Mixing
A question of probability.
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A box contains $N$ tickets numbered $1,2, cdots , N$. From this box $n$ tickets are drawn without replacement. What is the probability that the largest number drawn is $M$? What if the tickets were drawn with replacement?
For the case "without replacement" I found the answer to be $$frac n(M-1)!(N-n)! N!(M-n)!$$ and for the case "with replacement" I found the answer to be $$frac sum limits_r=0^n-1 n_P_ r M-1 choose r N^n.$$
Are they correct at all? Please check it.
Thank you very much.
probability probability-theory
asked Jul 21 at 5:59
Debabrata Chattopadhyay.
13611
add a comment |Â
up vote
1
down vote
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A box contains $N$ tickets numbered $1,2, cdots , N$. From this box $n$ tickets are drawn without replacement. What is the probability that the largest number drawn is $M$? What if the tickets were drawn with replacement?
For the case "without replacement" I found the answer to be $$frac n(M-1)!(N-n)! N!(M-n)!$$ and for the case "with replacement" I found the answer to be $$frac sum limits_r=0^n-1 n_P_ r M-1 choose r N^n.$$
Are they correct at all? Please check it.
Thank you very much.
probability probability-theory
asked Jul 21 at 5:59
Debabrata Chattopadhyay.
13611
What's $n_P_ r$?
– joriki
Jul 21 at 6:15
Answer on first question is correct. For second question you better use $P(X=M)=P(Xleq M)-P(Xleq M-1)$ where $X$ denotes largest number. I cannot recognize the notation that you use.
– drhab
Jul 21 at 6:16
$n_P_r$ means $frac n! (n-r)!$.
– Debabrata Chattopadhyay.
Jul 21 at 6:29
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
A box contains $N$ tickets numbered $1,2, cdots , N$. From this box $n$ tickets are drawn without replacement. What is the probability that the largest number drawn is $M$? What if the tickets were drawn with replacement?
For the case "without replacement" I found the answer to be $$frac n(M-1)!(N-n)! N!(M-n)!$$ and for the case "with replacement" I found the answer to be $$frac sum limits_r=0^n-1 n_P_ r M-1 choose r N^n.$$
Are they correct at all? Please check it.
Thank you very much.
probability probability-theory
asked Jul 21 at 5:59
Debabrata Chattopadhyay.
13611
A box contains $N$ tickets numbered $1,2, cdots , N$. From this box $n$ tickets are drawn without replacement. What is the probability that the largest number drawn is $M$? What if the tickets were drawn with replacement?
For the case "without replacement" I found the answer to be $$frac n(M-1)!(N-n)! N!(M-n)!$$ and for the case "with replacement" I found the answer to be $$frac sum limits_r=0^n-1 n_P_ r M-1 choose r N^n.$$
Are they correct at all? Please check it.
Thank you very much.
probability probability-theory
asked Jul 21 at 5:59
Debabrata Chattopadhyay.
13611
edited Jul 21 at 6:06
asked Jul 21 at 5:59
Debabrata Chattopadhyay.
13611
asked Jul 21 at 5:59
Debabrata Chattopadhyay.
13611
asked Jul 21 at 5:59
Debabrata Chattopadhyay.
13611
13611
What's $n_P_ r$?
– joriki
Jul 21 at 6:15
Answer on first question is correct. For second question you better use $P(X=M)=P(Xleq M)-P(Xleq M-1)$ where $X$ denotes largest number. I cannot recognize the notation that you use.
– drhab
Jul 21 at 6:16
$n_P_r$ means $frac n! (n-r)!$.
– Debabrata Chattopadhyay.
Jul 21 at 6:29
add a comment |Â
What's $n_P_ r$?
– joriki
Jul 21 at 6:15
Answer on first question is correct. For second question you better use $P(X=M)=P(Xleq M)-P(Xleq M-1)$ where $X$ denotes largest number. I cannot recognize the notation that you use.
– drhab
Jul 21 at 6:16
$n_P_r$ means $frac n! (n-r)!$.
– Debabrata Chattopadhyay.
