Conditional probability- what is wrong with my understanding
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From Ostaszewski, 2007:
An insurance policy covers two employees of a company. The policy will reimburse no more than one loss per employee per year. It reimburses the full amount of the loss up to a company wide maximum of $8000$. The probability of an employee incurring a loss is $40$%, and is independent of the other employee's losses. The amount of each loss is uniformly distributed on $(1000, 5000)$. Given that one of the employees has incurred a loss in excess of $2000$, determine the probability that losses will exceed reimbursements.
The way to solve this is to make a square with vertices $(1000,1000), (5000,1000), (5000,5000),$ and $(1000,5000)$.
Then, to find the condition that one of the employees incurred $text {loss} >2000$, one must find the area that goes from $2000$ to $5000$ on one of the axes, and from $1000$ to $5000$ on the other, for a total area of $3000cdot 4000$ in the denominator.
The probability that losses exceed reimbursement is $P(X+Y>8000)$, so the area is the triangle with vertices $(3000,5000), (5000,5000),$ and $(5000,3000)$, and so the numerator will be ${1over2}cdot 2000^2$.
What is confusing me is how to account for fact that the probability of an employee incurring a loss is $40$%. I would have thought that the total probability of at least one employee incurring a loss is $.40cdot.60 +.60cdot.40 + .40cdot.40$, and the probability that both employees incur a loss (which is necessary for losses to exceed reimbursements) is $.40 cdot .40$, and therefore the solution should be
$$.40 cdot .40cdot {1over2} cdot 2000^2 over (.40cdot.60 +.60cdot.40 + .40cdot.40)cdot 3000 cdot 4000$$
However the solution says to just do:
$$.40cdot {1over2} cdot 2000^2 over 3000 cdot 4000$$
What is wrong with how I wanted to do it?
probability statistics
asked Dec 31 '18 at 22:46
agbltagblt
348114
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add a comment |
$begingroup$
From Ostaszewski, 2007:
An insurance policy covers two employees of a company. The policy will reimburse no more than one loss per employee per year. It reimburses the full amount of the loss up to a company wide maximum of $8000$. The probability of an employee incurring a loss is $40$%, and is independent of the other employee's losses. The amount of each loss is uniformly distributed on $(1000, 5000)$. Given that one of the employees has incurred a loss in excess of $2000$, determine the probability that losses will exceed reimbursements.
The way to solve this is to make a square with vertices $(1000,1000), (5000,1000), (5000,5000),$ and $(1000,5000)$.
Then, to find the condition that one of the employees incurred $text {loss} >2000$, one must find the area that goes from $2000$ to $5000$ on one of the axes, and from $1000$ to $5000$ on the other, for a total area of $3000cdot 4000$ in the denominator.
The probability that losses exceed reimbursement is $P(X+Y>8000)$, so the area is the triangle with vertices $(3000,5000), (5000,5000),$ and $(5000,3000)$, and so the numerator will be ${1over2}cdot 2000^2$.
What is confusing me is how to account for fact that the probability of an employee incurring a loss is $40$%. I would have thought that the total probability of at least one employee incurring a loss is $.40cdot.60 +.60cdot.40 + .40cdot.40$, and the probability that both employees incur a loss (which is necessary for losses to exceed reimbursements) is $.40 cdot .40$, and therefore the solution should be
$$.40 cdot .40cdot {1over2} cdot 2000^2 over (.40cdot.60 +.60cdot.40 + .40cdot.40)cdot 3000 cdot 4000$$
However the solution says to just do:
$$.40cdot {1over2} cdot 2000^2 over 3000 cdot 4000$$
What is wrong with how I wanted to do it?
probability statistics
asked Dec 31 '18 at 22:46
agbltagblt
348114
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$P(A|B)=frac{P(Acap B)}{P(B)}$
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– John Douma
Dec 31 '18 at 23:30
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Is $P(B)$ not "given that at least one employee incurred a loss", and therefore should include where $X$ incurs a loss and not $Y$, where $Y$ incurs a loss and not $X$, and where both incur a loss?
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– agblt
Dec 31 '18 at 23:37
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$B$ is the event that one of $X$ or $Y$ incurs a loss in excess of $2000$.
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– John Douma
Dec 31 '18 at 23:49
add a comment |
$begingroup$
From Ostaszewski, 2007:
An insurance policy covers two employees of a company. The policy will reimburse no more than one loss per employee per year. It reimburses the full amount of the loss up to a company wide maximum of $8000$. The probability of an employee incurring a loss is $40$%, and is independent of the other employee's losses. The amount of each loss is uniformly distributed on $(1000, 5000)$. Given that one of the employees has incurred a loss in excess of $2000$, determine the probability that losses will exceed reimbursements.
The way to solve this is to make a square with vertices $(1000,1000), (5000,1000), (5000,5000),$ and $(1000,5000)$.
Then, to find the condition that one of the employees incurred $text {loss} >2000$, one must find the area that goes from $2000$ to $5000$ on one of the axes, and from $1000$ to $5000$ on the other, for a total area of $3000cdot 4000$ in the denominator.
The probability that losses exceed reimbursement is $P(X+Y>8000)$, so the area is the triangle with vertices $(3000,5000), (5000,5000),$ and $(5000,3000)$, and so the numerator will be ${1over2}cdot 2000^2$.
