Probability that the man is the first to die [duplicate]
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Then what is the probability that out of $n$ men $A_1,A_2,…,A_n$ each aged $x$ years,$A_1$ will die and will be the first to die [closed]
2 answers
If p is the probability that a man aged x will die in a year then the
probability that out of n men $A_1, A_2 ,A_3 ...A_n$, each aged x,
$A_1$ will die in an year and be the first to die is: ?
a) $1- (1- p)^{n-1}$
b)$(1-p)^n$
c) $dfrac 1n [1- (1-p)^n]$
d) $dfrac 1n (1-p)^n$
Attempt:
Order of dying matters.
Choose $r$ out of $n-1$ and arrange them (i.e. their deaths) in $r!$ ways.
So $P(E) = ptimes{^{n-1}C_0} (1- p)^{n-1}+ p times ptimes {^{n-1}C_1} (1-p)^{n-1} + ptimes p^2 times 2!times {^{n-1}C_2} (1- p)^{n-2}+...$
$implies P(E) = p sum_{r=0}^{n-1} {^{n-1}}C_r (1-p)^{n-1-r}p^r times r!$
But this doesn't match any of the options.
Could someone please tell me where I have gone wrong?
Edit: My prime focus is the mistake that I have made in solving which that question doesn't cover. therefore, please don't mark it as a duplicate
probability
marked as duplicate by JMoravitz, GNUSupporter 8964民主女神 地下教會, Rebellos, José Carlos Santos, Leucippus Nov 23 at 1:05
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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This question already has an answer here:
Then what is the probability that out of $n$ men $A_1,A_2,…,A_n$ each aged $x$ years,$A_1$ will die and will be the first to die [closed]
2 answers
If p is the probability that a man aged x will die in a year then the
probability that out of n men $A_1, A_2 ,A_3 ...A_n$, each aged x,
$A_1$ will die in an year and be the first to die is: ?
a) $1- (1- p)^{n-1}$
b)$(1-p)^n$
c) $dfrac 1n [1- (1-p)^n]$
d) $dfrac 1n (1-p)^n$
Attempt:
Order of dying matters.
Choose $r$ out of $n-1$ and arrange them (i.e. their deaths) in $r!$ ways.
So $P(E) = ptimes{^{n-1}C_0} (1- p)^{n-1}+ p times ptimes {^{n-1}C_1} (1-p)^{n-1} + ptimes p^2 times 2!times {^{n-1}C_2} (1- p)^{n-2}+...$
$implies P(E) = p sum_{r=0}^{n-1} {^{n-1}}C_r (1-p)^{n-1-r}p^r times r!$
But this doesn't match any of the options.
Could someone please tell me where I have gone wrong?
Edit: My prime focus is the mistake that I have made in solving which that question doesn't cover. therefore, please don't mark it as a duplicate
probability
marked as duplicate by JMoravitz, GNUSupporter 8964民主女神 地下教會, Rebellos, José Carlos Santos, Leucippus Nov 23 at 1:05
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
My prime focus is the mistake that I have made in solving which that question doesn't cover. therefore, please don't mark it as a duplicate.
– Abcd
Nov 22 at 16:15
@Abcd: If you look my answer below, you will notice your mistake is on $mathbb{P}[A_1 text{dies first} mid N=r]$.
– Daniel
Nov 22 at 16:17
@Daniel Didnt understand what you mean.
– Abcd
Nov 22 at 16:19
@Abcd: the actual value of $mathbb{P}[A_1 text{dies first} mid N=r]$ is $frac{1}{r+1}$, and you considered it being $r!$. The probability of $A_1$ being the first among $A_1, A_{i_1}, ldots, A_{i_r}$ is $frac{r!}{(r+1)!}$.
– Daniel
Nov 22 at 16:23
@Daniel $r!$ is for arrangement death of $A_2 , A3 ,etc$ not for $A_1$
– Abcd
Nov 22 at 16:27
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show 5 more comments
up vote
0
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up vote
0
down vote
favorite
This question already has an answer here:
Then what is the probability that out of $n$ men $A_1,A_2,…,A_n$ each aged $x$ years,$A_1$ will die and will be the first to die [closed]
2 answers
If p is the probability that a man aged x will die in a year then the
probability that out of n men $A_1, A_2 ,A_3 ...A_n$, each aged x,
$A_1$ will die in an year and be the first to die is: ?
a) $1- (1- p)^{n-1}$
b)$(1-p)^n$
c) $dfrac 1n [1- (1-p)^n]$
d) $dfrac 1n (1-p)^n$
Attempt:
Order of dying matters.