Jul 21 at 6:29
What's $n_P_ r$?
– joriki
Jul 21 at 6:15
What's $n_P_ r$?
– joriki
Jul 21 at 6:15
Answer on first question is correct. For second question you better use $P(X=M)=P(Xleq M)-P(Xleq M-1)$ where $X$ denotes largest number. I cannot recognize the notation that you use.
– drhab
Jul 21 at 6:16
Answer on first question is correct. For second question you better use $P(X=M)=P(Xleq M)-P(Xleq M-1)$ where $X$ denotes largest number. I cannot recognize the notation that you use.
– drhab
Jul 21 at 6:16
$n_P_r$ means $frac n! (n-r)!$.
– Debabrata Chattopadhyay.
Jul 21 at 6:29
$n_P_r$ means $frac n! (n-r)!$.
– Debabrata Chattopadhyay.
Jul 21 at 6:29
add a comment |Â
1 Answer
1
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It is better to stick to formal notation unless you define yours.
Suppose $X_i$ is the random variable denoting the number on the $i$th ticket drawn, $i=1,2,cdots,n$.
Denote $max (X_1,X_2,cdots,X_n)$ by $X_(n)$. We seek the probability mass function of $X_(n)$.
For sampling without replacement, the required probability is
beginalign
P(X_(n)=M)&=P(X_(n)le M)-P(X_(n)le M-1)
\&=fracbinomMnbinomNn-fracbinomM-1nbinomNn
\&=begincasesdfracbinomM-1n-1binomNn&,text if M=n,n+1,cdots,N\\quad0&,text otherwise endcases
endalign
No need to simplify further by breaking the binomial coefficient into factorials.
For sampling with replacement, the probability is simply
beginalign
P(X_(n)=M)&=P(X_(n)le M)-P(X_(n)le M-1)
\&=left(fracMNright)^n-left(fracM-1Nright)^n
\&=begincasesfrac1N^nleft(M^n-(M-1)^nright)&,text if M=1,2,cdots,N\\quad0&,text otherwise endcases
endalign
Here the random variables $X_1,X_2,cdots,X_n$ are independent unlike in the previous case.
answered Jul 21 at 7:51
StubbornAtom
3,78111134
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
It is better to stick to formal notation unless you define yours.
Suppose $X_i$ is the random variable denoting the number on the $i$th ticket drawn, $i=1,2,cdots,n$.
Denote $max (X_1,X_2,cdots,X_n)$ by $X_(n)$. We seek the probability mass function of $X_(n)$.
For sampling without replacement, the required probability is
beginalign
P(X_(n)=M)&=P(X_(n)le M)-P(X_(n)le M-1)
\&=fracbinomMnbinomNn-fracbinomM-1nbinomNn
\&=begincasesdfracbinomM-1n-1binomNn&,text if M=n,n+1,cdots,N\\quad0&,text otherwise endcases
endalign
No need to simplify further by breaking the binomial coefficient into factorials.
For sampling with replacement, the probability is simply
beginalign
P(X_(n)=M)&=P(X_(n)le M)-P(X_(n)le M-1)
\&=left(fracMNright)^n-left(fracM-1Nright)^n
\&=begincasesfrac1N^nleft(M^n-(M-1)^nright)&,text if M=1,2,cdots,N\\quad0&,text otherwise endcases
endalign
Here the random variables $X_1,X_2,cdots,X_n$ are independent unlike in the previous case.
answered Jul 21 at 7:51
StubbornAtom
3,78111134
add a comment |Â
up vote
0
down vote
It is better to stick to formal notation unless you define yours.
Suppose $X_i$ is the random variable denoting the number on the $i$th ticket drawn, $i=1,2,cdots,n$.
Denote $max (X_1,X_2,cdots,X_n)$ by $X_(n)$. We seek the probability mass function of $X_(n)$.
For sampling without replacement, the required probability is
beginalign
P(X_(n)=M)&=P(X_(n)le M)-P(X_(n)le M-1)
\&=fracbinomMnbinomNn-fracbinomM-1nbinomNn
\&=begincasesdfracbinomM-1n-1binomNn&,text if M=n,n+1,cdots,N\\quad0&,text otherwise endcases
endalign
No need to simplify further by breaking the binomial coefficient into factorials.
