Mixing
How do you calculate a probability given multiple probabilities?
$begingroup$
Assume that the following are the probabilities (P(S)) of runner's chance of finishing a race within S minutes
S 4 6 8 9 10
P(S) .10 .40 .20 .15 .15
If there are 50 runners, what is the probability that none will take more than 7 minutes?
What kind of problem is this and how do I approach it?
probability
asked Feb 6 '14 at 1:40
user3277633user3277633
1278
$endgroup$
add a comment |
$begingroup$
Assume that the following are the probabilities (P(S)) of runner's chance of finishing a race within S minutes
S 4 6 8 9 10
P(S) .10 .40 .20 .15 .15
If there are 50 runners, what is the probability that none will take more than 7 minutes?
What kind of problem is this and how do I approach it?
probability
asked Feb 6 '14 at 1:40
user3277633user3277633
1278
$endgroup$
$begingroup$
Splitting the possible outcomes into two groups (up to 7 minutes and less than 7 minutes) makes this a binomial distribution problem. The sections called "Specification" and "Example" on the Wikipedia entry will give you an overview. If you're not familiar with the factorial of a number ($n! = ncdot(n-1)cdot(n-2)cdotdotscdot(2)cdot(1)$), you're basically just multiplying a bunch of numbers together. The result can get pretty big pretty fast, but otherwise the calculation shouldn't get too hairy.
$endgroup$
– dmk
Feb 6 '14 at 3:37
add a comment |
$begingroup$
Assume that the following are the probabilities (P(S)) of runner's chance of finishing a race within S minutes
S 4 6 8 9 10
P(S) .10 .40 .20 .15 .15
If there are 50 runners, what is the probability that none will take more than 7 minutes?
What kind of problem is this and how do I approach it?
probability
asked Feb 6 '14 at 1:40
user3277633user3277633
1278
$endgroup$
Assume that the following are the probabilities (P(S)) of runner's chance of finishing a race within S minutes
S 4 6 8 9 10
P(S) .10 .40 .20 .15 .15
If there are 50 runners, what is the probability that none will take more than 7 minutes?
What kind of problem is this and how do I approach it?
probability
probability
asked Feb 6 '14 at 1:40
user3277633user3277633
1278
asked Feb 6 '14 at 1:40
user3277633user3277633
1278
edited Feb 6 '14 at 2:37
user940
asked Feb 6 '14 at 1:40
user3277633user3277633
1278
asked Feb 6 '14 at 1:40
user3277633user3277633
1278
asked Feb 6 '14 at 1:40
user3277633user3277633
1278
1278
$begingroup$
Splitting the possible outcomes into two groups (up to 7 minutes and less than 7 minutes) makes this a binomial distribution problem. The sections called "Specification" and "Example" on the Wikipedia entry will give you an overview. If you're not familiar with the factorial of a number ($n! = ncdot(n-1)cdot(n-2)cdotdotscdot(2)cdot(1)$), you're basically just multiplying a bunch of numbers together. The result can get pretty big pretty fast, but otherwise the calculation shouldn't get too hairy.
$endgroup$
– dmk
Feb 6 '14 at 3:37
add a comment |
$begingroup$
Splitting the possible outcomes into two groups (up to 7 minutes and less than 7 minutes) makes this a binomial distribution problem. The sections called "Specification" and "Example" on the Wikipedia entry will give you an overview. If you're not familiar with the factorial of a number ($n! = ncdot(n-1)cdot(n-2)cdotdotscdot(2)cdot(1)$), you're basically just multiplying a bunch of numbers together. The result can get pretty big pretty fast, but otherwise the calculation shouldn't get too hairy.
$endgroup$
– dmk
Feb 6 '14 at 3:37
$begingroup$
Splitting the possible outcomes into two groups (up to 7 minutes and less than 7 minutes) makes this a binomial distribution problem. The sections called "Specification" and "Example" on the Wikipedia entry will give you an overview. If you're not familiar with the factorial of a number ($n! = ncdot(n-1)cdot(n-2)cdotdotscdot(2)cdot(1)$), you're basically just multiplying a bunch of numbers together. The result can get pretty big pretty fast, but otherwise the calculation shouldn't get too hairy.