Choose $r$ out of $n-1$ and arrange them (i.e. their deaths) in $r!$ ways.
So $P(E) = ptimes{^{n-1}C_0} (1- p)^{n-1}+ p times ptimes {^{n-1}C_1} (1-p)^{n-1} + ptimes p^2 times 2!times {^{n-1}C_2} (1- p)^{n-2}+...$
$implies P(E) = p sum_{r=0}^{n-1} {^{n-1}}C_r (1-p)^{n-1-r}p^r times r!$
But this doesn't match any of the options.
Could someone please tell me where I have gone wrong?
Edit: My prime focus is the mistake that I have made in solving which that question doesn't cover. therefore, please don't mark it as a duplicate
probability
This question already has an answer here:
Then what is the probability that out of $n$ men $A_1,A_2,…,A_n$ each aged $x$ years,$A_1$ will die and will be the first to die [closed]
2 answers
If p is the probability that a man aged x will die in a year then the
probability that out of n men $A_1, A_2 ,A_3 ...A_n$, each aged x,
$A_1$ will die in an year and be the first to die is: ?
a) $1- (1- p)^{n-1}$
b)$(1-p)^n$
c) $dfrac 1n [1- (1-p)^n]$
d) $dfrac 1n (1-p)^n$
Attempt:
Order of dying matters.
Choose $r$ out of $n-1$ and arrange them (i.e. their deaths) in $r!$ ways.
So $P(E) = ptimes{^{n-1}C_0} (1- p)^{n-1}+ p times ptimes {^{n-1}C_1} (1-p)^{n-1} + ptimes p^2 times 2!times {^{n-1}C_2} (1- p)^{n-2}+...$
$implies P(E) = p sum_{r=0}^{n-1} {^{n-1}}C_r (1-p)^{n-1-r}p^r times r!$
But this doesn't match any of the options.
Could someone please tell me where I have gone wrong?
Edit: My prime focus is the mistake that I have made in solving which that question doesn't cover. therefore, please don't mark it as a duplicate
This question already has an answer here:
Then what is the probability that out of $n$ men $A_1,A_2,…,A_n$ each aged $x$ years,$A_1$ will die and will be the first to die [closed]
2 answers
probability
probability
edited Nov 22 at 16:27
asked Nov 22 at 15:45
Abcd
2,93221131
2,93221131
marked as duplicate by JMoravitz, GNUSupporter 8964民主女神 地下教會, Rebellos, José Carlos Santos, Leucippus Nov 23 at 1:05
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by JMoravitz, GNUSupporter 8964民主女神 地下教會, Rebellos, José Carlos Santos, Leucippus Nov 23 at 1:05
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
My prime focus is the mistake that I have made in solving which that question doesn't cover. therefore, please don't mark it as a duplicate.
– Abcd
Nov 22 at 16:15
@Abcd: If you look my answer below, you will notice your mistake is on $mathbb{P}[A_1 text{dies first} mid N=r]$.
– Daniel
Nov 22 at 16:17
@Daniel Didnt understand what you mean.
– Abcd
Nov 22 at 16:19
@Abcd: the actual value of $mathbb{P}[A_1 text{dies first} mid N=r]$ is $frac{1}{r+1}$, and you considered it being $r!$. The probability of $A_1$ being the first among $A_1, A_{i_1}, ldots, A_{i_r}$ is $frac{r!}{(r+1)!}$.
– Daniel
Nov 22 at 16:23
@Daniel $r!$ is for arrangement death of $A_2 , A3 ,etc$ not for $A_1$
– Abcd
Nov 22 at 16:27
|
show 5 more comments
My prime focus is the mistake that I have made in solving which that question doesn't cover. therefore, please don't mark it as a duplicate.
– Abcd
Nov 22 at 16:15
@Abcd: If you look my answer below, you will notice your mistake is on $mathbb{P}[A_1 text{dies first} mid N=r]$.
– Daniel
Nov 22 at 16:17
@Daniel Didnt understand what you mean.
