SOLVED BY MYSELF, YOU CAN CLOSE THE THREAD 
Hi guys,
I'm new to this forum, and some of you might find it annoying that I post a question straight off, but the assignment is due tomorrow and I have two different answers. I do promise to keep posting in threads (when I have something to say)
It's basic probability. Here goes:
Insurance company X sells a number of different policies, among which 60% are for drivers, 40% for homeowners and 20% for both types of policies. Let A1 be people with only a motor policy, A2 those with only a homeowner policy, A3 people with both types and A4 those with other types of policies.
(i) Assume a person is selected at random. What is the probability that this person belongs to A1/A2/A3/A4 ? (basically they want us to get probability of each separately).
(ii) Let B denote the event that a policyholder will renew at least one of the car or home insurance policies. Based on past experience, we can assume that
P(B|A1) = 0.6,
P(B|A2)= 0.7,
P(B|A3)= 0.8
Given that the person selected at random has a car or a home insurance policy, what is the probability that this person will renew at least one of those policies?
(It was I who made some bits italic and bold).
My Solution
(i) Basic Venn diagram, get
P(A1)=0.4
P(A2)=0.2
P(A3)=0.2
P(A4)=0.2
(ii) First of all, I assume that you're given that the person is allowed to have both policies (A3), because it says "at least one".
From the given information, I worked out
P(B ∩ A1)=0.24
P(B ∩ A2)=0.14
P(B ∩ A3)=0.16
Solution 1
I drew a tree diagram (numbers in brackets represent probabilities)
Then I just added the "total" probabilities to get 0.54
Solution 2
I said that I'm looking for P(B | (A1 U A2 U A3))
which equals to P(B ∩ (A1 U A2 U A3)) / P(A1 U A2 U A3)
which equals to (0.24 + 0.16 + 0.14) / (0.4 + 0.2 + 0.2)
which is 0.675
Which one is right? And what's the difference between the two approaches?
Thanks!
Hi guys,
I'm new to this forum, and some of you might find it annoying that I post a question straight off, but the assignment is due tomorrow and I have two different answers. I do promise to keep posting in threads (when I have something to say)
It's basic probability. Here goes:
Insurance company X sells a number of different policies, among which 60% are for drivers, 40% for homeowners and 20% for both types of policies. Let A1 be people with only a motor policy, A2 those with only a homeowner policy, A3 people with both types and A4 those with other types of policies.
(i) Assume a person is selected at random. What is the probability that this person belongs to A1/A2/A3/A4 ? (basically they want us to get probability of each separately).
(ii) Let B denote the event that a policyholder will renew at least one of the car or home insurance policies. Based on past experience, we can assume that
P(B|A1) = 0.6,
P(B|A2)= 0.7,
P(B|A3)= 0.8
Given that the person selected at random has a car or a home insurance policy, what is the probability that this person will renew at least one of those policies?
(It was I who made some bits italic and bold).
My Solution
(i) Basic Venn diagram, get
P(A1)=0.4
P(A2)=0.2
P(A3)=0.2
P(A4)=0.2
(ii) First of all, I assume that you're given that the person is allowed to have both policies (A3), because it says "at least one".
From the given information, I worked out
P(B ∩ A1)=0.24
P(B ∩ A2)=0.14
P(B ∩ A3)=0.16
Solution 1
I drew a tree diagram (numbers in brackets represent probabilities)
- A1 (0.4) ----> B (0.6) so total prob.= 0.24
- A2 (0.2) ----> B(0.7) so total prob. = 0.14
- A3 (0.2) ----> B(0.8) so total prob. = 0.16
Then I just added the "total" probabilities to get 0.54
Solution 2
I said that I'm looking for P(B | (A1 U A2 U A3))
which equals to P(B ∩ (A1 U A2 U A3)) / P(A1 U A2 U A3)
which equals to (0.24 + 0.16 + 0.14) / (0.4 + 0.2 + 0.2)
which is 0.675
Which one is right? And what's the difference between the two approaches?
Thanks!
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