Mixing
probability that he is NOT killed in $20$ years
Clash Royale CLAN TAG#URR8PPP
up vote
3
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Question
$textsuppose that the probability of being killed in a single flight is
P_c=frac10^-64 textbased$
$text on available statistics. Assume that different flights are independent. If a businessman$ $texttakes 20 flights per year, what is the probability that he is killed in a plane crash within the$ $textnext 20 years? (Let's assume that he will not die because of another $
$textreason within the next 20 years.)?$
My Approach
$$textprobability of NOT being killed in a single flight is
P_c=1-frac10^-64$$
There are $20$ flights per year .so Probability that he will not be dying in a year
$$=left(1-frac10^-64right)^20$$
probability that he is killed in a plane crash within the next $20$ years=probability that he is NOT killed in $20$ years and will be dying in next upcoming $20$years.
$$left(left(1-frac10^-64right)^20right)^20 times left(frac10^-64right)^20$$
Am i correct?
Bit confused.please help me
Thanks!
probability
edited Jul 24 at 17:30
saulspatz
10.4k21323
asked Jul 24 at 17:14
laura
1,238521
add a comment |Â
up vote
3
down vote
favorite
Question
$textsuppose that the probability of being killed in a single flight is
P_c=frac10^-64 textbased$
$text on available statistics. Assume that different flights are independent. If a businessman$ $texttakes 20 flights per year, what is the probability that he is killed in a plane crash within the$ $textnext 20 years? (Let's assume that he will not die because of another $
$textreason within the next 20 years.)?$
My Approach
$$textprobability of NOT being killed in a single flight is
P_c=1-frac10^-64$$
There are $20$ flights per year .so Probability that he will not be dying in a year
$$=left(1-frac10^-64right)^20$$
probability that he is killed in a plane crash within the next $20$ years=probability that he is NOT killed in $20$ years and will be dying in next upcoming $20$years.
$$left(left(1-frac10^-64right)^20right)^20 times left(frac10^-64right)^20$$
Am i correct?
Bit confused.please help me
Thanks!
probability
edited Jul 24 at 17:30
saulspatz
10.4k21323
asked Jul 24 at 17:14
laura
1,238521
2
Are you sure you don't want $P_c = dfrac10^-64$? Because $dfrac10^64 = 250,000$ is awfully big for a probability. Cheers!
– Robert Lewis
Jul 24 at 17:16
1
Appology! editing ! Thanks! :)
– laura
Jul 24 at 17:19
3
Where does that last factor of $left(frac10^-64right)^20$ come from? You were doing fine up to that point.
– saulspatz
Jul 24 at 17:24
4
In $20$ years, he takes $20 times 20 = 400$ flights. What are his odds of surviving $400$ flights? His probability of dying is $1$ minus that. (Actually, these numbers are small enough that you can essentially just multiply his probability of dying on any single flight by $400$.)
– Brian Tung
Jul 24 at 17:24
thanks all! @saulspatz actually i added last factor from the fact that he will surive first $20$ year and then die ..last factor i multiplied it with a probability of dying ..which indeed is wrong
– laura
Jul 24 at 17:50
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Question
$textsuppose that the probability of being killed in a single flight is
P_c=frac10^-64 textbased$
$text on available statistics. Assume that different flights are independent. If a businessman$ $texttakes 20 flights per year, what is the probability that he is killed in a plane crash within the$ $textnext 20 years? (Let's assume that he will not die because of another $
$textreason within the next 20 years.)?$
My Approach
$$textprobability of NOT being killed in a single flight is
P_c=1-frac10^-64$$
There are $20$ flights per year .so Probability that he will not be dying in a year
$$=left(1-frac10^-64right)^20$$
probability that he is killed in a plane crash within the next $20$ years=probability that he is NOT killed in $20$ years and will be dying in next upcoming $20$years.
$$left(left(1-frac10^-64right)^20right)^20 times left(frac10^-64right)^20$$
Am i correct?
Bit confused.please help me
Thanks!
probability
edited Jul 24 at 17:30
saulspatz
10.4k21323
asked Jul 24 at 17:14
laura
1,238521
Question
$textsuppose that the probability of being killed in a single flight is
P_c=frac10^-64 textbased$
$text on available statistics. Assume that different flights are independent. If a businessman$ $texttakes 20 flights per year, what is the probability that he is killed in a plane crash within the$ $textnext 20 years? (Let's assume that he will not die because of another $
$textreason within the next 20 years.)?$
My Approach
$$textprobability of NOT being killed in a single flight is
P_c=1-frac10^-64$$
There are $20$ flights per year .so Probability that he will not be dying in a year
$$=left(1-frac10^-64right)^20$$
probability that he is killed in a plane crash within the next $20$ years=probability that he is NOT killed in $20$ years and will be dying in next upcoming $20$years.
$$left(left(1-frac10^-64right)^20right)^20 times left(frac10^-64right)^20$$
Am i correct?