– Abcd
Nov 22 at 16:19
@Abcd: the actual value of $mathbb{P}[A_1 text{dies first} mid N=r]$ is $frac{1}{r+1}$, and you considered it being $r!$. The probability of $A_1$ being the first among $A_1, A_{i_1}, ldots, A_{i_r}$ is $frac{r!}{(r+1)!}$.
– Daniel
Nov 22 at 16:23
@Daniel $r!$ is for arrangement death of $A_2 , A3 ,etc$ not for $A_1$
– Abcd
Nov 22 at 16:27
My prime focus is the mistake that I have made in solving which that question doesn't cover. therefore, please don't mark it as a duplicate.
– Abcd
Nov 22 at 16:15
My prime focus is the mistake that I have made in solving which that question doesn't cover. therefore, please don't mark it as a duplicate.
– Abcd
Nov 22 at 16:15
@Abcd: If you look my answer below, you will notice your mistake is on $mathbb{P}[A_1 text{dies first} mid N=r]$.
– Daniel
Nov 22 at 16:17
@Abcd: If you look my answer below, you will notice your mistake is on $mathbb{P}[A_1 text{dies first} mid N=r]$.
– Daniel
Nov 22 at 16:17
@Daniel Didnt understand what you mean.
– Abcd
Nov 22 at 16:19
@Daniel Didnt understand what you mean.
– Abcd
Nov 22 at 16:19
@Abcd: the actual value of $mathbb{P}[A_1 text{dies first} mid N=r]$ is $frac{1}{r+1}$, and you considered it being $r!$. The probability of $A_1$ being the first among $A_1, A_{i_1}, ldots, A_{i_r}$ is $frac{r!}{(r+1)!}$.
– Daniel
Nov 22 at 16:23
@Abcd: the actual value of $mathbb{P}[A_1 text{dies first} mid N=r]$ is $frac{1}{r+1}$, and you considered it being $r!$. The probability of $A_1$ being the first among $A_1, A_{i_1}, ldots, A_{i_r}$ is $frac{r!}{(r+1)!}$.
– Daniel
Nov 22 at 16:23
@Daniel $r!$ is for arrangement death of $A_2 , A3 ,etc$ not for $A_1$
– Abcd
Nov 22 at 16:27
@Daniel $r!$ is for arrangement death of $A_2 , A3 ,etc$ not for $A_1$
– Abcd
Nov 22 at 16:27
|
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2 Answers
2
active
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up vote
2
down vote
accepted
Your expression is almost correct. Let $N$ be the number of deaths out of $A_2, ldots, A_n$. We know that $A_1$ must die. After that, we decompose the event with respect to $N$ (notice we are assuming all deaths are independent) and use that given $N = r$, the probability that the first to die is $A_1$ is $frac{1}{r+1}$ (here assuming that any order of deaths is equally likely). Thus:
begin{align*}
mathbb{P}[text{given event}]
&= mathbb{P}[A_1 text{dies}] cdot sum_{r=0}^{n-1} mathbb{P}[N=r] cdot mathbb{P}[A_1 text{dies first} mid N=r]\
&= p cdot sum_{r=0}^{n-1} binom{n-1}{r} p^r (1-p)^{n-1-r} cdot frac{1}{r+1}
end{align*}
To simplify the expression above, we notice that
$$
binom{n-1}{r}frac{1}{r+1} = frac1n binom{n}{r+1}.
$$
From the above, we can write
begin{align*}
mathbb{P}[text{given event}]
&= frac{1}{n} cdot sum_{r=0}^{n-1} binom{n}{r+1} p^{r+1} (1-p)^{n-1-r} \
&= frac1n [(p + (1-p))^n - (1-p)^n] = frac1n [1 - (1-p)^n].
end{align*}
add a comment |
up vote
1
down vote
They probability for a person aged $x$ to not die is $1-p$. Hence, the probability that none of them is $(1-p)^n$. Therefore, the probability that at least one of them will die is $1-(1-p)^n$. SInce every one of them has the same probability to be that first person (as all are identical), it being $A_1$ has probability ${1 over n}[1-(1-p)^n]$.
Hence, option $(c)$ is correct.
OK, but what's my mistake?
– Abcd
Nov 22 at 16:14
1
Your reasoning gives the same answer than mine. However, I got a bit confused here. We have $1 - (1-p)^n$ is the probability of someone dying (we have no control over who is dead and who is not). The question especifically states that $A_1$ must die. How is this reasoning ensuring it?