For sampling with replacement, the probability is simply
beginalign
P(X_(n)=M)&=P(X_(n)le M)-P(X_(n)le M-1)
\&=left(fracMNright)^n-left(fracM-1Nright)^n
\&=begincasesfrac1N^nleft(M^n-(M-1)^nright)&,text if M=1,2,cdots,N\\quad0&,text otherwise endcases
endalign
Here the random variables $X_1,X_2,cdots,X_n$ are independent unlike in the previous case.
answered Jul 21 at 7:51
StubbornAtom
3,78111134
add a comment |Â
up vote
0
down vote
up vote
0
down vote
It is better to stick to formal notation unless you define yours.
Suppose $X_i$ is the random variable denoting the number on the $i$th ticket drawn, $i=1,2,cdots,n$.
Denote $max (X_1,X_2,cdots,X_n)$ by $X_(n)$. We seek the probability mass function of $X_(n)$.
For sampling without replacement, the required probability is
beginalign
P(X_(n)=M)&=P(X_(n)le M)-P(X_(n)le M-1)
\&=fracbinomMnbinomNn-fracbinomM-1nbinomNn
\&=begincasesdfracbinomM-1n-1binomNn&,text if M=n,n+1,cdots,N\\quad0&,text otherwise endcases
endalign
No need to simplify further by breaking the binomial coefficient into factorials.
For sampling with replacement, the probability is simply
beginalign
P(X_(n)=M)&=P(X_(n)le M)-P(X_(n)le M-1)
\&=left(fracMNright)^n-left(fracM-1Nright)^n
\&=begincasesfrac1N^nleft(M^n-(M-1)^nright)&,text if M=1,2,cdots,N\\quad0&,text otherwise endcases
endalign
Here the random variables $X_1,X_2,cdots,X_n$ are independent unlike in the previous case.
answered Jul 21 at 7:51
StubbornAtom
3,78111134
It is better to stick to formal notation unless you define yours.
Suppose $X_i$ is the random variable denoting the number on the $i$th ticket drawn, $i=1,2,cdots,n$.
Denote $max (X_1,X_2,cdots,X_n)$ by $X_(n)$. We seek the probability mass function of $X_(n)$.
For sampling without replacement, the required probability is
beginalign
P(X_(n)=M)&=P(X_(n)le M)-P(X_(n)le M-1)
\&=fracbinomMnbinomNn-fracbinomM-1nbinomNn
\&=begincasesdfracbinomM-1n-1binomNn&,text if M=n,n+1,cdots,N\\quad0&,text otherwise endcases
endalign
No need to simplify further by breaking the binomial coefficient into factorials.
For sampling with replacement, the probability is simply
beginalign
P(X_(n)=M)&=P(X_(n)le M)-P(X_(n)le M-1)
\&=left(fracMNright)^n-left(fracM-1Nright)^n
\&=begincasesfrac1N^nleft(M^n-(M-1)^nright)&,text if M=1,2,cdots,N\\quad0&,text otherwise endcases
endalign
Here the random variables $X_1,X_2,cdots,X_n$ are independent unlike in the previous case.
answered Jul 21 at 7:51
StubbornAtom
3,78111134
edited Jul 21 at 8:06
answered Jul 21 at 7:51
StubbornAtom
3,78111134
answered Jul 21 at 7:51
StubbornAtom
3,78111134
answered Jul 21 at 7:51
StubbornAtom
3,78111134
3,78111134
add a comment |Â
add a comment |Â
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What's $n_P_ r$?
– joriki
Jul 21 at 6:15
Answer on first question is correct. For second question you better use $P(X=M)=P(Xleq M)-P(Xleq M-1)$ where $X$ denotes largest number. I cannot recognize the notation that you use.
– drhab
Jul 21 at 6:16
$n_P_r$ means $frac n! (n-r)!$.
– Debabrata Chattopadhyay.
Jul 21 at 6:29