Bit confused.please help me
Thanks!
probability
edited Jul 24 at 17:30
saulspatz
10.4k21323
asked Jul 24 at 17:14
laura
1,238521
edited Jul 24 at 17:30
saulspatz
10.4k21323
edited Jul 24 at 17:30
saulspatz
10.4k21323
edited Jul 24 at 17:30
saulspatz
10.4k21323
10.4k21323
asked Jul 24 at 17:14
laura
1,238521
asked Jul 24 at 17:14
laura
1,238521
asked Jul 24 at 17:14
laura
1,238521
1,238521
2
Are you sure you don't want $P_c = dfrac10^-64$? Because $dfrac10^64 = 250,000$ is awfully big for a probability. Cheers!
– Robert Lewis
Jul 24 at 17:16
1
Appology! editing ! Thanks! :)
– laura
Jul 24 at 17:19
3
Where does that last factor of $left(frac10^-64right)^20$ come from? You were doing fine up to that point.
– saulspatz
Jul 24 at 17:24
4
In $20$ years, he takes $20 times 20 = 400$ flights. What are his odds of surviving $400$ flights? His probability of dying is $1$ minus that. (Actually, these numbers are small enough that you can essentially just multiply his probability of dying on any single flight by $400$.)
– Brian Tung
Jul 24 at 17:24
thanks all! @saulspatz actually i added last factor from the fact that he will surive first $20$ year and then die ..last factor i multiplied it with a probability of dying ..which indeed is wrong
– laura
Jul 24 at 17:50
add a comment |Â
2
Are you sure you don't want $P_c = dfrac10^-64$? Because $dfrac10^64 = 250,000$ is awfully big for a probability. Cheers!
– Robert Lewis
Jul 24 at 17:16
1
Appology! editing ! Thanks! :)
– laura
Jul 24 at 17:19
3
Where does that last factor of $left(frac10^-64right)^20$ come from? You were doing fine up to that point.
– saulspatz
Jul 24 at 17:24
4
In $20$ years, he takes $20 times 20 = 400$ flights. What are his odds of surviving $400$ flights? His probability of dying is $1$ minus that. (Actually, these numbers are small enough that you can essentially just multiply his probability of dying on any single flight by $400$.)
– Brian Tung
Jul 24 at 17:24
thanks all! @saulspatz actually i added last factor from the fact that he will surive first $20$ year and then die ..last factor i multiplied it with a probability of dying ..which indeed is wrong
– laura
Jul 24 at 17:50
2
2
Are you sure you don't want $P_c = dfrac10^-64$? Because $dfrac10^64 = 250,000$ is awfully big for a probability. Cheers!
– Robert Lewis
Jul 24 at 17:16
Are you sure you don't want $P_c = dfrac10^-64$? Because $dfrac10^64 = 250,000$ is awfully big for a probability. Cheers!
– Robert Lewis
Jul 24 at 17:16
1
1
Appology! editing ! Thanks! :)
– laura
Jul 24 at 17:19
Appology! editing ! Thanks! :)
– laura
Jul 24 at 17:19
3
3
Where does that last factor of $left(frac10^-64right)^20$ come from? You were doing fine up to that point.
– saulspatz
Jul 24 at 17:24
Where does that last factor of $left(frac10^-64right)^20$ come from? You were doing fine up to that point.
– saulspatz
Jul 24 at 17:24
4
4
In $20$ years, he takes $20 times 20 = 400$ flights. What are his odds of surviving $400$ flights? His probability of dying is $1$ minus that. (Actually, these numbers are small enough that you can essentially just multiply his probability of dying on any single flight by $400$.)
– Brian Tung
Jul 24 at 17:24
In $20$ years, he takes $20 times 20 = 400$ flights. What are his odds of surviving $400$ flights? His probability of dying is $1$ minus that. (Actually, these numbers are small enough that you can essentially just multiply his probability of dying on any single flight by $400$.)
– Brian Tung
Jul 24 at 17:24
thanks all! @saulspatz actually i added last factor from the fact that he will surive first $20$ year and then die ..last factor i multiplied it with a probability of dying ..which indeed is wrong
– laura
Jul 24 at 17:50
thanks all! @saulspatz actually i added last factor from the fact that he will surive first $20$ year and then die ..last factor i multiplied it with a probability of dying ..which indeed is wrong
– laura
Jul 24 at 17:50
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
3
down vote
accepted
The total number of flights that he will take during the next $20$ years is $N=20times20=400$.