– Daniel
Nov 22 at 16:15
And have you accounted for the fact that A1 has to be the first to die
– Abcd
Nov 22 at 16:16
@Daniel The logic is that since all of the $n$ persons are identical and no one has any special properties, means that all have same probability to be that person.
– SinTan1729
Nov 22 at 16:17
1
@SinTan1729: I see the reasoning, and yet there is something misterious about it.
– Daniel
Nov 22 at 16:19
|
show 5 more comments
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Your expression is almost correct. Let $N$ be the number of deaths out of $A_2, ldots, A_n$. We know that $A_1$ must die. After that, we decompose the event with respect to $N$ (notice we are assuming all deaths are independent) and use that given $N = r$, the probability that the first to die is $A_1$ is $frac{1}{r+1}$ (here assuming that any order of deaths is equally likely). Thus:
begin{align*}
mathbb{P}[text{given event}]
&= mathbb{P}[A_1 text{dies}] cdot sum_{r=0}^{n-1} mathbb{P}[N=r] cdot mathbb{P}[A_1 text{dies first} mid N=r]\
&= p cdot sum_{r=0}^{n-1} binom{n-1}{r} p^r (1-p)^{n-1-r} cdot frac{1}{r+1}
end{align*}
To simplify the expression above, we notice that
$$
binom{n-1}{r}frac{1}{r+1} = frac1n binom{n}{r+1}.
$$
From the above, we can write
begin{align*}
mathbb{P}[text{given event}]
&= frac{1}{n} cdot sum_{r=0}^{n-1} binom{n}{r+1} p^{r+1} (1-p)^{n-1-r} \
&= frac1n [(p + (1-p))^n - (1-p)^n] = frac1n [1 - (1-p)^n].
end{align*}
add a comment |
up vote
2
down vote
accepted
Your expression is almost correct. Let $N$ be the number of deaths out of $A_2, ldots, A_n$. We know that $A_1$ must die. After that, we decompose the event with respect to $N$ (notice we are assuming all deaths are independent) and use that given $N = r$, the probability that the first to die is $A_1$ is $frac{1}{r+1}$ (here assuming that any order of deaths is equally likely). Thus:
begin{align*}
mathbb{P}[text{given event}]
&= mathbb{P}[A_1 text{dies}] cdot sum_{r=0}^{n-1} mathbb{P}[N=r] cdot mathbb{P}[A_1 text{dies first} mid N=r]\
&= p cdot sum_{r=0}^{n-1} binom{n-1}{r} p^r (1-p)^{n-1-r} cdot frac{1}{r+1}
end{align*}
To simplify the expression above, we notice that
$$
binom{n-1}{r}frac{1}{r+1} = frac1n binom{n}{r+1}.
$$
From the above, we can write
begin{align*}
mathbb{P}[text{given event}]
&= frac{1}{n} cdot sum_{r=0}^{n-1} binom{n}{r+1} p^{r+1} (1-p)^{n-1-r} \
&= frac1n [(p + (1-p))^n - (1-p)^n] = frac1n [1 - (1-p)^n].
end{align*}
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Your expression is almost correct. Let $N$ be the number of deaths out of $A_2, ldots, A_n$. We know that $A_1$ must die. After that, we decompose the event with respect to $N$ (notice we are assuming all deaths are independent) and use that given $N = r$, the probability that the first to die is $A_1$ is $frac{1}{r+1}$ (here assuming that any order of deaths is equally likely). Thus:
begin{align*}
mathbb{P}[text{given event}]
&= mathbb{P}[A_1 text{dies}] cdot sum_{r=0}^{n-1} mathbb{P}[N=r] cdot mathbb{P}[A_1 text{dies first} mid N=r]\
&= p cdot sum_{r=0}^{n-1} binom{n-1}{r} p^r (1-p)^{n-1-r} cdot frac{1}{r+1}
end{align*}
To simplify the expression above, we notice that
$$
binom{n-1}{r}frac{1}{r+1} = frac1n binom{n}{r+1}.
$$
From the above, we can write
begin{align*}
mathbb{P}[text{given event}]
&= frac{1}{n} cdot sum_{r=0}^{n-1} binom{n}{r+1} p^{r+1} (1-p)^{n-1-r} \
&= frac1n [(p + (1-p))^n - (1-p)^n] = frac1n [1 - (1-p)^n].
