Mixing
Earthquakes in another country Assume that in an other country the probability that during a year at least...
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Earthquakes in another country
Assume that in an other country the probability that during a year at least one
earthquake happens is 0,3 . What is the probability that during 5 years the number
of earthquakes is at least 3? Note: We need to apply "Poisson Distribution" for solving this problem.
Hi guys. I solved this problem with help of simple proportion. I said that if during 1 year probability is 0.3 , then during 5 is X and found X which is 1,5
. Thus probability is 1,5 . I just want to make sure that the result is right. Thanks in advance!
probability probability-theory probability-distributions conditional-probability
asked Jan 2 at 21:10
Murad Sh-ovMurad Sh-ov
113
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|
show 6 more comments
$begingroup$
Earthquakes in another country
Assume that in an other country the probability that during a year at least one
earthquake happens is 0,3 . What is the probability that during 5 years the number
of earthquakes is at least 3? Note: We need to apply "Poisson Distribution" for solving this problem.
Hi guys. I solved this problem with help of simple proportion. I said that if during 1 year probability is 0.3 , then during 5 is X and found X which is 1,5
. Thus probability is 1,5 . I just want to make sure that the result is right. Thanks in advance!
probability probability-theory probability-distributions conditional-probability
asked Jan 2 at 21:10
Murad Sh-ovMurad Sh-ov
113
$endgroup$
2
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How could the probability be greater than $1$?
$endgroup$
– lulu
Jan 2 at 21:12
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It's certainly wrong. Probabilities must be between 0 and 1
$endgroup$
– Ray Bern
Jan 2 at 21:14
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Note: as stated the problem really can't be solved. We don't even have enough information to compute the probability that there are at least $3$ earthquakes in a single year.
$endgroup$
– lulu
Jan 2 at 21:16
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@lulu that's right
$endgroup$
– Murad Sh-ov
Jan 2 at 21:17
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Hint: Let $X$ be the number of earthquakes that happen in 5 years. Then, $P{Xgeq 3}=1-P{X<3}$.
$endgroup$
– Ray Bern
Jan 2 at 21:18
|
show 6 more comments
$begingroup$
Earthquakes in another country
Assume that in an other country the probability that during a year at least one
earthquake happens is 0,3 . What is the probability that during 5 years the number
of earthquakes is at least 3? Note: We need to apply "Poisson Distribution" for solving this problem.
Hi guys. I solved this problem with help of simple proportion. I said that if during 1 year probability is 0.3 , then during 5 is X and found X which is 1,5
. Thus probability is 1,5 . I just want to make sure that the result is right. Thanks in advance!
probability probability-theory probability-distributions conditional-probability
asked Jan 2 at 21:10
Murad Sh-ovMurad Sh-ov
113
$endgroup$
Earthquakes in another country
Assume that in an other country the probability that during a year at least one
earthquake happens is 0,3 . What is the probability that during 5 years the number
of earthquakes is at least 3? Note: We need to apply "Poisson Distribution" for solving this problem.
Hi guys. I solved this problem with help of simple proportion. I said that if during 1 year probability is 0.3 , then during 5 is X and found X which is 1,5
. Thus probability is 1,5 . I just want to make sure that the result is right. Thanks in advance!
probability probability-theory probability-distributions conditional-probability
probability probability-theory probability-distributions conditional-probability
asked Jan 2 at 21:10
Murad Sh-ovMurad Sh-ov
113
asked Jan 2 at 21:10
Murad Sh-ovMurad Sh-ov
113
edited Jan 2 at 21:33
Murad Sh-ov
asked Jan 2 at 21:10
Murad Sh-ovMurad Sh-ov
113
asked Jan 2 at 21:10
Murad Sh-ovMurad Sh-ov
113
asked Jan 2 at 21:10
Murad Sh-ovMurad Sh-ov
113
113
2
$begingroup$
How could the probability be greater than $1$?
$endgroup$
– lulu
Jan 2 at 21:12
$begingroup$
It's certainly wrong. Probabilities must be between 0 and 1
$endgroup$
– Ray Bern
Jan 2 at 21:14
$begingroup$
Note: as stated the problem really can't be solved. We don't even have enough information to compute the probability that there are at least $3$ earthquakes in a single year.
$endgroup$
– lulu
Jan 2 at 21:16
$begingroup$
@lulu that's right
$endgroup$
– Murad Sh-ov
Jan 2 at 21:17
$begingroup$
Hint: Let $X$ be the number of earthquakes that happen in 5 years. Then, $P{Xgeq 3}=1-P{X<3}$.
$endgroup$
– Ray Bern
Jan 2 at 21:18
|
show 6 more comments
2
$begingroup$
How could the probability be greater than $1$?
$endgroup$
– lulu
Jan 2 at 21:12
$begingroup$
It's certainly wrong. Probabilities must be between 0 and 1
$endgroup$
– Ray Bern
Jan 2 at 21:14
$begingroup$
Note: as stated the problem really can't be solved. We don't even have enough information to compute the probability that there are at least $3$ earthquakes in a single year.
$endgroup$
– lulu
Jan 2 at 21:16
$begingroup$
@lulu that's right
$endgroup$
– Murad Sh-ov
Jan 2 at 21:17
$begingroup$
Hint: Let $X$ be the number of earthquakes that happen in 5 years. Then, $P{Xgeq 3}=1-P{X<3}$.