What is confusing me is how to account for fact that the probability of an employee incurring a loss is $40$%. I would have thought that the total probability of at least one employee incurring a loss is $.40cdot.60 +.60cdot.40 + .40cdot.40$, and the probability that both employees incur a loss (which is necessary for losses to exceed reimbursements) is $.40 cdot .40$, and therefore the solution should be
$$.40 cdot .40cdot {1over2} cdot 2000^2 over (.40cdot.60 +.60cdot.40 + .40cdot.40)cdot 3000 cdot 4000$$
However the solution says to just do:
$$.40cdot {1over2} cdot 2000^2 over 3000 cdot 4000$$
What is wrong with how I wanted to do it?
probability statistics
asked Dec 31 '18 at 22:46
agbltagblt
348114
$endgroup$
From Ostaszewski, 2007:
An insurance policy covers two employees of a company. The policy will reimburse no more than one loss per employee per year. It reimburses the full amount of the loss up to a company wide maximum of $8000$. The probability of an employee incurring a loss is $40$%, and is independent of the other employee's losses. The amount of each loss is uniformly distributed on $(1000, 5000)$. Given that one of the employees has incurred a loss in excess of $2000$, determine the probability that losses will exceed reimbursements.
The way to solve this is to make a square with vertices $(1000,1000), (5000,1000), (5000,5000),$ and $(1000,5000)$.
Then, to find the condition that one of the employees incurred $text {loss} >2000$, one must find the area that goes from $2000$ to $5000$ on one of the axes, and from $1000$ to $5000$ on the other, for a total area of $3000cdot 4000$ in the denominator.
The probability that losses exceed reimbursement is $P(X+Y>8000)$, so the area is the triangle with vertices $(3000,5000), (5000,5000),$ and $(5000,3000)$, and so the numerator will be ${1over2}cdot 2000^2$.
What is confusing me is how to account for fact that the probability of an employee incurring a loss is $40$%. I would have thought that the total probability of at least one employee incurring a loss is $.40cdot.60 +.60cdot.40 + .40cdot.40$, and the probability that both employees incur a loss (which is necessary for losses to exceed reimbursements) is $.40 cdot .40$, and therefore the solution should be
$$.40 cdot .40cdot {1over2} cdot 2000^2 over (.40cdot.60 +.60cdot.40 + .40cdot.40)cdot 3000 cdot 4000$$
However the solution says to just do:
$$.40cdot {1over2} cdot 2000^2 over 3000 cdot 4000$$
What is wrong with how I wanted to do it?
probability statistics
probability statistics
asked Dec 31 '18 at 22:46
agbltagblt
348114
asked Dec 31 '18 at 22:46
agbltagblt
348114
asked Dec 31 '18 at 22:46
agbltagblt
348114
asked Dec 31 '18 at 22:46
agbltagblt
348114
asked Dec 31 '18 at 22:46
agbltagblt
348114
348114
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$P(A|B)=frac{P(Acap B)}{P(B)}$
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– John Douma
Dec 31 '18 at 23:30
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Is $P(B)$ not "given that at least one employee incurred a loss", and therefore should include where $X$ incurs a loss and not $Y$, where $Y$ incurs a loss and not $X$, and where both incur a loss?
$endgroup$
– agblt
Dec 31 '18 at 23:37
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$B$ is the event that one of $X$ or $Y$ incurs a loss in excess of $2000$.
$endgroup$
– John Douma
Dec 31 '18 at 23:49
add a comment |
$begingroup$
$P(A|B)=frac{P(Acap B)}{P(B)}$
$endgroup$
– John Douma
Dec 31 '18 at 23:30
$begingroup$
Is $P(B)$ not "given that at least one employee incurred a loss", and therefore should include where $X$ incurs a loss and not $Y$, where $Y$ incurs a loss and not $X$, and where both incur a loss?
$endgroup$
– agblt
Dec 31 '18 at 23:37
$begingroup$
$B$ is the event that one of $X$ or $Y$ incurs a loss in excess of $2000$.
$endgroup$
– John Douma
Dec 31 '18 at 23:49
$begingroup$
$P(A|B)=frac{P(Acap B)}{P(B)}$
$endgroup$
– John Douma
Dec 31 '18 at 23:30
$begingroup$
$P(A|B)=frac{P(Acap B)}{P(B)}$
$endgroup$
– John Douma
Dec 31 '18 at 23:30
$begingroup$
Is $P(B)$ not "given that at least one employee incurred a loss", and therefore should include where $X$ incurs a loss and not $Y$, where $Y$ incurs a loss and not $X$, and where both incur a loss?
$endgroup$
– agblt
Dec 31 '18 at 23:37
$begingroup$
Is $P(B)$ not "given that at least one employee incurred a loss", and therefore should include where $X$ incurs a loss and not $Y$, where $Y$ incurs a loss and not $X$, and where both incur a loss?
$endgroup$
– agblt
Dec 31 '18 at 23:37
$begingroup$
$B$ is the event that one of $X$ or $Y$ incurs a loss in excess of $2000$.
$endgroup$
– John Douma
Dec 31 '18 at 23:49
$begingroup$
$B$ is the event that one of $X$ or $Y$ incurs a loss in excess of $2000$.
$endgroup$
– John Douma
Dec 31 '18 at 23:49
add a comment |
3 Answers
3
active
oldest
votes
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I think there is some ambiguity in the question. I see two interpretations:
Interpretation (1): (This is how I would have interpreted the question.)
The probability that at least one employee incurred a loss in excess of $2,000$ is
$P[L_1 > 2000 text{ or } L_2 > 2000] = 1-P[L_1 le 2000 text{ and } L_2 le 2000] =1-(0.6+{2000-1000 over 5000-1000}0.4)^2$.