Let $p_s$ be the probability that he survives a given single flight. Then we have
$$p_s=1-p_c$$
Since these flights are independent, the probability that he will survive all $N=400$ flights is $$P(mboxSurvive N flights )=p_stimes p_stimes......times p_s=p_s^N=(1-p_c)^N$$Let $A$ be the event that the businessman is killed in a plane crash within the next $20$ years. Then$$P(A)=1-(1-p_c)^N=9.9995times10^-5approx dfrac110000$$
answered Jul 24 at 17:46
Key Flex
4,193423
See my approach to this problem. I think both these methods achieve the same answer, but how can we prove the similarity of the 2 formulas?
– Rahul Goswami
Jul 24 at 18:13
add a comment |Â
up vote
1
down vote
There is another approach to this problem.
Consider $P(n)$ be the probability that the businessman dies in his $n$th flight.
$p_c$ is the probability of the person being killed in any flight.
Now, $P(i) = (1-p_c)^i-1 dot p_c$ i.e. he survived $i-1$ flights and died in $i$th flight.
He will take total $20 times 20$ flights in 20 years. The probability that he will die in any of these flights is $sum_i=1^20 times 20P(i)$.
Therefore, the required probability is $1 - sum_i=1^20 times 20P(i)$
answered Jul 24 at 18:04
Rahul Goswami
319114
1
(+1) for alternative approach. The difference is that you took that he dies for $n^th$ flight but I took that he survives for all $400$
– Key Flex
Jul 24 at 18:20
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
The total number of flights that he will take during the next $20$ years is $N=20times20=400$.
Let $p_s$ be the probability that he survives a given single flight. Then we have
$$p_s=1-p_c$$
Since these flights are independent, the probability that he will survive all $N=400$ flights is $$P(mboxSurvive N flights )=p_stimes p_stimes......times p_s=p_s^N=(1-p_c)^N$$Let $A$ be the event that the businessman is killed in a plane crash within the next $20$ years. Then$$P(A)=1-(1-p_c)^N=9.9995times10^-5approx dfrac110000$$
answered Jul 24 at 17:46
Key Flex
4,193423
See my approach to this problem. I think both these methods achieve the same answer, but how can we prove the similarity of the 2 formulas?
– Rahul Goswami
Jul 24 at 18:13
add a comment |Â
up vote
3
down vote
accepted
The total number of flights that he will take during the next $20$ years is $N=20times20=400$.
Let $p_s$ be the probability that he survives a given single flight. Then we have
$$p_s=1-p_c$$
Since these flights are independent, the probability that he will survive all $N=400$ flights is $$P(mboxSurvive N flights )=p_stimes p_stimes......times p_s=p_s^N=(1-p_c)^N$$Let $A$ be the event that the businessman is killed in a plane crash within the next $20$ years. Then$$P(A)=1-(1-p_c)^N=9.9995times10^-5approx dfrac110000$$
answered Jul 24 at 17:46
Key Flex
4,193423
See my approach to this problem. I think both these methods achieve the same answer, but how can we prove the similarity of the 2 formulas?
– Rahul Goswami
Jul 24 at 18:13
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
The total number of flights that he will take during the next $20$ years is $N=20times20=400$.
Let $p_s$ be the probability that he survives a given single flight. Then we have
$$p_s=1-p_c$$
Since these flights are independent, the probability that he will survive all $N=400$ flights is $$P(mboxSurvive N flights )=p_stimes p_stimes......times p_s=p_s^N=(1-p_c)^N$$Let $A$ be the event that the businessman is killed in a plane crash within the next $20$ years. Then$$P(A)=1-(1-p_c)^N=9.9995times10^-5approx dfrac110000$$
answered Jul 24 at 17:46
Key Flex
4,193423
The total number of flights that he will take during the next $20$ years is $N=20times20=400$.
Let $p_s$ be the probability that he survives a given single flight. Then we have
$$p_s=1-p_c$$
Since these flights are independent, the probability that he will survive all $N=400$ flights is $$P(mboxSurvive N flights )=p_stimes p_stimes......times p_s=p_s^N=(1-p_c)^N$$Let $A$ be the event that the businessman is killed in a plane crash within the next $20$ years. Then$$P(A)=1-(1-p_c)^N=9.9995times10^-5approx dfrac110000$$
answered Jul 24 at 17:46
Key Flex
4,193423
answered Jul 24 at 17:46
Key Flex
4,193423
answered Jul 24 at 17:46
Key Flex
4,193423
answered Jul 24 at 17:46
Key Flex
4,193423
4,193423
See my approach to this problem. I think both these methods achieve the same answer, but how can we prove the similarity of the 2 formulas?
– Rahul Goswami
Jul 24 at 18:13
add a comment |Â
See my approach to this problem. I think both these methods achieve the same answer, but how can we prove the similarity of the 2 formulas?
– Rahul Goswami
Jul 24 at 18:13
See my approach to this problem. I think both these methods achieve the same answer, but how can we prove the similarity of the 2 formulas?
