Probability Law of addition

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Thomas74

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Good evening, I'm working on my homework, but since I'm not really sure about what I have done, I want somebody to check it for me.
Thank you!

The problem is here :
1. You run accident insurance. Have 3 types of clients: athletes, clerks, and workers. The proportion in the portfolio is MM% athletes, DD% workers and the rest is clerks. The probability of an accident during a year is 0.1 for athletes, 0.05 for workers, and 0.02 for clerks.
a. What is the probability that a randomly chosen policyholder will suffer an accident?
b. Policy-holder suffered an accident. What is the probability, that he was an athlete?
c. Policy-holder did not suffer an accident. What is the probability, that he was an athlete?

2. Manager of a company is pursuing(independently) three possible contracts. She assumes that the first will be awarded probability 0.5, the second with probability 0.3, and the last with a probability of 0.4. The profits from the contracts are 100,000;60,000 and 40,000, respectively.
a. Assume total profit. Find out all the possible values and their probabilities and cumulative distribution functions.
b.Find out expected profit and its variance.

and my work will be posted as a picture below, please have a comment on it.
Thank you!
 

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How do you expect to get a numerical answer if you do not know what the relative frequencies of the various types of workers are?
 
Your question 2 probability doesn't add up to 100%? 0.5 + 0.4 + 0.3 = 1.2 >1. Are you sure you're writing down the question correctly? Unless the probability of winning the 2nd and 3rd contract is conditioned on winning the previous contracts.
 
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BigBeachBananas said:
Your question 2 probability doesn't add up to 100%? 0.5 + 0.4 + 0.3 = 1.2 >1. Are you sure you're writing down the question correctly? Unless the probability of winning the 2nd and 3rd contract is conditioned on winning the previous contracts.
Click to expand...
They didn't mention that it's conditional or dependent; it said that it is independent. That's why I'm wondering did I use the correct formula. Here is the original question
 

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Thomas74 said:
They didn't give the proportion of the Athletes, workers, and clerks. Here's the original file.
Click to expand...
Question 1) Notice your birthday and month at the top DD and MM. I see you wrote 18/01, meaning athletes are 18%, workers are 1%, and clerks are 100%-18%-1%=71%. Now, we have enough information to solve. Question 1 are conditional probabilities questions. What's the formula for conditional probability?
Question 2) You have to account for all the possibilities of winning contracts. For example,
Possibility#1: win 1st, 2nd, and 3rd
Possibility#2: lose 1st, win 2nd, lose 3rd, etc...
Hint: there's a total of 8 possibilities of winning/losing the contracts.
Can you write down all the possibilities? This question can be solved using multinomial distribution.
 
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BigBeachBananas said:
Question 1) Notice your birthday and month at the top DD and MM. I see you wrote 18/01, meaning athletes are 18%, workers are 1%, and clerks are 100%-18%-1%=71%. Now, we have enough information to solve. Question 1 are conditional probabilities questions. What's the formula for conditional probability?
Question 2) You have to account for all the possibilities of winning contracts. For example,
Possibility#1: win 1st, 2nd, and 3rd
Possibility#2: lose 1st, win 2nd, lose 3rd, etc...
Hint: there's a total of 8 possibilities of winning/losing the contracts.
Can you write down all the possibilities? This question can be solved using multinomial distribution.
Click to expand...
Thank you, I didn't notice that, I'll try it and report back soon! ;)
 
BigBeachBananas said:
I think I swapped the distribution. Athletes are MM% = 1%, Workers are DD = 18%, Clerks = 71%
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Thank you @BigBeachBananas, I did the first one, and about the last one, I don't know where to put the number into the Multinomial formula, but here I wrote all of the possibilities.
 

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Thomas74 said:
Thank you @BigBeachBananas, I did the first one, and about the last one, I don't know where to put the number into the Multinomial formula, but here I wrote all of the possibilities.
Click to expand...
Question1, I find how you define your probabilities to be unclear. I would define them as follow:
[math]\text{Pr(A) = .01 , Pr(W)=.18,Pr(C)=.71}\\ \text{Pr(A | Accident) =.1 , Pr(W|Accident)=.05 Pr(C|Accident)=.02 }[/math]Your calculation for 1a) and 1b) are correct, however, 1c) is incorrect. The question is asking you to find Pr(A |Not in Accident) [math]\text{Pr(A| Not in Accident)} = \frac{\text{Pr(A} \cap \text{Not in Accident)}}{\text{Pr(Not in Accident)}}=\frac{\text{Pr(A) - Pr(A} \cap \text{Accident)}}{1-\text{Pr(Accident)}}[/math]Question 2, good job on writing out the possibilities, however, your expected value is incorrect. You have to account for all the possilbities. Can you fill out the table below?
Scenario​
Pr(Scenario)​
Profit given the scenario​
WWW0.5(.3)(.4) = .06100,000+60,000+40,000 = 200,000
WWL
WLW
LWW
LLW
LWL
WLL
LLL
 
