Mixing
Number of Injury claims per month
$begingroup$
The number of injury claims per month is given by $N$ where
$\$
$
P(N=n) =dfrac{1}{(n+1)(n+2)}
$
where $0 leq n$
Determine the probability of at least one claim during a particular month given that there have been at most four claims during that month.
I have this so far: You have 5 conditions, the 1st is no claims in a month, the 2nd is the 1 claim and so on. I've deicded that the easiest way to do this is to use the compliement that is, $PR($No claim| at most 4 claims$)$
Condition 1: $PR;(No; claim; | ;No; claim)=dfrac{1}{4}$
Condition 2: $PR;(No ; claim; | ;1 ; claim)= dfrac{1}{12}$
Condition 3: $PR;(No ; claim ; | ; 2; claims)= dfrac{1}{24} $
Condition 4: $PR;(No ; claim ; | ; 3 ; claims)= dfrac{1}{40}$
Condition 5: $PR;(No ; claim ; | ; 4 ; claims) = dfrac{1}{60}$
The sum of the probabilities will be my demoniator. I dont think I am going about this the right way since the answer is now here near what I got.
probability combinatorics statistics
asked Nov 30 '13 at 3:52
adamadam
799722
$endgroup$
add a comment |
$begingroup$
The number of injury claims per month is given by $N$ where
$\$
$
P(N=n) =dfrac{1}{(n+1)(n+2)}
$
where $0 leq n$
Determine the probability of at least one claim during a particular month given that there have been at most four claims during that month.
I have this so far: You have 5 conditions, the 1st is no claims in a month, the 2nd is the 1 claim and so on. I've deicded that the easiest way to do this is to use the compliement that is, $PR($No claim| at most 4 claims$)$
Condition 1: $PR;(No; claim; | ;No; claim)=dfrac{1}{4}$
Condition 2: $PR;(No ; claim; | ;1 ; claim)= dfrac{1}{12}$
Condition 3: $PR;(No ; claim ; | ; 2; claims)= dfrac{1}{24} $
Condition 4: $PR;(No ; claim ; | ; 3 ; claims)= dfrac{1}{40}$
Condition 5: $PR;(No ; claim ; | ; 4 ; claims) = dfrac{1}{60}$
The sum of the probabilities will be my demoniator. I dont think I am going about this the right way since the answer is now here near what I got.
probability combinatorics statistics
asked Nov 30 '13 at 3:52
adamadam
799722
$endgroup$
$begingroup$
This is a SOA practice problem. :) I'd be happy to spoil it for you, but are you sure you can't figure it out from the solutions they provide?
$endgroup$
– nomen
Nov 30 '13 at 5:06
$begingroup$
Oh yes I found it, it was actually quite simple. I overlooked the definition of conditional probability.
$endgroup$
– adam
Nov 30 '13 at 5:08
add a comment |
$begingroup$
The number of injury claims per month is given by $N$ where
$\$
$
P(N=n) =dfrac{1}{(n+1)(n+2)}
$
where $0 leq n$
Determine the probability of at least one claim during a particular month given that there have been at most four claims during that month.
I have this so far: You have 5 conditions, the 1st is no claims in a month, the 2nd is the 1 claim and so on. I've deicded that the easiest way to do this is to use the compliement that is, $PR($No claim| at most 4 claims$)$
Condition 1: $PR;(No; claim; | ;No; claim)=dfrac{1}{4}$
Condition 2: $PR;(No ; claim; | ;1 ; claim)= dfrac{1}{12}$
Condition 3: $PR;(No ; claim ; | ; 2; claims)= dfrac{1}{24} $
Condition 4: $PR;(No ; claim ; | ; 3 ; claims)= dfrac{1}{40}$
Condition 5: $PR;(No ; claim ; | ; 4 ; claims) = dfrac{1}{60}$
The sum of the probabilities will be my demoniator. I dont think I am going about this the right way since the answer is now here near what I got.
probability combinatorics statistics
asked Nov 30 '13 at 3:52
adamadam
799722
$endgroup$
The number of injury claims per month is given by $N$ where
$\$
$
P(N=n) =dfrac{1}{(n+1)(n+2)}
$
where $0 leq n$
Determine the probability of at least one claim during a particular month given that there have been at most four claims during that month.
