Nathanael Skrepek
Jan 27 at 17:08
$begingroup$
No, hypergeometric distribution is not right. But why you don´t use the binomial distribution? And you can use the converse probability. $P(Xgeq 2)=1-P(X=1)-P(X=0)geq 0.9$, where $Xsim Bin(n, 0.8)$ Note that the inequality cannot be solved algebraically. Just try out some values for $n$.
$endgroup$
– callculus
Jan 27 at 17:21
add a comment |
0
0
0
$begingroup$
A footballer has a $0.8$ chance of scoring a penalty. How many
penalties does he need to take to score at least two times with a
$0.9$ probability?
I tried the negative binomial method ($r=2$), so that becomes $binom{m-1}{1}cdot 0.8^2cdot 0.2^{m-2}$. Or should I use hypergeometrical?
probability
edited Jan 27 at 17:32
callculus
18.6k31428
asked Jan 27 at 16:57
NiccoNicco
12
$endgroup$
A footballer has a $0.8$ chance of scoring a penalty. How many
penalties does he need to take to score at least two times with a
$0.9$ probability?
I tried the negative binomial method ($r=2$), so that becomes $binom{m-1}{1}cdot 0.8^2cdot 0.2^{m-2}$. Or should I use hypergeometrical?
probability
probability
edited Jan 27 at 17:32
callculus
18.6k31428
asked Jan 27 at 16:57
NiccoNicco
12
edited Jan 27 at 17:32
callculus
18.6k31428
asked Jan 27 at 16:57
NiccoNicco
12
edited Jan 27 at 17:32
callculus
18.6k31428
edited Jan 27 at 17:32
callculus
18.6k31428
edited Jan 27 at 17:32
callculus
18.6k31428
18.6k31428
asked Jan 27 at 16:57
NiccoNicco
12
asked Jan 27 at 16:57
NiccoNicco
12
asked Jan 27 at 16:57
NiccoNicco
12
12
$begingroup$
What are your thoughts? I don´t think that you get an answer if you don´t show a minimum of effort. See also here: How to ask a good question.
$endgroup$
– callculus
Jan 27 at 17:01
$begingroup$
The answer is $4$. However, if you want a more detailed answer you should put a little effort into your question.
$endgroup$
– Nathanael Skrepek
Jan 27 at 17:08
$begingroup$
No, hypergeometric distribution is not right. But why you don´t use the binomial distribution? And you can use the converse probability. $P(Xgeq 2)=1-P(X=1)-P(X=0)geq 0.9$, where $Xsim Bin(n, 0.8)$ Note that the inequality cannot be solved algebraically. Just try out some values for $n$.
$endgroup$
– callculus
Jan 27 at 17:21
add a comment |
$begingroup$
What are your thoughts? I don´t think that you get an answer if you don´t show a minimum of effort. See also here: How to ask a good question.
$endgroup$
– callculus
Jan 27 at 17:01
$begingroup$
The answer is $4$. However, if you want a more detailed answer you should put a little effort into your question.
$endgroup$
– Nathanael Skrepek
Jan 27 at 17:08
$begingroup$
No, hypergeometric distribution is not right. But why you don´t use the binomial distribution? And you can use the converse probability. $P(Xgeq 2)=1-P(X=1)-P(X=0)geq 0.9$, where $Xsim Bin(n, 0.8)$ Note that the inequality cannot be solved algebraically. Just try out some values for $n$.
$endgroup$
– callculus
Jan 27 at 17:21
$begingroup$
What are your thoughts? I don´t think that you get an answer if you don´t show a minimum of effort. See also here: How to ask a good question.
$endgroup$
– callculus
Jan 27 at 17:01
$begingroup$
What are your thoughts? I don´t think that you get an answer if you don´t show a minimum of effort. See also here: How to ask a good question.
$endgroup$
– callculus
Jan 27 at 17:01
$begingroup$
The answer is $4$. However, if you want a more detailed answer you should put a little effort into your question.
$endgroup$
– Nathanael Skrepek
Jan 27 at 17:08
$begingroup$
The answer is $4$. However, if you want a more detailed answer you should put a little effort into your question.
$endgroup$
– Nathanael Skrepek
Jan 27 at 17:08
$begingroup$
No, hypergeometric distribution is not right. But why you don´t use the binomial distribution? And you can use the converse probability. $P(Xgeq 2)=1-P(X=1)-P(X=0)geq 0.9$, where $Xsim Bin(n, 0.8)$ Note that the inequality cannot be solved algebraically. Just try out some values for $n$.
$endgroup$
– callculus
Jan 27 at 17:21
$begingroup$
No, hypergeometric distribution is not right. But why you don´t use the binomial distribution? And you can use the converse probability. $P(Xgeq 2)=1-P(X=1)-P(X=0)geq 0.9$, where $Xsim Bin(n, 0.8)$ Note that the inequality cannot be solved algebraically. Just try out some values for $n$.
$endgroup$
– callculus
Jan 27 at 17:21
add a comment |
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$begingroup$
What are your thoughts? I don´t think that you get an answer if you don´t show a minimum of effort. See also here: How to ask a good question.
$endgroup$
– callculus
Jan 27 at 17:01
$begingroup$
The answer is $4$. However, if you want a more detailed answer you should put a little effort into your question.
$endgroup$
– Nathanael Skrepek
Jan 27 at 17:08
$begingroup$
No, hypergeometric distribution is not right. But why you don´t use the binomial distribution? And you can use the converse probability. $P(Xgeq 2)=1-P(X=1)-P(X=0)geq 0.9$, where $Xsim Bin(n, 0.8)$ Note that the inequality cannot be solved algebraically. Just try out some values for $n$.
$endgroup$
– callculus
Jan 27 at 17:21