$endgroup$
– Ray Bern
Jan 2 at 21:18
2
2
$begingroup$
How could the probability be greater than $1$?
$endgroup$
– lulu
Jan 2 at 21:12
$begingroup$
How could the probability be greater than $1$?
$endgroup$
– lulu
Jan 2 at 21:12
$begingroup$
It's certainly wrong. Probabilities must be between 0 and 1
$endgroup$
– Ray Bern
Jan 2 at 21:14
$begingroup$
It's certainly wrong. Probabilities must be between 0 and 1
$endgroup$
– Ray Bern
Jan 2 at 21:14
$begingroup$
Note: as stated the problem really can't be solved. We don't even have enough information to compute the probability that there are at least $3$ earthquakes in a single year.
$endgroup$
– lulu
Jan 2 at 21:16
$begingroup$
Note: as stated the problem really can't be solved. We don't even have enough information to compute the probability that there are at least $3$ earthquakes in a single year.
$endgroup$
– lulu
Jan 2 at 21:16
$begingroup$
@lulu that's right
$endgroup$
– Murad Sh-ov
Jan 2 at 21:17
$begingroup$
@lulu that's right
$endgroup$
– Murad Sh-ov
Jan 2 at 21:17
$begingroup$
Hint: Let $X$ be the number of earthquakes that happen in 5 years. Then, $P{Xgeq 3}=1-P{X<3}$.
$endgroup$
– Ray Bern
Jan 2 at 21:18
$begingroup$
Hint: Let $X$ be the number of earthquakes that happen in 5 years. Then, $P{Xgeq 3}=1-P{X<3}$.
$endgroup$
– Ray Bern
Jan 2 at 21:18
|
show 6 more comments
1 Answer
1
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oldest
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If you want to use a Poisson distribution, the expected number in five years is $1.5$. That is the $lambda$ parameter in the distribution, so the probability of $n$ earthquakes in $5$ years is
$$P(n)=frac {1.5^ne^{-1.5}}{n!}$$Now compute the probability of $0$ to $2$ earthquakes and subtract from $1$ to get the chance of at least $3$ earthquakes.
answered Jan 2 at 22:10
Ross MillikanRoss Millikan
293k23197371
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add a comment |
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1 Answer
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1 Answer
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oldest
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votes
$begingroup$
If you want to use a Poisson distribution, the expected number in five years is $1.5$. That is the $lambda$ parameter in the distribution, so the probability of $n$ earthquakes in $5$ years is
$$P(n)=frac {1.5^ne^{-1.5}}{n!}$$Now compute the probability of $0$ to $2$ earthquakes and subtract from $1$ to get the chance of at least $3$ earthquakes.
answered Jan 2 at 22:10
Ross MillikanRoss Millikan
293k23197371
$endgroup$
add a comment |
$begingroup$
If you want to use a Poisson distribution, the expected number in five years is $1.5$. That is the $lambda$ parameter in the distribution, so the probability of $n$ earthquakes in $5$ years is
$$P(n)=frac {1.5^ne^{-1.5}}{n!}$$Now compute the probability of $0$ to $2$ earthquakes and subtract from $1$ to get the chance of at least $3$ earthquakes.
answered Jan 2 at 22:10
Ross MillikanRoss Millikan
293k23197371
$endgroup$
add a comment |
$begingroup$
If you want to use a Poisson distribution, the expected number in five years is $1.5$. That is the $lambda$ parameter in the distribution, so the probability of $n$ earthquakes in $5$ years is
$$P(n)=frac {1.5^ne^{-1.5}}{n!}$$Now compute the probability of $0$ to $2$ earthquakes and subtract from $1$ to get the chance of at least $3$ earthquakes.
answered Jan 2 at 22:10
Ross MillikanRoss Millikan
293k23197371
$endgroup$
If you want to use a Poisson distribution, the expected number in five years is $1.5$. That is the $lambda$ parameter in the distribution, so the probability of $n$ earthquakes in $5$ years is
$$P(n)=frac {1.5^ne^{-1.5}}{n!}$$Now compute the probability of $0$ to $2$ earthquakes and subtract from $1$ to get the chance of at least $3$ earthquakes.
answered Jan 2 at 22:10
Ross MillikanRoss Millikan
293k23197371
answered Jan 2 at 22:10
Ross MillikanRoss Millikan
293k23197371
answered Jan 2 at 22:10
Ross MillikanRoss Millikan
293k23197371
answered Jan 2 at 22:10
Ross MillikanRoss Millikan
293k23197371
293k23197371
add a comment |
add a comment |
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2
$begingroup$
How could the probability be greater than $1$?
$endgroup$
– lulu
Jan 2 at 21:12
$begingroup$
It's certainly wrong. Probabilities must be between 0 and 1
$endgroup$
– Ray Bern
Jan 2 at 21:14
$begingroup$
Note: as stated the problem really can't be solved. We don't even have enough information to compute the probability that there are at least $3$ earthquakes in a single year.
$endgroup$
– lulu
Jan 2 at 21:16
$begingroup$
@lulu that's right
$endgroup$
– Murad Sh-ov
Jan 2 at 21:17
$begingroup$
Hint: Let $X$ be the number of earthquakes that happen in 5 years. Then, $P{Xgeq 3}=1-P{X<3}$.
$endgroup$
– Ray Bern
Jan 2 at 21:18