S2 January 2015 IAL Q1) and Q7)

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φM23
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#1
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#1
1) For 1(b), Why isn't it P(X=1)? Also, I understand it is 2 days to the mean is doubled, but couldn't using X~Po(1.6) also include exactly 2 days? So why is this correct for less than 2 days?

2) For 1(c), please could someone explain this step by step?

3) For 7), where does Var(Y)=(4n/25) come from?

Probability is my worst topic and this is really bugging me.

Thanks for any help.
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ghostwalker
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#2
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#2
(Original post by PhyM23)
1) For 1(b), Why isn't it P(X=1)?
X=1 is the probability that only one car was caught in the first 48 hours. There could be any number >= 1, and it still satisfies the criterion of catching a car within 48 hours.

Also, I understand it is 2 days to the mean is doubled, but couldn't using X~Po(1.6) also include exactly 2 days? So why is this correct for less than 2 days?
The parameter lambda is a continuous function (multiple) of the number of days.
What would lambda be if the distribution was for 2 days minus 1 hour?
Or two days minus 1 minute? Or minus 1 second?, Etc.

Also, although the Poisson is a discrete distribution, it's modelling a continous time period. What's the probability of a car arriving at exactly 2 days? Zero!

Don't feel I've really dealt with that bit properly, but hope it helps.

2) For 1(c), please could someone explain this step by step?
Conditional probabilty.

P(1 caught one day 1 AND 3 caught on day 2 | 4 caught over 2 days.)

The first two parts (red) refer to a single day period - hence lambda = 0.8.
Last past (green) refers to a 2 day period - hence lambda = 1.6.

Then standard conditional probability.

= P(1 caught one day 1 AND 3 caught on day 2 ) / P( 4 caught over 2 days.)

The two days are independent, so


= P(1 caught one day 1) P( 3 caught on day 2 ) / P( 4 caught over 2 days.)

3) For 7), where does Var(Y)=(4n/25) come from?

Probability is my worst topic and this is really bugging me.

Thanks for any help.
The binomial, which you are approximating, has a variance of npq, or np(1-p)
Hence n x (1/5) x (4/5) = 4n/25.
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φM23
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#3
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#3
(Original post by ghostwalker)
X=1 is the probability that only one car was caught in the first 48 hours. There could be any number >= 1, and it still satisfies the criterion of catching a car within 48 hours.



The parameter lambda is a continuous function (multiple) of the number of days.
What would lambda be if the distribution was for 2 days minus 1 hour?
Or two days minus 1 minute? Or minus 1 second?, Etc.

Also, although the Poisson is a discrete distribution, it's modelling a continous time period. What's the probability of a car arriving at exactly 2 days? Zero!

Don't feel I've really dealt with that bit properly, but hope it helps.



Conditional probabilty.

P(1 caught one day 1 AND 3 caught on day 2 | 4 caught over 2 days.)

The first two parts (red) refer to a single day period - hence lambda = 0.8.
Last past (green) refers to a 2 day period - hence lambda = 1.6.

Then standard conditional probability.

= P(1 caught one day 1 AND 3 caught on day 2 ) / P( 4 caught over 2 days.)

The two days are independent, so


= P(1 caught one day 1) P( 3 caught on day 2 ) / P( 4 caught over 2 days.)



The binomial, which you are approximating, has a variance of npq, or np(1-p)
Hence n x (1/5) x (4/5) = 4n/25.
Thank you very much for such a detailed reply.

1) This makes sense and your explanation of the second part is exactly what I was looking for.

2) This makes more sense, but why isn't P(4 caught over 2 days) in the numerator too?

3) Ah yes I should have spotted that :facepalm:
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Zacken
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#4
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#4
(Original post by PhyM23)
2) This makes more sense, but why isn't P(4 caught over 2 days) in the numerator too?
Look up conditional probability in the S1 section of your formula booklet:

P(A | B) = \frac{P(A \cap B)}{P(B)}
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φM23
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#5
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(Original post by Zacken)
Look up conditional probability in the S1 section of your formula booklet:

P(A | B) = \frac{P(A \cap B)}{P(B)}
That's what's confusing me.

Let P(1 caught on day 1) = P(A)
P(3 caught on day 2) = P(B)
P(4 caught over 2 days) = P(C)

P(A and B|C) = P(A and B and C) / P(C)

Why isn't it this?
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Zacken
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#6
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(Original post by PhyM23)
That's what's confusing me.

Let P(1 caught on day 1) = P(A)
P(3 caught on day 2) = P(B)
P(4 caught over 2 days) = P(C)

P(A and B|C) = P(A and B and C) / P(C)

Why isn't it this?
Ah, I see what you mean.

It's just that A and B and C is (1 caught on day 1 and 3 caught on day 2 and 4 caught over 2 days).

But (1 caught on day 1 and 3 caught on day 2) is already (4 caught over 2 days). So A \cap B \cap C = A \cap B.

Kind of like \{1, 2, 3\} \cap \{3, 4, 5\} \cap \{1, 2, 3, 4, 5, 6, 7,...\} = \{1, 2, 3\} \cap \{3, 4, 5\}

Edit: see below, ghostwalker explained it better.
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ghostwalker
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#7
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(Original post by PhyM23)
That's what's confusing me.

Let P(1 caught on day 1) = P(A)
P(3 caught on day 2) = P(B)
P(4 caught over 2 days) = P(C)

P(A and B|C) = P(A and B and C) / P(C)

Why isn't it this?
You're quite right. I missed a step out.

P(A and B|C) = P(A and B and C) / P(C)

The event "A and B" (i.e. 1 on day 1, and 3 on day 2), implies 4 over 2 days, i.e. C.

So, the event "A and B" is a subset of event "C".

A\cap B \subseteq C


Hence P(A and B and C) = P(A and B).

Edit: ninja'd
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φM23
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#8
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#8
(Original post by Zacken)
Ah, I see what you mean.

It's just that A and B and C is (1 caught on day 1 and 3 caught on day 2 and 4 caught over 2 days).

But (1 caught on day 1 and 3 caught on day 2) is already (4 caught over 2 days). So A \cap B \cap C = A \cap B.

Kind of like \{1, 2, 3\} \cap \{3, 4, 5\} \cap \{1, 2, 3, 4, 5, 6, 7,...\} = \{1, 2, 3\} \cap \{3, 4, 5\}

Edit: see below, ghostwalker explained it better.
(Original post by ghostwalker)
You're quite right. I missed a step out.

P(A and B|C) = P(A and B and C) / P(C)

The event "A and B" (i.e. 1 on day 1, and 3 on day 2), implies 4 over 2 days, i.e. C.

So, the event "A and B" is a subset of event "C".

A\cap B \subseteq C


Hence P(A and B and C) = P(A and B).

Edit: ninja'd
Haha

This is exactly what I was looking for; thank you both for your help
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