The probability that the loss exceeds $8,000$ can be computed graphically as
$P[L_1+L_2 > 8000] ={1 over 2} P[L_1 > 3000]^2={1 over 2} ({5000-3000 over 5000-1000}0.4)^2$.
Note that the second event implies the first.
Hence
$P[L_1+L_2 > 8000 | L_1 > 2000 text{ or } L_2 > 2000] = {P[L_1+L_2 > 8000 ] over P[L_1 > 2000 text{ or } L_2 > 2000] } = {2 over 51}$.
Interpretation (2): (This matches your solution above.)
We have $P[L_1 > 2000] = 1-(0.6+{2000-1000 over 5000-1000}0.4)$. Again, this event
implies the first so
$P[L_1+L_2 > 8000 | L_1 > 2000] = {P[L_1+L_2 > 8000 ] over P[L_1 > 2000 ]} = {1 over 15}$.
answered Jan 1 at 2:18
copper.hatcopper.hat
128k559161
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The first interpretation makes sense to me, and is indeed how I was understanding, although I got the probabilities wrong. I am not sure if I understand the second interpretation, however. Do the words "Given that one of the employees has incurred a loss..." not mean the same thing as "Given that at least one of the employees has incurred a loss..."?
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– agblt
Jan 1 at 4:51
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@agblt: Unfortunately, probability questions often expose a gap between the precision of mathematical statements and the English language. On can interpret 'Given that one...' as meaning that we know that a specific one of the employees has incurred... I picked $L_1$ as the specific employee. From a 'practical' standpoint this interpretation is a bit weak as if we knew this to be true, we would likely know the specific amount that $L_1$ has incurred also.
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– copper.hat
Jan 1 at 9:18
add a comment |
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You are given the condition that one employee incurs a loss so the $0.4$ probability only applies once for the other employee.
In terms of graphical representation, the red triangle represents the condition of exceeding $8000$, while the whole rectangle the total possible outcomes.
$P$ is therefore the area of the red triangle divided by the area of the rectangle times the probably of the second employee incurring a loss.
This is $$P = frac{0.5cdot 2000^2}{3000cdot 4000}cdot 0.4 = .06667$$
answered Jan 1 at 1:10
Phil HPhil H
4,2682312
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Maybe what is confusing me is this: What exactly is the probability that one of the employees has incurred a loss in excess of $2000$? I am getting $(.40cdot.60 +.60cdot.40 + .40cdot.40)cdot{{ 3000 cdot 4000}over 5000^2}$. Do you have something different?
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– agblt
Jan 1 at 1:22
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I don't see how the probability of one employee incurring a loss greater than 2000 comes into it. The given condition of this (one employee incurring a loss > 2000) just starts the vertical height of the rectangle at 2000 instead of 1000.
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– Phil H
Jan 1 at 1:35
add a comment |
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This is not intended as a complete answer (you already have a good one) but is too much to put in a comment.
The question indeed is ambiguous. If the employees are named Ray and Sam, the intended answer could be elicited by asking, "Given that Ray has incurred a loss in excess of $2000,$ determine the probability that losses will exceed reimbursements."
Alternatively, it could be "Given that Sam has ... ."
Your interpretation is, "Given that at least one of the employees has ... ."
I think we agree that "one of the employees" could be Ray, or could be Sam, or could be "at least one of the employees."
But if it is "at least one," we must determine the probabilities $P(Amid B)$
and $P(Acap B),$ where $A$ is the probability that losses exceed $8000,$
and $B$ is the probability that at least one employee incurs a loss exceeding $2000.$
But interpreting the question is only half the battle. The rest is to derive a numeric result correctly from that interpretation.
Your calculation does not explicitly write either probability, but if we multiply both your numerator and denominator by $1/4000^2,$ we have the same final result,
and your numerator is $P(A) = frac{1}{50},$ which is correct.
So in effect you are saying that
$$
P(B) stackrel?=
frac{(.4cdot.6 +.6cdot.4 + .4cdot.4) cdot 3000 cdot 4000}{4000^2}
=(.4cdot.6 +.6cdot.4 + .4cdot.4)cdotfrac34 = frac{12}{25}.
$$
This is correct regarding the case that Ray incurs a loss exceeding $2000$ and Sam incurs no loss and the case that Sam incurs a loss exceeding $2000$ and Ray incurs no loss. Each of those events has probability $.4cdot.6cdotfrac34 = frac{9}{50}.$
But the probability that both employees incur losses, at least one of which exceeds $2000,$ is not $.4cdot.4cdotfrac34 = frac{3}{25}.$
That's the probability that both incurred losses and Ray's loss exceeded $2000.$
The probability that both employees incur losses and that at least one of the losses exceeds $2000$ is the probability that both employees incur losses and they do not both incur losses less than $2000.$ That probability is
$.4cdot.4cdotleft(1 - left(frac14right)^2right) = frac{3}{20}.$
So
$$
P(B) = .4cdot.6cdotfrac34 + .6cdot.4cdotfrac34
+ .4cdot.4cdotleft(1 - left(frac14right)^2right)
= frac{51}{100}.
$$
And that is why your interpretation should give the result $frac{2}{51},$
not the $frac{1}{24}$ that your calculation would give.
answered Jan 1 at 14:58
David KDavid K
55.4k344120
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I think there is some ambiguity in the question. I see two interpretations:
Interpretation (1): (This is how I would have interpreted the question.)