end{align*}
Your expression is almost correct. Let $N$ be the number of deaths out of $A_2, ldots, A_n$. We know that $A_1$ must die. After that, we decompose the event with respect to $N$ (notice we are assuming all deaths are independent) and use that given $N = r$, the probability that the first to die is $A_1$ is $frac{1}{r+1}$ (here assuming that any order of deaths is equally likely). Thus:
begin{align*}
mathbb{P}[text{given event}]
&= mathbb{P}[A_1 text{dies}] cdot sum_{r=0}^{n-1} mathbb{P}[N=r] cdot mathbb{P}[A_1 text{dies first} mid N=r]\
&= p cdot sum_{r=0}^{n-1} binom{n-1}{r} p^r (1-p)^{n-1-r} cdot frac{1}{r+1}
end{align*}
To simplify the expression above, we notice that
$$
binom{n-1}{r}frac{1}{r+1} = frac1n binom{n}{r+1}.
$$
From the above, we can write
begin{align*}
mathbb{P}[text{given event}]
&= frac{1}{n} cdot sum_{r=0}^{n-1} binom{n}{r+1} p^{r+1} (1-p)^{n-1-r} \
&= frac1n [(p + (1-p))^n - (1-p)^n] = frac1n [1 - (1-p)^n].
end{align*}
answered Nov 22 at 16:08
Daniel
1,516210
1,516210
add a comment |
add a comment |
up vote
1
down vote
They probability for a person aged $x$ to not die is $1-p$. Hence, the probability that none of them is $(1-p)^n$. Therefore, the probability that at least one of them will die is $1-(1-p)^n$. SInce every one of them has the same probability to be that first person (as all are identical), it being $A_1$ has probability ${1 over n}[1-(1-p)^n]$.
Hence, option $(c)$ is correct.
OK, but what's my mistake?
– Abcd
Nov 22 at 16:14
1
Your reasoning gives the same answer than mine. However, I got a bit confused here. We have $1 - (1-p)^n$ is the probability of someone dying (we have no control over who is dead and who is not). The question especifically states that $A_1$ must die. How is this reasoning ensuring it?
– Daniel
Nov 22 at 16:15
And have you accounted for the fact that A1 has to be the first to die
– Abcd
Nov 22 at 16:16
@Daniel The logic is that since all of the $n$ persons are identical and no one has any special properties, means that all have same probability to be that person.
– SinTan1729
Nov 22 at 16:17
1
@SinTan1729: I see the reasoning, and yet there is something misterious about it.
– Daniel
Nov 22 at 16:19
|
show 5 more comments
up vote
1
down vote
They probability for a person aged $x$ to not die is $1-p$. Hence, the probability that none of them is $(1-p)^n$. Therefore, the probability that at least one of them will die is $1-(1-p)^n$. SInce every one of them has the same probability to be that first person (as all are identical), it being $A_1$ has probability ${1 over n}[1-(1-p)^n]$.
Hence, option $(c)$ is correct.
OK, but what's my mistake?
– Abcd
Nov 22 at 16:14
1
Your reasoning gives the same answer than mine. However, I got a bit confused here. We have $1 - (1-p)^n$ is the probability of someone dying (we have no control over who is dead and who is not). The question especifically states that $A_1$ must die. How is this reasoning ensuring it?
– Daniel
Nov 22 at 16:15
And have you accounted for the fact that A1 has to be the first to die
– Abcd
Nov 22 at 16:16
@Daniel The logic is that since all of the $n$ persons are identical and no one has any special properties, means that all have same probability to be that person.
– SinTan1729
Nov 22 at 16:17
1
@SinTan1729: I see the reasoning, and yet there is something misterious about it.
– Daniel
Nov 22 at 16:19
|
show 5 more comments
up vote
1
down vote
up vote
1
down vote
They probability for a person aged $x$ to not die is $1-p$. Hence, the probability that none of them is $(1-p)^n$. Therefore, the probability that at least one of them will die is $1-(1-p)^n$. SInce every one of them has the same probability to be that first person (as all are identical), it being $A_1$ has probability ${1 over n}[1-(1-p)^n]$.
Hence, option $(c)$ is correct.