$endgroup$
– dmk
Feb 6 '14 at 3:37
$begingroup$
Splitting the possible outcomes into two groups (up to 7 minutes and less than 7 minutes) makes this a binomial distribution problem. The sections called "Specification" and "Example" on the Wikipedia entry will give you an overview. If you're not familiar with the factorial of a number ($n! = ncdot(n-1)cdot(n-2)cdotdotscdot(2)cdot(1)$), you're basically just multiplying a bunch of numbers together. The result can get pretty big pretty fast, but otherwise the calculation shouldn't get too hairy.
$endgroup$
– dmk
Feb 6 '14 at 3:37
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
First you identify the probability of one person finishing in less tha 7 minutes which would be if the finish in 4 or 6 minutes, the probability of this is .5
P(S<7) = P(4)+P(6) = 0.5
Now this is only for one person, so the next person would have also the same probability, so for both of them to finish in less than 7 minutes would be (.5)(.5) = .25 and so on
so for the fifty runners to finish in less than 7 mins it would be
(.5)^50 = 0.00000000000000088817842
Which means it's pretty unlikely that it happens
To better understand how unlikely it is, is like throwing a coin 50 times in a row and have the same result every time
answered Feb 6 '14 at 1:54
Mark EMark E
11016
$endgroup$
$begingroup$
how would i change the formula if I just want to find what's the prob of exactly one runner who ran under 7 minute?
$endgroup$
– user3277633
Feb 6 '14 at 3:09
$begingroup$
The prob of one runner who ran under 7 mins is the first we calculated:"P(S<7) = P(4)+P(6) = 0.5"
$endgroup$
– Mark E
Feb 6 '14 at 3:17
$begingroup$
But some runners may finish in less than 7 minutes and be in the $6 < t < 8$ group...
$endgroup$
– Richard Ambler
Jan 25 at 4:31
add a comment |
$begingroup$
I think that the original problem is not worded very carefully. For example, if a runner makes it within 4 minutes then they also make it within 10 minutes, right?
It might be clearer to express the given information in terms of the running time for an individual runner $t$ as:
$$begin{align}
P(t<0)&=0\
P(0<t<4)&=0.10\
P(4 < t < 6)&=0.40\
P(6 < t < 8)&=0.20\
P(8 < t < 9)&=0.15\
P(9 < t < 10)&=0.15\
P(t>10)&=0
end{align}$$
(With $t$ a continuous variable so $P(t=k)=0$ for any $k$ in the domain.)
Unfortunately, if we're interested in finding $P(t<7)$, we're going to have to make an assumption about how the probability is distributed in the $6<t<8$ region of the domain in order to make an educated guess.
For example, we might simply assume $P(6<t<7)=P(7<t<8)=0.10$. In this case, $P(t<7)=0.10+0.40+0.10=0.60$.
Now that we have the probability that a single runner will make it in under 7 minutes we can move on to the problem that all of 50 runners will make it under 7 minutes.
As mentioned in a previous answer, the number of runners who make it under 7 minutes (assuming independence, etc.) would be modeled using the binomial distribution:
$$Xsimtext B(50, 0.60)Rightarrow P(X=x)={50choose x}(0.60)^x(0.40)^{50-x}$$
where $X$ represents the number of runners who make it in under 7 minutes where the probability that any runner will make it in under 7 minutes is $0.60$.
For $x=50$ (the number of runners who make it under 7 minutes is all 50), this gives us:
$$P(X=50)=(0.60)^{50}=8.1times 10^{-12}$$
(which is pretty darn close to impossible).
answered Jan 25 at 5:05
Richard AmblerRichard Ambler
1,308615
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
First you identify the probability of one person finishing in less tha 7 minutes which would be if the finish in 4 or 6 minutes, the probability of this is .5
P(S<7) = P(4)+P(6) = 0.5
Now this is only for one person, so the next person would have also the same probability, so for both of them to finish in less than 7 minutes would be (.5)(.5) = .25 and so on
so for the fifty runners to finish in less than 7 mins it would be
(.5)^50 = 0.00000000000000088817842
Which means it's pretty unlikely that it happens
To better understand how unlikely it is, is like throwing a coin 50 times in a row and have the same result every time
answered Feb 6 '14 at 1:54
Mark EMark E
11016
$endgroup$
$begingroup$
how would i change the formula if I just want to find what's the prob of exactly one runner who ran under 7 minute?
$endgroup$
– user3277633
Feb 6 '14 at 3:09
$begingroup$
The prob of one runner who ran under 7 mins is the first we calculated:"P(S<7) = P(4)+P(6) = 0.5"
$endgroup$
– Mark E
Feb 6 '14 at 3:17
$begingroup$
But some runners may finish in less than 7 minutes and be in the $6 < t < 8$ group...