The probability that at least one employee incurred a loss in excess of $2,000$ is
$P[L_1 > 2000 text{ or } L_2 > 2000] = 1-P[L_1 le 2000 text{ and } L_2 le 2000] =1-(0.6+{2000-1000 over 5000-1000}0.4)^2$.
The probability that the loss exceeds $8,000$ can be computed graphically as
$P[L_1+L_2 > 8000] ={1 over 2} P[L_1 > 3000]^2={1 over 2} ({5000-3000 over 5000-1000}0.4)^2$.
Note that the second event implies the first.
Hence
$P[L_1+L_2 > 8000 | L_1 > 2000 text{ or } L_2 > 2000] = {P[L_1+L_2 > 8000 ] over P[L_1 > 2000 text{ or } L_2 > 2000] } = {2 over 51}$.
Interpretation (2): (This matches your solution above.)
We have $P[L_1 > 2000] = 1-(0.6+{2000-1000 over 5000-1000}0.4)$. Again, this event
implies the first so
$P[L_1+L_2 > 8000 | L_1 > 2000] = {P[L_1+L_2 > 8000 ] over P[L_1 > 2000 ]} = {1 over 15}$.
answered Jan 1 at 2:18
copper.hatcopper.hat
128k559161
$endgroup$
$begingroup$
The first interpretation makes sense to me, and is indeed how I was understanding, although I got the probabilities wrong. I am not sure if I understand the second interpretation, however. Do the words "Given that one of the employees has incurred a loss..." not mean the same thing as "Given that at least one of the employees has incurred a loss..."?
$endgroup$
– agblt
Jan 1 at 4:51
$begingroup$
@agblt: Unfortunately, probability questions often expose a gap between the precision of mathematical statements and the English language. On can interpret 'Given that one...' as meaning that we know that a specific one of the employees has incurred... I picked $L_1$ as the specific employee. From a 'practical' standpoint this interpretation is a bit weak as if we knew this to be true, we would likely know the specific amount that $L_1$ has incurred also.
$endgroup$
– copper.hat
Jan 1 at 9:18
add a comment |
$begingroup$
I think there is some ambiguity in the question. I see two interpretations:
Interpretation (1): (This is how I would have interpreted the question.)
The probability that at least one employee incurred a loss in excess of $2,000$ is
$P[L_1 > 2000 text{ or } L_2 > 2000] = 1-P[L_1 le 2000 text{ and } L_2 le 2000] =1-(0.6+{2000-1000 over 5000-1000}0.4)^2$.
The probability that the loss exceeds $8,000$ can be computed graphically as
$P[L_1+L_2 > 8000] ={1 over 2} P[L_1 > 3000]^2={1 over 2} ({5000-3000 over 5000-1000}0.4)^2$.
Note that the second event implies the first.
Hence
$P[L_1+L_2 > 8000 | L_1 > 2000 text{ or } L_2 > 2000] = {P[L_1+L_2 > 8000 ] over P[L_1 > 2000 text{ or } L_2 > 2000] } = {2 over 51}$.
Interpretation (2): (This matches your solution above.)
We have $P[L_1 > 2000] = 1-(0.6+{2000-1000 over 5000-1000}0.4)$. Again, this event
implies the first so
$P[L_1+L_2 > 8000 | L_1 > 2000] = {P[L_1+L_2 > 8000 ] over P[L_1 > 2000 ]} = {1 over 15}$.
answered Jan 1 at 2:18
copper.hatcopper.hat
128k559161
$endgroup$
$begingroup$
The first interpretation makes sense to me, and is indeed how I was understanding, although I got the probabilities wrong. I am not sure if I understand the second interpretation, however. Do the words "Given that one of the employees has incurred a loss..." not mean the same thing as "Given that at least one of the employees has incurred a loss..."?
$endgroup$
– agblt
Jan 1 at 4:51
$begingroup$
@agblt: Unfortunately, probability questions often expose a gap between the precision of mathematical statements and the English language. On can interpret 'Given that one...' as meaning that we know that a specific one of the employees has incurred... I picked $L_1$ as the specific employee. From a 'practical' standpoint this interpretation is a bit weak as if we knew this to be true, we would likely know the specific amount that $L_1$ has incurred also.
$endgroup$
– copper.hat
Jan 1 at 9:18
add a comment |
$begingroup$
I think there is some ambiguity in the question. I see two interpretations:
Interpretation (1): (This is how I would have interpreted the question.)
The probability that at least one employee incurred a loss in excess of $2,000$ is
$P[L_1 > 2000 text{ or } L_2 > 2000] = 1-P[L_1 le 2000 text{ and } L_2 le 2000] =1-(0.6+{2000-1000 over 5000-1000}0.4)^2$.
The probability that the loss exceeds $8,000$ can be computed graphically as
$P[L_1+L_2 > 8000] ={1 over 2} P[L_1 > 3000]^2={1 over 2} ({5000-3000 over 5000-1000}0.4)^2$.
Note that the second event implies the first.
Hence
$P[L_1+L_2 > 8000 | L_1 > 2000 text{ or } L_2 > 2000] = {P[L_1+L_2 > 8000 ] over P[L_1 > 2000 text{ or } L_2 > 2000] } = {2 over 51}$.
Interpretation (2): (This matches your solution above.)
We have $P[L_1 > 2000] = 1-(0.6+{2000-1000 over 5000-1000}0.4)$. Again, this event
implies the first so
$P[L_1+L_2 > 8000 | L_1 > 2000] = {P[L_1+L_2 > 8000 ] over P[L_1 > 2000 ]} = {1 over 15}$.
answered Jan 1 at 2:18
copper.hatcopper.hat
128k559161
$endgroup$
I think there is some ambiguity in the question. I see two interpretations:
Interpretation (1): (This is how I would have interpreted the question.)
