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2009 s2 q13

2009 s2 q13.png
First part. Could someone help I'm confused

If we let the random variable XX describe the number of failed engines in a single launch of Andover, then

E(cost)=KP(X=1)+4KP(X=2)+4KP(X=3)+4KP(X=4)E(cost)=K P(X=1) + 4K P(X=2) + 4K P(X=3) + 4K P(X=4).
Is this correct?

I'm confused by how to find the individual probabilities.
Is P(X=1)=(41)pP(X=1)=\binom{4}{1}p or (41)pq3\binom{4}{1}pq^{3} or pq3pq^{3} or none of them? This question is really confusing me...
(edited 2 years ago)
Reply 1
Original post by username53983805
2009 s2 q13.png
First part. Could someone help I'm confused

If we let the random variable XX describe the number of failed engines in a single launch of Andover, then

E(cost)=KP(X=1)+4KP(X=2)+4KP(X=3)+4KP(X=4)E(cost)=K P(X=1) + 4K P(X=2) + 4K P(X=3) + 4K P(X=4).
Is this correct?

I'm confused by how to find the individual probabilities.
Is P(X=1)=(41)pP(X=1)=\binom{4}{1}p or (41)pq3\binom{4}{1}pq^{3} or pq3pq^{3} or none of them? This question is really confusing me...

An engine has a probability of p of failing and q=1-p of working correctly.

There are 4 engines. So the probability of exactly one engine failing is 4pq^3 and the corresponding expected cost is 4Kpq^3.

The probability of all working is q^4, so the probability of 2 or more failing is
1 - 4pq^3 - q^4 = 6p^2q^2 + 4p^3q + p^4
WIth an expected cost of 4K(6p^2q^2 + 4p^3q + p^4). Combining the two gives the total cost.

There isn't really anything beyond basic binomial. You could do it in one line by simply combining the binomial with the cost and write down the expected cost as
0*q^4 + K*4pq^3 + 4K*(6p^2q^2 + 4p^3q + p^4)
(edited 2 years ago)
Original post by mqb2766
An engine has a probability of p of failing and q=1-p of working correctly.

There are 4 engines. So the probability of exactly one engine failing is 4pq^3 and the corresponding expected cost is 4Kpq^3.

The probability of all working is q^4, so the probability of 2 or more failing is
1 - 4pq^3 - q^4 = 6p^2q^2 + 4p^3q + p^4
WIth an expected cost of 4K(6p^2q^2 + 4p^3q + p^4). Combining the two gives the total cost.

There isn't really anything beyond basic binomial. You could do it in one line by simply combining the binomial with the cost and write down the expected cost as
0*q^4 + K*4pq^3 + 4K*(6p^2q^2 + 4p^3q + p^4)

Thank you

The only thing I'm confused about is that it mentions the expected cost, and that's why I thought to consider the expectation value for engines failing and adjust it accordingly so that I could instead consider the cost, but it seems that we aren't using the idea of expectation anywhere here even though they mention expected cost. I don't see why this is?
Reply 3
Original post by username53983805
Thank you

The only thing I'm confused about is that it mentions the expected cost, and that's why I thought to consider the expectation value for engines failing and adjust it accordingly so that I could instead consider the cost, but it seems that we aren't using the idea of expectation anywhere here even though they mention expected cost. I don't see why this is?

Thought you were happy with the basic idea of expected cost and were asking about the binomial part.

The cost C is a random variable as it depends on the random event engines failing. It has a finite number of outcomes 0, K, 4K and we want to get its expected value which you can imagine as a weighted value, depending on the probability of each event occurring. So as you said
E(C) = 0*p(no failures) + K*p(1 engine failure) + 4K*p(2 or more failures)
https://en.wikipedia.org/wiki/Expected_value
(edited 2 years ago)
Original post by mqb2766
Thought you were happy with the basic idea of expected cost and were asking about the binomial part.

The cost C is a random variable as it depends on the random event engines failing. It has a finite number of outcomes 0, K, 4K and we want to get its expected value which you can imagine as a weighted value, depending on the probability of each event occurring. So as you said
E(C) = 0*p(no failures) + K*p(1 engine failure) + 4K*p(2 or more failures)
https://en.wikipedia.org/wiki/Expected_value

I'll try to rephrase my question (as I know it wasn't very clear what I meant and I'm now starting to confuse myself on why my original E(cost) expression is correct):

So I know that E(X)=∑i=1∞i⋅P(X=i)E(X)=\sum_{i=1}^{\infty} i\cdot P(X=i).
If, like I did in OP, let XX describe the number of failed engines in a single launch of Andover, then we are concerned with
P(X=1)+2⋅P(X=2)+3⋅P(X=3)+4⋅P(X=4)P(X=1)+2\cdot P(X=2)+3\cdot P(X=3)+4\cdot P(X=4).

My question is, as this gives the expected number of failed engines in a single launch, why could I not just slightly tweak this to find the E(cost)E(cost)?
I.E. why could I not say E(cost)=KP(X=1)+4K[2⋅]E(cost)=KP(X=1)+4K[2\cdot]? This may be a silly question to ask (and may be down to the fact I had a go at the question a few days ago now) but I just wanted to know why
(edited 2 years ago)
Reply 5
Original post by username53983805
I'll try to rephrase my question (as I know it wasn't very clear what I meant and I'm now starting to confuse myself on why my original E(cost) expression is correct):

So I know that E(X)=∑i=1∞i⋅P(X=i)E(X)=\sum_{i=1}^{\infty} i\cdot P(X=i).
If, like I did in OP, let XX describe the number of failed engines in a single launch of Andover, then we are concerned with
P(X=1)+2⋅P(X=2)+3⋅P(X=3)+4⋅P(X=4)P(X=1)+2\cdot P(X=2)+3\cdot P(X=3)+4\cdot P(X=4).

My question is, as this gives the expected number of failed engines in a single launch, why could I not just slightly tweak this to find the E(cost)E(cost)?
I.E. why could I not say E(cost)=KP(X=1)+4K(2⋅P(X=2)+3⋅P(X=3)+4⋅P(X=4))E(cost)=KP(X=1)+4K(2\cdot P(X=2)+3\cdot P(X=3)+4\cdot P(X=4))? This may be a silly question to ask (and may be down to the fact I had a go at the question a few days ago now) but I just wanted to know why

Cost is a discrete random variable with outcome, probability:
0 q^4
K 4pq^3
4K (6p^2q^2 + ...)
It is wrong to base E(C) it on the expected number of engines failed.

Your expression weighs the cost of 4 engines failing higher than 2 engines failing. However, they both have the same cost of 4K.
(edited 2 years ago)
Original post by mqb2766
Cost is a discrete random variable with outcome, probability:
0 q^4
K 4pq^3
4K (6p^2q^2 + ...)
It is wrong to base E(C) it on the expected number of engines failed as the outcomes 0, K, 4K are different. If the outcomes were all the same you could do this.

Your expression weighs the cost of 4 engines failing higher than 2 engines failing. However, they both have the same cost of 4K.

Ohh yes of course I see

Thank you for your help:smile:

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