$endgroup$
– Richard Ambler
Jan 25 at 4:31
add a comment |
$begingroup$
First you identify the probability of one person finishing in less tha 7 minutes which would be if the finish in 4 or 6 minutes, the probability of this is .5
P(S<7) = P(4)+P(6) = 0.5
Now this is only for one person, so the next person would have also the same probability, so for both of them to finish in less than 7 minutes would be (.5)(.5) = .25 and so on
so for the fifty runners to finish in less than 7 mins it would be
(.5)^50 = 0.00000000000000088817842
Which means it's pretty unlikely that it happens
To better understand how unlikely it is, is like throwing a coin 50 times in a row and have the same result every time
answered Feb 6 '14 at 1:54
Mark EMark E
11016
$endgroup$
$begingroup$
how would i change the formula if I just want to find what's the prob of exactly one runner who ran under 7 minute?
$endgroup$
– user3277633
Feb 6 '14 at 3:09
$begingroup$
The prob of one runner who ran under 7 mins is the first we calculated:"P(S<7) = P(4)+P(6) = 0.5"
$endgroup$
– Mark E
Feb 6 '14 at 3:17
$begingroup$
But some runners may finish in less than 7 minutes and be in the $6 < t < 8$ group...
$endgroup$
– Richard Ambler
Jan 25 at 4:31
add a comment |
$begingroup$
First you identify the probability of one person finishing in less tha 7 minutes which would be if the finish in 4 or 6 minutes, the probability of this is .5
P(S<7) = P(4)+P(6) = 0.5
Now this is only for one person, so the next person would have also the same probability, so for both of them to finish in less than 7 minutes would be (.5)(.5) = .25 and so on
so for the fifty runners to finish in less than 7 mins it would be
(.5)^50 = 0.00000000000000088817842
Which means it's pretty unlikely that it happens
To better understand how unlikely it is, is like throwing a coin 50 times in a row and have the same result every time
answered Feb 6 '14 at 1:54
Mark EMark E
11016
$endgroup$
First you identify the probability of one person finishing in less tha 7 minutes which would be if the finish in 4 or 6 minutes, the probability of this is .5
P(S<7) = P(4)+P(6) = 0.5
Now this is only for one person, so the next person would have also the same probability, so for both of them to finish in less than 7 minutes would be (.5)(.5) = .25 and so on
so for the fifty runners to finish in less than 7 mins it would be
(.5)^50 = 0.00000000000000088817842
Which means it's pretty unlikely that it happens
To better understand how unlikely it is, is like throwing a coin 50 times in a row and have the same result every time
answered Feb 6 '14 at 1:54
Mark EMark E
11016
answered Feb 6 '14 at 1:54
Mark EMark E
11016
answered Feb 6 '14 at 1:54
Mark EMark E
11016
answered Feb 6 '14 at 1:54
Mark EMark E
11016
11016
$begingroup$
how would i change the formula if I just want to find what's the prob of exactly one runner who ran under 7 minute?
$endgroup$
– user3277633
Feb 6 '14 at 3:09
$begingroup$
The prob of one runner who ran under 7 mins is the first we calculated:"P(S<7) = P(4)+P(6) = 0.5"
$endgroup$
– Mark E
Feb 6 '14 at 3:17
$begingroup$
But some runners may finish in less than 7 minutes and be in the $6 < t < 8$ group...
$endgroup$
– Richard Ambler
Jan 25 at 4:31
add a comment |
$begingroup$
how would i change the formula if I just want to find what's the prob of exactly one runner who ran under 7 minute?
$endgroup$
– user3277633
Feb 6 '14 at 3:09
$begingroup$
The prob of one runner who ran under 7 mins is the first we calculated:"P(S<7) = P(4)+P(6) = 0.5"
$endgroup$
– Mark E
Feb 6 '14 at 3:17
$begingroup$
But some runners may finish in less than 7 minutes and be in the $6 < t < 8$ group...
$endgroup$
– Richard Ambler
Jan 25 at 4:31
$begingroup$
how would i change the formula if I just want to find what's the prob of exactly one runner who ran under 7 minute?
$endgroup$
– user3277633
Feb 6 '14 at 3:09
$begingroup$
how would i change the formula if I just want to find what's the prob of exactly one runner who ran under 7 minute?