The probability that at least one employee incurred a loss in excess of $2,000$ is
$P[L_1 > 2000 text{ or } L_2 > 2000] = 1-P[L_1 le 2000 text{ and } L_2 le 2000] =1-(0.6+{2000-1000 over 5000-1000}0.4)^2$.
The probability that the loss exceeds $8,000$ can be computed graphically as
$P[L_1+L_2 > 8000] ={1 over 2} P[L_1 > 3000]^2={1 over 2} ({5000-3000 over 5000-1000}0.4)^2$.
Note that the second event implies the first.
Hence
$P[L_1+L_2 > 8000 | L_1 > 2000 text{ or } L_2 > 2000] = {P[L_1+L_2 > 8000 ] over P[L_1 > 2000 text{ or } L_2 > 2000] } = {2 over 51}$.
Interpretation (2): (This matches your solution above.)
We have $P[L_1 > 2000] = 1-(0.6+{2000-1000 over 5000-1000}0.4)$. Again, this event
implies the first so
$P[L_1+L_2 > 8000 | L_1 > 2000] = {P[L_1+L_2 > 8000 ] over P[L_1 > 2000 ]} = {1 over 15}$.
answered Jan 1 at 2:18
copper.hatcopper.hat
128k559161
answered Jan 1 at 2:18
copper.hatcopper.hat
128k559161
answered Jan 1 at 2:18
copper.hatcopper.hat
128k559161
answered Jan 1 at 2:18
copper.hatcopper.hat
128k559161
128k559161
$begingroup$
The first interpretation makes sense to me, and is indeed how I was understanding, although I got the probabilities wrong. I am not sure if I understand the second interpretation, however. Do the words "Given that one of the employees has incurred a loss..." not mean the same thing as "Given that at least one of the employees has incurred a loss..."?
$endgroup$
– agblt
Jan 1 at 4:51
$begingroup$
@agblt: Unfortunately, probability questions often expose a gap between the precision of mathematical statements and the English language. On can interpret 'Given that one...' as meaning that we know that a specific one of the employees has incurred... I picked $L_1$ as the specific employee. From a 'practical' standpoint this interpretation is a bit weak as if we knew this to be true, we would likely know the specific amount that $L_1$ has incurred also.
$endgroup$
– copper.hat
Jan 1 at 9:18
add a comment |
$begingroup$
The first interpretation makes sense to me, and is indeed how I was understanding, although I got the probabilities wrong. I am not sure if I understand the second interpretation, however. Do the words "Given that one of the employees has incurred a loss..." not mean the same thing as "Given that at least one of the employees has incurred a loss..."?
$endgroup$
– agblt
Jan 1 at 4:51
$begingroup$
@agblt: Unfortunately, probability questions often expose a gap between the precision of mathematical statements and the English language. On can interpret 'Given that one...' as meaning that we know that a specific one of the employees has incurred... I picked $L_1$ as the specific employee. From a 'practical' standpoint this interpretation is a bit weak as if we knew this to be true, we would likely know the specific amount that $L_1$ has incurred also.
$endgroup$
– copper.hat
Jan 1 at 9:18
$begingroup$
The first interpretation makes sense to me, and is indeed how I was understanding, although I got the probabilities wrong. I am not sure if I understand the second interpretation, however. Do the words "Given that one of the employees has incurred a loss..." not mean the same thing as "Given that at least one of the employees has incurred a loss..."?
$endgroup$
– agblt
Jan 1 at 4:51
$begingroup$
The first interpretation makes sense to me, and is indeed how I was understanding, although I got the probabilities wrong. I am not sure if I understand the second interpretation, however. Do the words "Given that one of the employees has incurred a loss..." not mean the same thing as "Given that at least one of the employees has incurred a loss..."?
$endgroup$
– agblt
Jan 1 at 4:51
$begingroup$
@agblt: Unfortunately, probability questions often expose a gap between the precision of mathematical statements and the English language. On can interpret 'Given that one...' as meaning that we know that a specific one of the employees has incurred... I picked $L_1$ as the specific employee. From a 'practical' standpoint this interpretation is a bit weak as if we knew this to be true, we would likely know the specific amount that $L_1$ has incurred also.
$endgroup$
– copper.hat
Jan 1 at 9:18
$begingroup$
@agblt: Unfortunately, probability questions often expose a gap between the precision of mathematical statements and the English language. On can interpret 'Given that one...' as meaning that we know that a specific one of the employees has incurred... I picked $L_1$ as the specific employee. From a 'practical' standpoint this interpretation is a bit weak as if we knew this to be true, we would likely know the specific amount that $L_1$ has incurred also.
$endgroup$
– copper.hat
Jan 1 at 9:18
add a comment |
$begingroup$
You are given the condition that one employee incurs a loss so the $0.4$ probability only applies once for the other employee.
In terms of graphical representation, the red triangle represents the condition of exceeding $8000$, while the whole rectangle the total possible outcomes.
$P$ is therefore the area of the red triangle divided by the area of the rectangle times the probably of the second employee incurring a loss.
This is $$P = frac{0.5cdot 2000^2}{3000cdot 4000}cdot 0.4 = .06667$$
answered Jan 1 at 1:10
Phil HPhil H
4,2682312
$endgroup$
$begingroup$
Maybe what is confusing me is this: What exactly is the probability that one of the employees has incurred a loss in excess of $2000$? I am getting $(.40cdot.60 +.60cdot.40 + .40cdot.40)cdot{{ 3000 cdot 4000}over 5000^2}$. Do you have something different?
