# S1

#1
A computer company provides an insurance policy for one of its systems. If the system fails during the first year, the policy pays £3,000; if it fails during the second year, the policy pays £2,000; and if it fails during the third year, the policy pays £1,000. The policy terminates after the first payout or after 3 years, whichever is the sooner. If the system has not failed at the beginning of a year, the probability that it will fail at some point during that year is 0.1. How much should the company charge for the insurance policy so that, on average, its net gain per policy is £100?

Well I did (0.1*3000) +.... +... and got =600
0
4 years ago
#2
Remember that to pay out £2,000, the system must have not failed in the first year, and also failed in the second year. Similarly for the £1,000 in the third year.
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#3
(Original post by ThomH97)
Remember that to pay out £2,000, the system must have not failed in the first year, and also failed in the second year. Similarly for the £1,000 in the third year.
Oh yea, ill work those out now , thanks
0
#4
(Original post by ThomH97)
Remember that to pay out £2,000, the system must have not failed in the first year, and also failed in the second year. Similarly for the £1,000 in the third year.
Okay to pay out 2000 the system must have not failed in the first year which is 90% chance? So I get 180 pounds and for 1000 pounds I have got 81 pounds which gives me 561 in total , then I add 100?
1
4 years ago
#5
(Original post by Yiuyu)
Okay to pay out 2000 the system must have not failed in the first year which is 90% chance? So I get 180 pounds and for 1000 pounds I have got 81 pounds which gives me 561 in total , then I add 100?
I'd agree with that
1
#6
(Original post by ThomH97)
I'd agree with that
I have another question thanks
If X is geometrically distributed with p=0.8
where Y= X, X<3
or 2 otherwise
Find E(y)
0
4 years ago
#7
(Original post by Yiuyu)
I have another question thanks
If X is geometrically distributed with p=0.8
where Y= X, X<3
or 2 otherwise
Find E(y)
What values can Y take? And with what probability?

Though I am a bit confused by 'Y=2 otherwise' since Y already inherits a probability for 2 from X. I would guess at just adding the remaining probability onto P(X=2).
0
#8
(Original post by ThomH97)
What values can Y take? And with what probability?

Though I am a bit confused by 'Y=2 otherwise' since Y already inherits a probability for 2 from X. I would guess at just adding the remaining probability onto P(X=2).
Anything less than 2 with p=0.8?
0
4 years ago
#9
(Original post by Yiuyu)
Anything less than 2 with p=0.8?
P(Y=a) = P(X=a) for a<3.

Then, because these won't all add to 1, stick the extra onto P(Y=2)
0
#10
(Original post by ThomH97)
P(Y=a) = P(X=a) for a<3.

Then, because these won't all add to 1, stick the extra onto P(Y=2)
Yeah I meant less than or equal to 2?
So for (X=0) p =0.8 , what eqn am I using again?
0
4 years ago
#11
(Original post by Yiuyu)
Yeah I meant less than or equal to 2?
So for (X=0) p =0.8 , what eqn am I using again?
P(X=a) = (1-p)^a p

(I am assuming you are using the geometric distribution that permits the variable to be 0)
0
#12
(Original post by ThomH97)
P(X=a) = (1-p)^a p

(I am assuming you are using the geometric distribution that permits the variable to be 0)
But the answer is 1.2? So Idk how we could get that?
0
4 years ago
#13
(Original post by Yiuyu)
But the answer is 1.2? So Idk how we could get that?
In that case it's the 'other' geometric distribution, where the formula is P(X=a) = (1-p)^(a-1) p. I don't know which exam board you're doing, but this will be consistent within them.

Weird question though, was there no other information?
0
#14
(Original post by ThomH97)
In that case it's the 'other' geometric distribution, where the formula is P(X=a) = (1-p)^(a-1) p. I don't know which exam board you're doing, but this will be consistent within them.

Weird question though, was there no other information?
But that unfortunately gives me an answer higher than 1.2?

Literally thats all they have gave me
0
4 years ago
#15
(Original post by Yiuyu)
But that unfortunately gives me an answer higher than 1.2?

Literally thats all they have gave me
With the 'new' formula, what is P(Y=1)? Everything else goes into P(Y=2) because it's a weird question asking you to transform just one value. Then you can work out E(Y) in the usual way.
0
#16
(Original post by ThomH97)
With the 'new' formula, what is P(Y=1)? Everything else goes into P(Y=2) because it's a weird question asking you to transform just one value. Then you can work out E(Y) in the usual way.
P(Y=1) is 0.8
and P (Y=2) =0.16 right?
I get 0.96 for the total?
0
4 years ago
#17
(Original post by Yiuyu)
P(Y=1) is 0.8
and P (Y=2) =0.16 right?
I get 0.96 for the total?
This is the weird bit that I'm guessing as to what they mean. You have 0.96 probability so far, so you need to add the remaining 0.04 to P(Y=2) to get 0.2.

Then E(Y) gets to 1.2. Remember E(Z) = sum over all z (z times P(Z=z))

Forgive the notation XD
0
#18
(Original post by ThomH97)
This is the weird bit that I'm guessing as to what they mean. You have 0.96 probability so far, so you need to add the remaining 0.04 to P(Y=2) to get 0.2.

Then E(Y) gets to 1.2. Remember E(Z) = sum over all z (z times P(Z=z))

Forgive the notation XD
So I just randomly add 0.04? and where does the extra 0.2 come from again?
0
4 years ago
#19
(Original post by Yiuyu)
So I just randomly add 0.04? and where does the extra 0.2 come from again?
They've restricted the values Y can take to be 1 and 2. You've found P(Y=1) from the geometric formula, so the remaining probability must all be for Y=2 (all probabilities sum to 1).
0
#20
(Original post by ThomH97)
They've restricted the values Y can take to be 1 and 2. You've found P(Y=1) from the geometric formula, so the remaining probability must all be for Y=2 (all probabilities sum to 1).
Sorry I dont understand it but its fine , its okay dont worry
0
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