I have this so far: You have 5 conditions, the 1st is no claims in a month, the 2nd is the 1 claim and so on. I've deicded that the easiest way to do this is to use the compliement that is, $PR($No claim| at most 4 claims$)$
Condition 1: $PR;(No; claim; | ;No; claim)=dfrac{1}{4}$
Condition 2: $PR;(No ; claim; | ;1 ; claim)= dfrac{1}{12}$
Condition 3: $PR;(No ; claim ; | ; 2; claims)= dfrac{1}{24} $
Condition 4: $PR;(No ; claim ; | ; 3 ; claims)= dfrac{1}{40}$
Condition 5: $PR;(No ; claim ; | ; 4 ; claims) = dfrac{1}{60}$
The sum of the probabilities will be my demoniator. I dont think I am going about this the right way since the answer is now here near what I got.
probability combinatorics statistics
probability combinatorics statistics
asked Nov 30 '13 at 3:52
adamadam
799722
asked Nov 30 '13 at 3:52
adamadam
799722
asked Nov 30 '13 at 3:52
adamadam
799722
asked Nov 30 '13 at 3:52
adamadam
799722
asked Nov 30 '13 at 3:52
adamadam
799722
799722
$begingroup$
This is a SOA practice problem. :) I'd be happy to spoil it for you, but are you sure you can't figure it out from the solutions they provide?
$endgroup$
– nomen
Nov 30 '13 at 5:06
$begingroup$
Oh yes I found it, it was actually quite simple. I overlooked the definition of conditional probability.
$endgroup$
– adam
Nov 30 '13 at 5:08
add a comment |
$begingroup$
This is a SOA practice problem. :) I'd be happy to spoil it for you, but are you sure you can't figure it out from the solutions they provide?
$endgroup$
– nomen
Nov 30 '13 at 5:06
$begingroup$
Oh yes I found it, it was actually quite simple. I overlooked the definition of conditional probability.
$endgroup$
– adam
Nov 30 '13 at 5:08
$begingroup$
This is a SOA practice problem. :) I'd be happy to spoil it for you, but are you sure you can't figure it out from the solutions they provide?
$endgroup$
– nomen
Nov 30 '13 at 5:06
$begingroup$
This is a SOA practice problem. :) I'd be happy to spoil it for you, but are you sure you can't figure it out from the solutions they provide?
$endgroup$
– nomen
Nov 30 '13 at 5:06
$begingroup$
Oh yes I found it, it was actually quite simple. I overlooked the definition of conditional probability.
$endgroup$
– adam
Nov 30 '13 at 5:08
$begingroup$
Oh yes I found it, it was actually quite simple. I overlooked the definition of conditional probability.