$endgroup$
– agblt
Jan 1 at 1:22
$begingroup$
I don't see how the probability of one employee incurring a loss greater than 2000 comes into it. The given condition of this (one employee incurring a loss > 2000) just starts the vertical height of the rectangle at 2000 instead of 1000.
$endgroup$
– Phil H
Jan 1 at 1:35
add a comment |
$begingroup$
You are given the condition that one employee incurs a loss so the $0.4$ probability only applies once for the other employee.
In terms of graphical representation, the red triangle represents the condition of exceeding $8000$, while the whole rectangle the total possible outcomes.
$P$ is therefore the area of the red triangle divided by the area of the rectangle times the probably of the second employee incurring a loss.
This is $$P = frac{0.5cdot 2000^2}{3000cdot 4000}cdot 0.4 = .06667$$
answered Jan 1 at 1:10
Phil HPhil H
4,2682312
$endgroup$
$begingroup$
Maybe what is confusing me is this: What exactly is the probability that one of the employees has incurred a loss in excess of $2000$? I am getting $(.40cdot.60 +.60cdot.40 + .40cdot.40)cdot{{ 3000 cdot 4000}over 5000^2}$. Do you have something different?
$endgroup$
– agblt
Jan 1 at 1:22
$begingroup$
I don't see how the probability of one employee incurring a loss greater than 2000 comes into it. The given condition of this (one employee incurring a loss > 2000) just starts the vertical height of the rectangle at 2000 instead of 1000.
$endgroup$
– Phil H
Jan 1 at 1:35
add a comment |
$begingroup$
You are given the condition that one employee incurs a loss so the $0.4$ probability only applies once for the other employee.
In terms of graphical representation, the red triangle represents the condition of exceeding $8000$, while the whole rectangle the total possible outcomes.
$P$ is therefore the area of the red triangle divided by the area of the rectangle times the probably of the second employee incurring a loss.
This is $$P = frac{0.5cdot 2000^2}{3000cdot 4000}cdot 0.4 = .06667$$
answered Jan 1 at 1:10
Phil HPhil H
4,2682312
$endgroup$
You are given the condition that one employee incurs a loss so the $0.4$ probability only applies once for the other employee.
In terms of graphical representation, the red triangle represents the condition of exceeding $8000$, while the whole rectangle the total possible outcomes.
$P$ is therefore the area of the red triangle divided by the area of the rectangle times the probably of the second employee incurring a loss.
This is $$P = frac{0.5cdot 2000^2}{3000cdot 4000}cdot 0.4 = .06667$$
answered Jan 1 at 1:10
Phil HPhil H
4,2682312
answered Jan 1 at 1:10
Phil HPhil H
4,2682312
answered Jan 1 at 1:10
Phil HPhil H
4,2682312
answered Jan 1 at 1:10
Phil HPhil H
4,2682312
4,2682312
$begingroup$
Maybe what is confusing me is this: What exactly is the probability that one of the employees has incurred a loss in excess of $2000$? I am getting $(.40cdot.60 +.60cdot.40 + .40cdot.40)cdot{{ 3000 cdot 4000}over 5000^2}$. Do you have something different?
$endgroup$
– agblt
Jan 1 at 1:22
$begingroup$
I don't see how the probability of one employee incurring a loss greater than 2000 comes into it. The given condition of this (one employee incurring a loss > 2000) just starts the vertical height of the rectangle at 2000 instead of 1000.
$endgroup$
– Phil H
Jan 1 at 1:35
add a comment |
$begingroup$
Maybe what is confusing me is this: What exactly is the probability that one of the employees has incurred a loss in excess of $2000$? I am getting $(.40cdot.60 +.60cdot.40 + .40cdot.40)cdot{{ 3000 cdot 4000}over 5000^2}$. Do you have something different?
$endgroup$
– agblt
Jan 1 at 1:22
$begingroup$
I don't see how the probability of one employee incurring a loss greater than 2000 comes into it. The given condition of this (one employee incurring a loss > 2000) just starts the vertical height of the rectangle at 2000 instead of 1000.
$endgroup$
– Phil H
Jan 1 at 1:35
$begingroup$
Maybe what is confusing me is this: What exactly is the probability that one of the employees has incurred a loss in excess of $2000$? I am getting $(.40cdot.60 +.60cdot.40 + .40cdot.40)cdot{{ 3000 cdot 4000}over 5000^2}$. Do you have something different?
$endgroup$
– agblt
Jan 1 at 1:22
$begingroup$
Maybe what is confusing me is this: What exactly is the probability that one of the employees has incurred a loss in excess of $2000$? I am getting $(.40cdot.60 +.60cdot.40 + .40cdot.40)cdot{{ 3000 cdot 4000}over 5000^2}$. Do you have something different?
$endgroup$
– agblt
Jan 1 at 1:22
$begingroup$
I don't see how the probability of one employee incurring a loss greater than 2000 comes into it. The given condition of this (one employee incurring a loss > 2000) just starts the vertical height of the rectangle at 2000 instead of 1000.
$endgroup$
– Phil H
Jan 1 at 1:35
$begingroup$
I don't see how the probability of one employee incurring a loss greater than 2000 comes into it. The given condition of this (one employee incurring a loss > 2000) just starts the vertical height of the rectangle at 2000 instead of 1000.