– Rahul Goswami
Jul 24 at 18:13
See my approach to this problem. I think both these methods achieve the same answer, but how can we prove the similarity of the 2 formulas?
– Rahul Goswami
Jul 24 at 18:13
add a comment |Â
up vote
1
down vote
There is another approach to this problem.
Consider $P(n)$ be the probability that the businessman dies in his $n$th flight.
$p_c$ is the probability of the person being killed in any flight.
Now, $P(i) = (1-p_c)^i-1 dot p_c$ i.e. he survived $i-1$ flights and died in $i$th flight.
He will take total $20 times 20$ flights in 20 years. The probability that he will die in any of these flights is $sum_i=1^20 times 20P(i)$.
Therefore, the required probability is $1 - sum_i=1^20 times 20P(i)$
answered Jul 24 at 18:04
Rahul Goswami
319114
1
(+1) for alternative approach. The difference is that you took that he dies for $n^th$ flight but I took that he survives for all $400$
– Key Flex
Jul 24 at 18:20
add a comment |Â
up vote
1
down vote
There is another approach to this problem.
Consider $P(n)$ be the probability that the businessman dies in his $n$th flight.
$p_c$ is the probability of the person being killed in any flight.
Now, $P(i) = (1-p_c)^i-1 dot p_c$ i.e. he survived $i-1$ flights and died in $i$th flight.
He will take total $20 times 20$ flights in 20 years. The probability that he will die in any of these flights is $sum_i=1^20 times 20P(i)$.
Therefore, the required probability is $1 - sum_i=1^20 times 20P(i)$
answered Jul 24 at 18:04
Rahul Goswami
319114
1
(+1) for alternative approach. The difference is that you took that he dies for $n^th$ flight but I took that he survives for all $400$
– Key Flex
Jul 24 at 18:20
add a comment |Â
up vote
1
down vote
up vote
1
down vote
There is another approach to this problem.
Consider $P(n)$ be the probability that the businessman dies in his $n$th flight.
$p_c$ is the probability of the person being killed in any flight.
Now, $P(i) = (1-p_c)^i-1 dot p_c$ i.e. he survived $i-1$ flights and died in $i$th flight.
He will take total $20 times 20$ flights in 20 years. The probability that he will die in any of these flights is $sum_i=1^20 times 20P(i)$.
Therefore, the required probability is $1 - sum_i=1^20 times 20P(i)$
answered Jul 24 at 18:04
Rahul Goswami
319114
There is another approach to this problem.
Consider $P(n)$ be the probability that the businessman dies in his $n$th flight.
$p_c$ is the probability of the person being killed in any flight.
Now, $P(i) = (1-p_c)^i-1 dot p_c$ i.e. he survived $i-1$ flights and died in $i$th flight.
He will take total $20 times 20$ flights in 20 years. The probability that he will die in any of these flights is $sum_i=1^20 times 20P(i)$.
Therefore, the required probability is $1 - sum_i=1^20 times 20P(i)$
answered Jul 24 at 18:04
Rahul Goswami
319114
edited Jul 24 at 18:10
answered Jul 24 at 18:04
Rahul Goswami
319114
answered Jul 24 at 18:04
Rahul Goswami
319114
answered Jul 24 at 18:04
Rahul Goswami
319114
319114
1
(+1) for alternative approach. The difference is that you took that he dies for $n^th$ flight but I took that he survives for all $400$
– Key Flex
Jul 24 at 18:20
add a comment |Â
1
(+1) for alternative approach. The difference is that you took that he dies for $n^th$ flight but I took that he survives for all $400$
– Key Flex
Jul 24 at 18:20
1
1
(+1) for alternative approach. The difference is that you took that he dies for $n^th$ flight but I took that he survives for all $400$
– Key Flex
Jul 24 at 18:20
(+1) for alternative approach. The difference is that you took that he dies for $n^th$ flight but I took that he survives for all $400$
– Key Flex
Jul 24 at 18:20
add a comment |Â
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2
Are you sure you don't want $P_c = dfrac10^-64$? Because $dfrac10^64 = 250,000$ is awfully big for a probability. Cheers!
– Robert Lewis
Jul 24 at 17:16
1
Appology! editing ! Thanks! :)
– laura
Jul 24 at 17:19
3
Where does that last factor of $left(frac10^-64right)^20$ come from? You were doing fine up to that point.
– saulspatz
Jul 24 at 17:24
4
In $20$ years, he takes $20 times 20 = 400$ flights. What are his odds of surviving $400$ flights? His probability of dying is $1$ minus that. (Actually, these numbers are small enough that you can essentially just multiply his probability of dying on any single flight by $400$.)
– Brian Tung
Jul 24 at 17:24
thanks all! @saulspatz actually i added last factor from the fact that he will surive first $20$ year and then die ..last factor i multiplied it with a probability of dying ..which indeed is wrong
– laura
Jul 24 at 17:50