$endgroup$
– user3277633
Feb 6 '14 at 3:09
$begingroup$
The prob of one runner who ran under 7 mins is the first we calculated:"P(S<7) = P(4)+P(6) = 0.5"
$endgroup$
– Mark E
Feb 6 '14 at 3:17
$begingroup$
The prob of one runner who ran under 7 mins is the first we calculated:"P(S<7) = P(4)+P(6) = 0.5"
$endgroup$
– Mark E
Feb 6 '14 at 3:17
$begingroup$
But some runners may finish in less than 7 minutes and be in the $6 < t < 8$ group...
$endgroup$
– Richard Ambler
Jan 25 at 4:31
$begingroup$
But some runners may finish in less than 7 minutes and be in the $6 < t < 8$ group...
$endgroup$
– Richard Ambler
Jan 25 at 4:31
add a comment |
$begingroup$
I think that the original problem is not worded very carefully. For example, if a runner makes it within 4 minutes then they also make it within 10 minutes, right?
It might be clearer to express the given information in terms of the running time for an individual runner $t$ as:
$$begin{align}
P(t<0)&=0\
P(0<t<4)&=0.10\
P(4 < t < 6)&=0.40\
P(6 < t < 8)&=0.20\
P(8 < t < 9)&=0.15\
P(9 < t < 10)&=0.15\
P(t>10)&=0
end{align}$$
(With $t$ a continuous variable so $P(t=k)=0$ for any $k$ in the domain.)
Unfortunately, if we're interested in finding $P(t<7)$, we're going to have to make an assumption about how the probability is distributed in the $6<t<8$ region of the domain in order to make an educated guess.
For example, we might simply assume $P(6<t<7)=P(7<t<8)=0.10$. In this case, $P(t<7)=0.10+0.40+0.10=0.60$.
Now that we have the probability that a single runner will make it in under 7 minutes we can move on to the problem that all of 50 runners will make it under 7 minutes.
As mentioned in a previous answer, the number of runners who make it under 7 minutes (assuming independence, etc.) would be modeled using the binomial distribution:
$$Xsimtext B(50, 0.60)Rightarrow P(X=x)={50choose x}(0.60)^x(0.40)^{50-x}$$
where $X$ represents the number of runners who make it in under 7 minutes where the probability that any runner will make it in under 7 minutes is $0.60$.
For $x=50$ (the number of runners who make it under 7 minutes is all 50), this gives us:
$$P(X=50)=(0.60)^{50}=8.1times 10^{-12}$$
(which is pretty darn close to impossible).
answered Jan 25 at 5:05
Richard AmblerRichard Ambler
1,308615
$endgroup$
add a comment |
$begingroup$
I think that the original problem is not worded very carefully. For example, if a runner makes it within 4 minutes then they also make it within 10 minutes, right?
It might be clearer to express the given information in terms of the running time for an individual runner $t$ as:
$$begin{align}
P(t<0)&=0\
P(0<t<4)&=0.10\
P(4 < t < 6)&=0.40\
P(6 < t < 8)&=0.20\
P(8 < t < 9)&=0.15\
P(9 < t < 10)&=0.15\
P(t>10)&=0
end{align}$$
(With $t$ a continuous variable so $P(t=k)=0$ for any $k$ in the domain.)
Unfortunately, if we're interested in finding $P(t<7)$, we're going to have to make an assumption about how the probability is distributed in the $6<t<8$ region of the domain in order to make an educated guess.
For example, we might simply assume $P(6<t<7)=P(7<t<8)=0.10$. In this case, $P(t<7)=0.10+0.40+0.10=0.60$.
Now that we have the probability that a single runner will make it in under 7 minutes we can move on to the problem that all of 50 runners will make it under 7 minutes.
As mentioned in a previous answer, the number of runners who make it under 7 minutes (assuming independence, etc.) would be modeled using the binomial distribution:
$$Xsimtext B(50, 0.60)Rightarrow P(X=x)={50choose x}(0.60)^x(0.40)^{50-x}$$
where $X$ represents the number of runners who make it in under 7 minutes where the probability that any runner will make it in under 7 minutes is $0.60$.
For $x=50$ (the number of runners who make it under 7 minutes is all 50), this gives us:
$$P(X=50)=(0.60)^{50}=8.1times 10^{-12}$$
(which is pretty darn close to impossible).
answered Jan 25 at 5:05
Richard AmblerRichard Ambler
1,308615
$endgroup$
add a comment |
$begingroup$
I think that the original problem is not worded very carefully. For example, if a runner makes it within 4 minutes then they also make it within 10 minutes, right?