$endgroup$
– adam
Nov 30 '13 at 5:08
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I'll enumerate the mass function:
$$P(n) =
begin{cases}
frac{1}{2} text{ if } n = 0\
frac{1}{6} text{ if } n = 1\
frac{1}{12}text{ if } n = 2\
frac{1}{20}text{ if } n = 3\
frac{1}{30}text{ if } n = 4\
end{cases}
$$
This is a sufficiently "deep" enough enumeration, since, recalling the definition of the conditional probability,
$$
P(Xgeq 1 | X leq 4) = frac{P(1 leq X leq 4)}{P(Xleq 4)}\
= frac{ frac{1}{6} + frac{1}{12} + frac{1}{20} + frac{1}{30} }{ frac{1}{2} + frac{1}{6} + frac{1}{12} + frac{1}{20} + frac{1}{30} }
$$
answered Nov 30 '13 at 5:16
nomennomen
1,7431020
$endgroup$
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I'll enumerate the mass function:
$$P(n) =
begin{cases}
frac{1}{2} text{ if } n = 0\
frac{1}{6} text{ if } n = 1\
frac{1}{12}text{ if } n = 2\
frac{1}{20}text{ if } n = 3\
frac{1}{30}text{ if } n = 4\
end{cases}
$$
This is a sufficiently "deep" enough enumeration, since, recalling the definition of the conditional probability,
$$
P(Xgeq 1 | X leq 4) = frac{P(1 leq X leq 4)}{P(Xleq 4)}\
= frac{ frac{1}{6} + frac{1}{12} + frac{1}{20} + frac{1}{30} }{ frac{1}{2} + frac{1}{6} + frac{1}{12} + frac{1}{20} + frac{1}{30} }
$$
answered Nov 30 '13 at 5:16
nomennomen
1,7431020
$endgroup$
add a comment |
$begingroup$
I'll enumerate the mass function:
$$P(n) =
begin{cases}
frac{1}{2} text{ if } n = 0\
frac{1}{6} text{ if } n = 1\
frac{1}{12}text{ if } n = 2\
frac{1}{20}text{ if } n = 3\
frac{1}{30}text{ if } n = 4\
end{cases}
$$
This is a sufficiently "deep" enough enumeration, since, recalling the definition of the conditional probability,
$$
P(Xgeq 1 | X leq 4) = frac{P(1 leq X leq 4)}{P(Xleq 4)}\
= frac{ frac{1}{6} + frac{1}{12} + frac{1}{20} + frac{1}{30} }{ frac{1}{2} + frac{1}{6} + frac{1}{12} + frac{1}{20} + frac{1}{30} }
$$
answered Nov 30 '13 at 5:16
nomennomen
1,7431020
$endgroup$
add a comment |
$begingroup$
I'll enumerate the mass function:
$$P(n) =
begin{cases}
frac{1}{2} text{ if } n = 0\
frac{1}{6} text{ if } n = 1\
frac{1}{12}text{ if } n = 2\
frac{1}{20}text{ if } n = 3\
frac{1}{30}text{ if } n = 4\
end{cases}
$$
This is a sufficiently "deep" enough enumeration, since, recalling the definition of the conditional probability,
$$
P(Xgeq 1 | X leq 4) = frac{P(1 leq X leq 4)}{P(Xleq 4)}\
= frac{ frac{1}{6} + frac{1}{12} + frac{1}{20} + frac{1}{30} }{ frac{1}{2} + frac{1}{6} + frac{1}{12} + frac{1}{20} + frac{1}{30} }
$$
answered Nov 30 '13 at 5:16
nomennomen
1,7431020
$endgroup$
I'll enumerate the mass function:
$$P(n) =
begin{cases}
frac{1}{2} text{ if } n = 0\
frac{1}{6} text{ if } n = 1\
frac{1}{12}text{ if } n = 2\
frac{1}{20}text{ if } n = 3\
frac{1}{30}text{ if } n = 4\
end{cases}
$$
This is a sufficiently "deep" enough enumeration, since, recalling the definition of the conditional probability,
$$
P(Xgeq 1 | X leq 4) = frac{P(1 leq X leq 4)}{P(Xleq 4)}\
= frac{ frac{1}{6} + frac{1}{12} + frac{1}{20} + frac{1}{30} }{ frac{1}{2} + frac{1}{6} + frac{1}{12} + frac{1}{20} + frac{1}{30} }
$$
answered Nov 30 '13 at 5:16
nomennomen
1,7431020
answered Nov 30 '13 at 5:16
nomennomen
1,7431020
answered Nov 30 '13 at 5:16
nomennomen
1,7431020
answered Nov 30 '13 at 5:16
nomennomen
1,7431020
1,7431020
add a comment |
add a comment |
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$begingroup$
This is a SOA practice problem. :) I'd be happy to spoil it for you, but are you sure you can't figure it out from the solutions they provide?
$endgroup$
– nomen
Nov 30 '13 at 5:06
$begingroup$
Oh yes I found it, it was actually quite simple. I overlooked the definition of conditional probability.
$endgroup$
– adam
Nov 30 '13 at 5:08