They probability for a person aged $x$ to not die is $1-p$. Hence, the probability that none of them is $(1-p)^n$. Therefore, the probability that at least one of them will die is $1-(1-p)^n$. SInce every one of them has the same probability to be that first person (as all are identical), it being $A_1$ has probability ${1 over n}[1-(1-p)^n]$.
Hence, option $(c)$ is correct.
answered Nov 22 at 16:07
SinTan1729
2,399622
2,399622
OK, but what's my mistake?
– Abcd
Nov 22 at 16:14
1
Your reasoning gives the same answer than mine. However, I got a bit confused here. We have $1 - (1-p)^n$ is the probability of someone dying (we have no control over who is dead and who is not). The question especifically states that $A_1$ must die. How is this reasoning ensuring it?
– Daniel
Nov 22 at 16:15
And have you accounted for the fact that A1 has to be the first to die
– Abcd
Nov 22 at 16:16
@Daniel The logic is that since all of the $n$ persons are identical and no one has any special properties, means that all have same probability to be that person.
– SinTan1729
Nov 22 at 16:17
1
@SinTan1729: I see the reasoning, and yet there is something misterious about it.
– Daniel
Nov 22 at 16:19
|
show 5 more comments
OK, but what's my mistake?
– Abcd
Nov 22 at 16:14
1
Your reasoning gives the same answer than mine. However, I got a bit confused here. We have $1 - (1-p)^n$ is the probability of someone dying (we have no control over who is dead and who is not). The question especifically states that $A_1$ must die. How is this reasoning ensuring it?
– Daniel
Nov 22 at 16:15
And have you accounted for the fact that A1 has to be the first to die
– Abcd
Nov 22 at 16:16
@Daniel The logic is that since all of the $n$ persons are identical and no one has any special properties, means that all have same probability to be that person.
– SinTan1729
Nov 22 at 16:17
1
@SinTan1729: I see the reasoning, and yet there is something misterious about it.
– Daniel
Nov 22 at 16:19
OK, but what's my mistake?
– Abcd
Nov 22 at 16:14
OK, but what's my mistake?
– Abcd
Nov 22 at 16:14
1
1
Your reasoning gives the same answer than mine. However, I got a bit confused here. We have $1 - (1-p)^n$ is the probability of someone dying (we have no control over who is dead and who is not). The question especifically states that $A_1$ must die. How is this reasoning ensuring it?
– Daniel
Nov 22 at 16:15
Your reasoning gives the same answer than mine. However, I got a bit confused here. We have $1 - (1-p)^n$ is the probability of someone dying (we have no control over who is dead and who is not). The question especifically states that $A_1$ must die. How is this reasoning ensuring it?
– Daniel
Nov 22 at 16:15
And have you accounted for the fact that A1 has to be the first to die
– Abcd
Nov 22 at 16:16
And have you accounted for the fact that A1 has to be the first to die
– Abcd
Nov 22 at 16:16
@Daniel The logic is that since all of the $n$ persons are identical and no one has any special properties, means that all have same probability to be that person.
– SinTan1729
Nov 22 at 16:17
@Daniel The logic is that since all of the $n$ persons are identical and no one has any special properties, means that all have same probability to be that person.
– SinTan1729
Nov 22 at 16:17
1
1
@SinTan1729: I see the reasoning, and yet there is something misterious about it.
– Daniel
Nov 22 at 16:19
@SinTan1729: I see the reasoning, and yet there is something misterious about it.
– Daniel
Nov 22 at 16:19
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My prime focus is the mistake that I have made in solving which that question doesn't cover. therefore, please don't mark it as a duplicate.
– Abcd
Nov 22 at 16:15
@Abcd: If you look my answer below, you will notice your mistake is on $mathbb{P}[A_1 text{dies first} mid N=r]$.
– Daniel
Nov 22 at 16:17
@Daniel Didnt understand what you mean.
– Abcd
Nov 22 at 16:19
@Abcd: the actual value of $mathbb{P}[A_1 text{dies first} mid N=r]$ is $frac{1}{r+1}$, and you considered it being $r!$. The probability of $A_1$ being the first among $A_1, A_{i_1}, ldots, A_{i_r}$ is $frac{r!}{(r+1)!}$.
– Daniel
Nov 22 at 16:23
@Daniel $r!$ is for arrangement death of $A_2 , A3 ,etc$ not for $A_1$
– Abcd
Nov 22 at 16:27