It might be clearer to express the given information in terms of the running time for an individual runner $t$ as:
$$begin{align}
P(t<0)&=0\
P(0<t<4)&=0.10\
P(4 < t < 6)&=0.40\
P(6 < t < 8)&=0.20\
P(8 < t < 9)&=0.15\
P(9 < t < 10)&=0.15\
P(t>10)&=0
end{align}$$
(With $t$ a continuous variable so $P(t=k)=0$ for any $k$ in the domain.)
Unfortunately, if we're interested in finding $P(t<7)$, we're going to have to make an assumption about how the probability is distributed in the $6<t<8$ region of the domain in order to make an educated guess.
For example, we might simply assume $P(6<t<7)=P(7<t<8)=0.10$. In this case, $P(t<7)=0.10+0.40+0.10=0.60$.
Now that we have the probability that a single runner will make it in under 7 minutes we can move on to the problem that all of 50 runners will make it under 7 minutes.
As mentioned in a previous answer, the number of runners who make it under 7 minutes (assuming independence, etc.) would be modeled using the binomial distribution:
$$Xsimtext B(50, 0.60)Rightarrow P(X=x)={50choose x}(0.60)^x(0.40)^{50-x}$$
where $X$ represents the number of runners who make it in under 7 minutes where the probability that any runner will make it in under 7 minutes is $0.60$.
For $x=50$ (the number of runners who make it under 7 minutes is all 50), this gives us:
$$P(X=50)=(0.60)^{50}=8.1times 10^{-12}$$
(which is pretty darn close to impossible).
answered Jan 25 at 5:05
Richard AmblerRichard Ambler
1,308615
$endgroup$
I think that the original problem is not worded very carefully. For example, if a runner makes it within 4 minutes then they also make it within 10 minutes, right?
It might be clearer to express the given information in terms of the running time for an individual runner $t$ as:
$$begin{align}
P(t<0)&=0\
P(0<t<4)&=0.10\
P(4 < t < 6)&=0.40\
P(6 < t < 8)&=0.20\
P(8 < t < 9)&=0.15\
P(9 < t < 10)&=0.15\
P(t>10)&=0
end{align}$$
(With $t$ a continuous variable so $P(t=k)=0$ for any $k$ in the domain.)
Unfortunately, if we're interested in finding $P(t<7)$, we're going to have to make an assumption about how the probability is distributed in the $6<t<8$ region of the domain in order to make an educated guess.
For example, we might simply assume $P(6<t<7)=P(7<t<8)=0.10$. In this case, $P(t<7)=0.10+0.40+0.10=0.60$.
Now that we have the probability that a single runner will make it in under 7 minutes we can move on to the problem that all of 50 runners will make it under 7 minutes.
As mentioned in a previous answer, the number of runners who make it under 7 minutes (assuming independence, etc.) would be modeled using the binomial distribution:
$$Xsimtext B(50, 0.60)Rightarrow P(X=x)={50choose x}(0.60)^x(0.40)^{50-x}$$
where $X$ represents the number of runners who make it in under 7 minutes where the probability that any runner will make it in under 7 minutes is $0.60$.
For $x=50$ (the number of runners who make it under 7 minutes is all 50), this gives us:
$$P(X=50)=(0.60)^{50}=8.1times 10^{-12}$$
(which is pretty darn close to impossible).
answered Jan 25 at 5:05
Richard AmblerRichard Ambler
1,308615
answered Jan 25 at 5:05
Richard AmblerRichard Ambler
1,308615
answered Jan 25 at 5:05
Richard AmblerRichard Ambler
1,308615
answered Jan 25 at 5:05
Richard AmblerRichard Ambler
1,308615
1,308615
add a comment |
add a comment |
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$begingroup$
Splitting the possible outcomes into two groups (up to 7 minutes and less than 7 minutes) makes this a binomial distribution problem. The sections called "Specification" and "Example" on the Wikipedia entry will give you an overview. If you're not familiar with the factorial of a number ($n! = ncdot(n-1)cdot(n-2)cdotdotscdot(2)cdot(1)$), you're basically just multiplying a bunch of numbers together. The result can get pretty big pretty fast, but otherwise the calculation shouldn't get too hairy.
$endgroup$
– dmk
Feb 6 '14 at 3:37