$endgroup$
– Phil H
Jan 1 at 1:35
add a comment |
$begingroup$
This is not intended as a complete answer (you already have a good one) but is too much to put in a comment.
The question indeed is ambiguous. If the employees are named Ray and Sam, the intended answer could be elicited by asking, "Given that Ray has incurred a loss in excess of $2000,$ determine the probability that losses will exceed reimbursements."
Alternatively, it could be "Given that Sam has ... ."
Your interpretation is, "Given that at least one of the employees has ... ."
I think we agree that "one of the employees" could be Ray, or could be Sam, or could be "at least one of the employees."
But if it is "at least one," we must determine the probabilities $P(Amid B)$
and $P(Acap B),$ where $A$ is the probability that losses exceed $8000,$
and $B$ is the probability that at least one employee incurs a loss exceeding $2000.$
But interpreting the question is only half the battle. The rest is to derive a numeric result correctly from that interpretation.
Your calculation does not explicitly write either probability, but if we multiply both your numerator and denominator by $1/4000^2,$ we have the same final result,
and your numerator is $P(A) = frac{1}{50},$ which is correct.
So in effect you are saying that
$$
P(B) stackrel?=
frac{(.4cdot.6 +.6cdot.4 + .4cdot.4) cdot 3000 cdot 4000}{4000^2}
=(.4cdot.6 +.6cdot.4 + .4cdot.4)cdotfrac34 = frac{12}{25}.
$$
This is correct regarding the case that Ray incurs a loss exceeding $2000$ and Sam incurs no loss and the case that Sam incurs a loss exceeding $2000$ and Ray incurs no loss. Each of those events has probability $.4cdot.6cdotfrac34 = frac{9}{50}.$
But the probability that both employees incur losses, at least one of which exceeds $2000,$ is not $.4cdot.4cdotfrac34 = frac{3}{25}.$
That's the probability that both incurred losses and Ray's loss exceeded $2000.$
The probability that both employees incur losses and that at least one of the losses exceeds $2000$ is the probability that both employees incur losses and they do not both incur losses less than $2000.$ That probability is
$.4cdot.4cdotleft(1 - left(frac14right)^2right) = frac{3}{20}.$
So
$$
P(B) = .4cdot.6cdotfrac34 + .6cdot.4cdotfrac34
+ .4cdot.4cdotleft(1 - left(frac14right)^2right)
= frac{51}{100}.
$$
And that is why your interpretation should give the result $frac{2}{51},$
not the $frac{1}{24}$ that your calculation would give.
answered Jan 1 at 14:58
David KDavid K
55.4k344120
$endgroup$
add a comment |
$begingroup$
This is not intended as a complete answer (you already have a good one) but is too much to put in a comment.
The question indeed is ambiguous. If the employees are named Ray and Sam, the intended answer could be elicited by asking, "Given that Ray has incurred a loss in excess of $2000,$ determine the probability that losses will exceed reimbursements."
Alternatively, it could be "Given that Sam has ... ."
Your interpretation is, "Given that at least one of the employees has ... ."
I think we agree that "one of the employees" could be Ray, or could be Sam, or could be "at least one of the employees."
But if it is "at least one," we must determine the probabilities $P(Amid B)$
and $P(Acap B),$ where $A$ is the probability that losses exceed $8000,$
and $B$ is the probability that at least one employee incurs a loss exceeding $2000.$
But interpreting the question is only half the battle. The rest is to derive a numeric result correctly from that interpretation.
Your calculation does not explicitly write either probability, but if we multiply both your numerator and denominator by $1/4000^2,$ we have the same final result,
and your numerator is $P(A) = frac{1}{50},$ which is correct.
So in effect you are saying that
$$
P(B) stackrel?=
frac{(.4cdot.6 +.6cdot.4 + .4cdot.4) cdot 3000 cdot 4000}{4000^2}
=(.4cdot.6 +.6cdot.4 + .4cdot.4)cdotfrac34 = frac{12}{25}.
$$
This is correct regarding the case that Ray incurs a loss exceeding $2000$ and Sam incurs no loss and the case that Sam incurs a loss exceeding $2000$ and Ray incurs no loss. Each of those events has probability $.4cdot.6cdotfrac34 = frac{9}{50}.$
But the probability that both employees incur losses, at least one of which exceeds $2000,$ is not $.4cdot.4cdotfrac34 = frac{3}{25}.$
That's the probability that both incurred losses and Ray's loss exceeded $2000.$
The probability that both employees incur losses and that at least one of the losses exceeds $2000$ is the probability that both employees incur losses and they do not both incur losses less than $2000.$ That probability is
$.4cdot.4cdotleft(1 - left(frac14right)^2right) = frac{3}{20}.$
So
$$
P(B) = .4cdot.6cdotfrac34 + .6cdot.4cdotfrac34
+ .4cdot.4cdotleft(1 - left(frac14right)^2right)
= frac{51}{100}.
$$
And that is why your interpretation should give the result $frac{2}{51},$
not the $frac{1}{24}$ that your calculation would give.
answered Jan 1 at 14:58
David KDavid K
55.4k344120
$endgroup$
add a comment |
$begingroup$
This is not intended as a complete answer (you already have a good one) but is too much to put in a comment.
The question indeed is ambiguous. If the employees are named Ray and Sam, the intended answer could be elicited by asking, "Given that Ray has incurred a loss in excess of $2000,$ determine the probability that losses will exceed reimbursements."
Alternatively, it could be "Given that Sam has ... ."
