Pakora99
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Doing an S2 question and I don't get this
Biased 4 sided spinner can only land on 1,2,3,4
Random Variable X is the number that the spinner lands on after a single spin
P(X=q) = P(X=q+2) for q=1.2
Given P(X=2) = 0.35


(a) Find the complete probability distribution of q


(b) spinner spinned 50 times
find probability more than half land on number 4

The random variable
12
Y= ----
X
Find P (Y-X<=4)
Find P( Y - X less or equal to 4)
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Pakora99
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ghostwalker
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(Original post by Pakora99)
bump
Insufficient information to resolve the problem - unless it's to be left in terms of a parameter.

Please link the original question.
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Pakora99
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(Original post by ghostwalker)
Insufficient information to resolve the problem - unless it's to be left in terms of a parameter.

Please link the original question.
sorry bout that forgot to add this bit Given P(X=2) = 0.35
updated it at the top as well
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ghostwalker
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(Original post by Pakora99)
sorry bout that forgot to add this bit Given P(X=2) = 0.35
updated it at the top as well
From the question, then:

P(X=2) = P(X=4) = 0.35

P(X=1) = P(X=3) = ?

And use the fact that the total probability must be 1, to work out the unknown value.
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Pakora99
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(Original post by ghostwalker)
From the question, then:

P(X=2) = P(X=4) = 0.35

P(X=1) = P(X=3) = ?

And use the fact that the total probability must be 1, to work out the unknown value.
P(X=2) = P(X=4) = 0.35 how do you know 2 and 4 are related
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Pakora99
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(Original post by ghostwalker)
From the question, then:

P(X=2) = P(X=4) = 0.35

P(X=1) = P(X=3) = ?

And use the fact that the total probability must be 1, to work out the unknown value.
so P(X=1) = 0.15
P(X=3) =0.15
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ghostwalker
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(Original post by Pakora99)
P(X=2) = P(X=4) = 0.35 how do you know 2 and 4 are related
Question tells you.

"P(X=q) = P(X=q+2) for q=1.2"

Setting q=2

P(X=2) = P(X=4)
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Pakora99
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(Original post by ghostwalker)
Question tells you.

"P(X=q) = P(X=q+2) for q=1.2"

Setting q=2

P(X=2) = P(X=4)
Ahh Thank You!!!
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ghostwalker
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(Original post by Pakora99)
so P(X=1) = 0.15
P(X=3) =0.15
Yes.

Can you do part b? If not, what distribution do you thnk it is?
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Pakora99
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(Original post by ghostwalker)
Yes.

Can you do part b? If not, what distribution do you thnk it is?
Looks like binomial distribution so I've done
1-P(X<25) = P(X>25)
1-P(X{less or equal} 25) = P(X>25)
let x be 0.35
0.989956 for P(X<25)
1-0.989956=0.01
not too sure on this bit if I've done It right
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ghostwalker
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(Original post by Pakora99)
Looks like binomial distribution so I've done
1-P(X<25) = P(X>25)
1-P(X{less or equal} 25) = P(X>25)
Yes, should be \leq

let x be 0.35
p=0.35
Probability of success on a single spin.

0.989956 for P(X<25)
1-0.989956=0.01
not too sure on this bit if I've done It right
Yep.
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Pakora99
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(Original post by ghostwalker)
Yes, should be \leq



p=0.35
Probability of success on a single spin.



Yep.
y=12/X
FindP(Y−X≤4)
I was thinking about finding in the 50 spins where X is less or equal to 4
But how would I do the Y bit?
Sorry about no latex I can't figure out how to use it
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ghostwalker
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(Original post by Pakora99)
y=12/X
FindP(Y−X≤4)
I was thinking about finding in the 50 spins where X is less or equal to 4
But how would I do the Y bit?
Sorry about no latex I can't figure out how to use it
The 50 spins don't come into it, unless there's something you've not posted.

You're only dealing with 1 spin, with r.v. X

So, you just need to check when (12/X)- X \leq 4. There are only 4 values, with their corresponding probabilities.
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Pakora99
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(Original post by ghostwalker)
The 50 spins don't come into it, unless there's something you've not posted.

You're only dealing with 1 spin, with r.v. X

So, you just need to check when (12/X)- X \leq 4. There are only 4 values, with their corresponding probabilities.
If P(X<=4) then
all values less than it is 0.65 and including is equal to 1
cuz P(X=4) is 0.35 all values less is 0.65
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ghostwalker
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(Original post by Pakora99)
If P(X<=4) then
all values less than it is 0.65 and including is equal to 1
cuz P(X=4) is 0.35 all values less is 0.65
OK, I think I know what you mean, but so what!

You're interested in P(12/X-X<= 4), not P(X<= 4)

You need to evaluate that function over the domain of X, i.e. 1,2,3,4. And check if that function is <=4.
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Pakora99
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(Original post by ghostwalker)
OK, I think I know what you mean, but so what!

You're interested in P(12/X-X<= 4), not P(X<= 4)

You need to evaluate that function over the domain of X, i.e. 1,2,3,4. And check if that function is <=4.
I don't know where to start
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ghostwalker
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(Original post by Pakora99)
I don't know where to start
IF X=1, then (12/X)-X = (12/1)-1 = 11.

This is >4, so we don't include P(X=1) in the sum of the probability.

Etc.
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