myii
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N/A
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RDKGames
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(Original post by myii)
guyss how can I approach this question?

'A Binomial process over 20 trials with a fixed probability p has a probability of ten successes being 0.0543 and fifteen successes being 0.1457. Find the probability of one trial p??'
P(X=10) = \dfrac{20!}{10!10!}p^{10}(1-p)^{10} = 0.0543

You can solve this equation for two values of p \in [0,1], but only one of them will fit the second condition you must have:

P(X=15) = \dfrac{20!}{15!5!}p^{15}(1-p)^{5} = 0.1457
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myii
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(Original post by RDKGames)
P(X=10) = \dfrac{20!}{10!10!}p^{10}(1-p)^{10} = 0.0543

You can solve this equation for two values of p \in [0,1], but only one of them will fit the second condition you must have:

P(X=15) = \dfrac{20!}{15!5!}p^{15}(1-p)^{5} = 0.1457
Ahhhh okay, i understand hence do I used both equations and solve for p?
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RDKGames
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(Original post by myii)
Ahhhh okay, i understand hence do I used both equations and solve for p?
You can use both simultaneously if you want...

What I am suggesting is just using the first one to solve for two value of p, and then just check which one fits the second equation without solving it.
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myii
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(Original post by RDKGames)
You can use both simultaneously if you want...

What I am suggesting is just using the first one to solve for two value of p, and then just check which one fits the second equation without solving it.
Ahhhhh i get it.. lemme try that out.
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myii
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(Original post by RDKGames)
You can use both simultaneously if you want...

What I am suggesting is just using the first one to solve for two value of p, and then just check which one fits the second equation without solving it.
so basically do i need to rearrange the first one in the form of p = something?
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Dina_acqua16
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(Original post by myii)
so basically do i need to rearrange the first one in the form of p = something?
Basically yeah you do
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myii
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(Original post by RDKGames)
You can use both simultaneously if you want...

What I am suggesting is just using the first one to solve for two value of p, and then just check which one fits the second equation without solving it.
i got 0.2 as p when i rearrange for p in the first one? is that correct?
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myii
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(Original post by Dina_acqua16)
Basically yeah you do
yeahh i got 0.2? do you think this is correct?
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RDKGames
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(Original post by myii)
yeahh i got 0.2? do you think this is correct?
Clearly not. Sub it into the original eqn. to check.

Look;

p^{10}(1-p)^{10} is precisely just (p-p^2)^{10}.

Use this in the equation, we get that

p-p^2 = \pm \sqrt[10]{\dfrac{0.0543 \cdot 10! 10!}{20!}}

and just solve this quadratic for two values of p?
Last edited by RDKGames; 8 months ago
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mqb2766
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You can easily check, by subbing back in the equations.

Why not post your working / validation?
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myii
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(Original post by RDKGames)
Clearly not. Sub it into the original eqn. to check.

Look;

p^{10}(1-p)^{10} is precisely just (p-p^2)^{10}.

Use this in the equation, we get that

p-p^2 = \pm \sqrt[10]{\dfrac{0.0543 \cdot 10! 10!}{20!}}

and just solve this quadratic for two values of p?
ahhh okay i got 1/2 - 1/root of 5 or 1/2 + 1/root of 5?
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mqb2766
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(Original post by myii)
ahhh okay i got 1/2 - 1/root of 5 or 1/2 + 1/root of 5?
Again, check it? But not right ...
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myii
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(Original post by mqb2766)
Again, check it? But not right ...
ahhhh i got 0.997281.. and 2.7180..x10^-3?
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mqb2766
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(Original post by myii)
ahhhh i got 0.997281.. and 2.7180..x10^-3?
Are you checking any of these?
Post some working?
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myii
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(Original post by mqb2766)
Are you checking any of these?
Post some working?
well im checking my workings into the second condition of P(X=15) but i don't think im getting the correct p values when i solve the quadratic, i can use the quadratic formula right? Also, you know 10!10! is it 10! times 10!?

also lol my working is so messy but i will post it when i get some sort of close answer
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mqb2766
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(Original post by myii)
well im checking my workings into the second condition of P(X=15) but i don't think im getting the correct p values when i solve the quadratic, i can use the quadratic formula right? Also, you know 10!10! is it 10! times 10!?

also lol my working is so messy but i will post it when i get some sort of close answer
You're just solving a quadratic. Can't be that messy?

Edit - yes, you can use the quadratic formula, but you can see the answer directly. Yes 10!10! is 10!*10! - you must know this to form the quadratic?
Last edited by mqb2766; 8 months ago
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mqb2766
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(Original post by myii)
Attachment 862004
here are my workings? idk if im doing it correct
The value on the right hand side as you go from line 1 to line 2 is wrong. You seem to be dividing by 10? Can you break up the steps?
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myii
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(Original post by mqb2766)
The value on the right hand side as you go from line 1 to line 2 is wrong. You seem to be dividing by 10? Can you break up the steps?
which value is it the values in the quadratic formula equation?
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mqb2766
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(Original post by myii)
which value is it the values in the quadratic formula equation?
p*(1-p) = ....
What is the numerical value of the right hand side?
(0.0543*10!*10!/20!)^(1/10)
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