Your interpretation is, "Given that at least one of the employees has ... ."
I think we agree that "one of the employees" could be Ray, or could be Sam, or could be "at least one of the employees."
But if it is "at least one," we must determine the probabilities $P(Amid B)$
and $P(Acap B),$ where $A$ is the probability that losses exceed $8000,$
and $B$ is the probability that at least one employee incurs a loss exceeding $2000.$
But interpreting the question is only half the battle. The rest is to derive a numeric result correctly from that interpretation.
Your calculation does not explicitly write either probability, but if we multiply both your numerator and denominator by $1/4000^2,$ we have the same final result,
and your numerator is $P(A) = frac{1}{50},$ which is correct.
So in effect you are saying that
$$
P(B) stackrel?=
frac{(.4cdot.6 +.6cdot.4 + .4cdot.4) cdot 3000 cdot 4000}{4000^2}
=(.4cdot.6 +.6cdot.4 + .4cdot.4)cdotfrac34 = frac{12}{25}.
$$
This is correct regarding the case that Ray incurs a loss exceeding $2000$ and Sam incurs no loss and the case that Sam incurs a loss exceeding $2000$ and Ray incurs no loss. Each of those events has probability $.4cdot.6cdotfrac34 = frac{9}{50}.$
But the probability that both employees incur losses, at least one of which exceeds $2000,$ is not $.4cdot.4cdotfrac34 = frac{3}{25}.$
That's the probability that both incurred losses and Ray's loss exceeded $2000.$
The probability that both employees incur losses and that at least one of the losses exceeds $2000$ is the probability that both employees incur losses and they do not both incur losses less than $2000.$ That probability is
$.4cdot.4cdotleft(1 - left(frac14right)^2right) = frac{3}{20}.$
So
$$
P(B) = .4cdot.6cdotfrac34 + .6cdot.4cdotfrac34
+ .4cdot.4cdotleft(1 - left(frac14right)^2right)
= frac{51}{100}.
$$
And that is why your interpretation should give the result $frac{2}{51},$
not the $frac{1}{24}$ that your calculation would give.
answered Jan 1 at 14:58
David KDavid K
55.4k344120
$endgroup$
This is not intended as a complete answer (you already have a good one) but is too much to put in a comment.
The question indeed is ambiguous. If the employees are named Ray and Sam, the intended answer could be elicited by asking, "Given that Ray has incurred a loss in excess of $2000,$ determine the probability that losses will exceed reimbursements."
Alternatively, it could be "Given that Sam has ... ."
Your interpretation is, "Given that at least one of the employees has ... ."
I think we agree that "one of the employees" could be Ray, or could be Sam, or could be "at least one of the employees."
But if it is "at least one," we must determine the probabilities $P(Amid B)$
and $P(Acap B),$ where $A$ is the probability that losses exceed $8000,$
and $B$ is the probability that at least one employee incurs a loss exceeding $2000.$
But interpreting the question is only half the battle. The rest is to derive a numeric result correctly from that interpretation.
Your calculation does not explicitly write either probability, but if we multiply both your numerator and denominator by $1/4000^2,$ we have the same final result,
and your numerator is $P(A) = frac{1}{50},$ which is correct.
So in effect you are saying that
$$
P(B) stackrel?=
frac{(.4cdot.6 +.6cdot.4 + .4cdot.4) cdot 3000 cdot 4000}{4000^2}
=(.4cdot.6 +.6cdot.4 + .4cdot.4)cdotfrac34 = frac{12}{25}.
$$
This is correct regarding the case that Ray incurs a loss exceeding $2000$ and Sam incurs no loss and the case that Sam incurs a loss exceeding $2000$ and Ray incurs no loss. Each of those events has probability $.4cdot.6cdotfrac34 = frac{9}{50}.$
But the probability that both employees incur losses, at least one of which exceeds $2000,$ is not $.4cdot.4cdotfrac34 = frac{3}{25}.$
That's the probability that both incurred losses and Ray's loss exceeded $2000.$
The probability that both employees incur losses and that at least one of the losses exceeds $2000$ is the probability that both employees incur losses and they do not both incur losses less than $2000.$ That probability is
$.4cdot.4cdotleft(1 - left(frac14right)^2right) = frac{3}{20}.$
So
$$
P(B) = .4cdot.6cdotfrac34 + .6cdot.4cdotfrac34
+ .4cdot.4cdotleft(1 - left(frac14right)^2right)
= frac{51}{100}.
$$
And that is why your interpretation should give the result $frac{2}{51},$
not the $frac{1}{24}$ that your calculation would give.
answered Jan 1 at 14:58
David KDavid K
55.4k344120
answered Jan 1 at 14:58
David KDavid K
55.4k344120
answered Jan 1 at 14:58
David KDavid K
55.4k344120
answered Jan 1 at 14:58
David KDavid K
55.4k344120
55.4k344120
add a comment |
add a comment |
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$begingroup$
$P(A|B)=frac{P(Acap B)}{P(B)}$
$endgroup$
– John Douma
Dec 31 '18 at 23:30
$begingroup$
Is $P(B)$ not "given that at least one employee incurred a loss", and therefore should include where $X$ incurs a loss and not $Y$, where $Y$ incurs a loss and not $X$, and where both incur a loss?
$endgroup$
– agblt
Dec 31 '18 at 23:37
$begingroup$
$B$ is the event that one of $X$ or $Y$ incurs a loss in excess of $2000$.
$endgroup$
– John Douma
Dec 31 '18 at 23:49