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BigBeachBananas said:
Question1, I find how you define your probabilities to be unclear. I would define them as follow:
[math]\text{Pr(A) = .01 , Pr(W)=.18,Pr(C)=.71}\\ \text{Pr(A | Accident) =.1 , Pr(W|Accident)=.05 Pr(C|Accident)=.02 }[/math]Your calculation for 1a) and 1b) are correct, however, 1c) is incorrect. The question is asking you to find Pr(A |Not in Accident) [math]\text{Pr(A| Not in Accident)} = \frac{\text{Pr(A} \cap \text{Not in Accident)}}{\text{Pr(Not in Accident)}}=\frac{\text{Pr(A) - Pr(A} \cap \text{Accident)}}{1-\text{Pr(Accident)}}[/math]
Click to expand...
[math]\text{Pr(A| Not in Accident)}=\frac{\text{Pr(A) - Pr(A} \cap \text{Accident)}}{1-\text{Pr(Accident)}}=\frac{\text{(.01-.001)}}{\text{(1-.0242)}}= .0092[/math]
BigBeachBananas said:
Question 2, good job on writing out the possibilities, however, your expected value is incorrect. You have to account for all the possilbities. Can you fill out the table below?
Scenario​
Pr(Scenario)​
Profit given the scenario​
WWW0.5(.3)(.4) = .06100,000+60,000+40,000 = 200,000
WWL0.5(.3)(.6) = .09100,000+60,000= 160,000
WLW0.5(.7)(.4) = .14100,000+40,000 = 140,000
LWW0.5(.3)(.4) = .0660,000+40,000 = 100,000
LLW0.5(.7)(.4) = .1440,000
LWL0.5(.3)(.6) = .0960,000
WLL0.5(.7)(.6) = .21100,000
LLL0.5(.7)(.6) = .210
Click to expand...

Thank you @BigBeachBananas, what should I do next to calculate variance.
 
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Question 2, the probabilities aren't correct, so not ready to calculate the variance yet. Let's look at WWL as an example.
Currently, you have Pr(WWL)=(.5)*(.3)=(.5)*(.3)*(1). What you're saying is the probability of losing the 3rd contract is 100%.
Instead, it should be Pr(WWL)=.5*.3*(1-.4). Can you redo the probablities?
 
BigBeachBananas said:
Question 2, the probabilities aren't correct, so not ready to calculate the variance yet. Let's look at WWL as an example.
Currently, you have Pr(WWL)=(.5)*(.3)=(.5)*(.3)*(1). What you're saying is the probability of losing the 3rd contract is 100%.
Instead, it should be Pr(WWL)=.5*.3*(1-.4). Can you redo the probablities?
Click to expand...
Ok, I understand; I've revised it on the previous chat :)
 
Thomas74 said:
Ok, I understand; I've revised it on the previous chat :)
Click to expand...
Correct, as a check the sum of the probabilities must equal 1. Now you're ready to calculate your expected and variance of the profit i.e E[Profit] and Var[Profit].
[math]\text{E[Profit]}=\sum_{i=1}^{8} Profit_i*{Pr(Profit_i)} \\ Var[Profit] = E[Profit^2]-(E[Profit])^2[/math]
 
BigBeachBananas said:
Correct, as a check the sum of the probabilities must equal 1. Now you're ready to calculate your expected and variance of the profit i.e E[Profit] and Var[Profit].
[math]\text{E[Profit]}=\sum_{i=1}^{8} Profit_i*{Pr(Profit_i)} \\ Var[Profit] = E[Profit^2]-(E[Profit])^2[/math]
Click to expand...
Pr(Profit-i)Profit-iSum
.06200,00012,000
.09160,00014,400
.14140,00019,600
.06100,0006,000
.1440,0005,600
.0960,0005,400
.21100,00021,000
.2100

E[Profit]= 84,000
Variance = (10,696,000,000)-(84,000)^2= 3,640,000
 
BigBeachBananas said:
No problem, you still need to do the cumulative distribution, you already have the probability mass function from the table. Let me know if you need help
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If I have it, correctly, I think my CDF would look as follows: I've another HW but I've to try it myself first, thank you @BigBeachBananas :):thumbup:

Pr(X)F(X)
.06.06
.09.15
.14.29
.06.35
.14.49
.09.58
.21.79
.211
1-
 
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