camfanclash
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Size:  454.7 KBCan I have some help on the second part of this question please? I figured out the first part but am very confused about the second part.

I'm guessing it's something to do with manipulating the probability distribution on the left, but how?
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username2538449
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Check khanacademy.org
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camfanclash
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The number of eggs laid per pair of birds can be considered as a discrete random variable X.
P(X=0) = 0.2
P(X=x) = k(4x-x^2) for x=1,2,3,4
P(X=x) = 0 otherwise

Probability of at least one egg surviving depending on number of eggs in nest (x):
Where x=1, probability of survival is 0.8
Where x = 2, probability of survival is 0.6
Where x = 3 probability of survival is 0.4

Find the probability distribution of number of chicks surviving per pair of adults in a table
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ghostwalker
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(Original post by camfanclash)
The number of eggs laid per pair of birds can be considered as a discrete random variable X.
P(X=0) = 0.2
P(X=x) = k(4x-x^2) for x=1,2,3,4
P(X=x) = 0 otherwise

Probability of at least one egg surviving depending on number of eggs in nest (x):
Where x=1, probability of survival is 0.8
Where x = 2, probability of survival is 0.6
Where x = 3 probability of survival is 0.4

Find the probability distribution of number of chicks surviving per pair of adults in a table
As you've written it, there is insufficient information.

E.g. If two eggs are laid, you have the probability of at least one surviving, but you don't have the probability of exactly 1 surviving, or exactly 2 surviving, so you can't work out the distribution of the number of chicks surviving.

Can you link/scan/photo the original question?
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camfanclash
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(Original post by ghostwalker)
As you've written it, there is insufficient information.

E.g. If two eggs are laid, you have the probability of at least one surviving, but you don't have the probability of exactly 1 surviving, or exactly 2 surviving, so you can't work out the distribution of the number of chicks surviving.

Can you link/scan/photo the original question?
Yeah I put itin the original post now.
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ghostwalker
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(Original post by camfanclash)
Yeah I put itin the original post now.
OK. That final table is the probability of an egg surviving to become a chick, for each egg. Not the probabilty of at least one surviving.
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camfanclash
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(Original post by ghostwalker)
OK. That final table is the probability of an egg surviving to become a chick, for each egg. Not the probabilty of at least one surviving.

So, what have you done, any thoughts, and where are you stuck.
So I've completed the first part and got:

When x=0, P(X=x) = 0.2
When x=1, P(X=x) = 0.24
When x=2, P(X=x) = 0.32
When x=3, P(X=x) = 0.24

I'm probably still struggling to interpret the last table. When there are three eggs, isn't the probability of an egg surviving the same as the probability of at least one surviving? Or is it only one egg surviving?
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ghostwalker
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(Original post by camfanclash)
So I've completed the first part and got:

When x=0, P(X=x) = 0.2
When x=1, P(X=x) = 0.24
When x=2, P(X=x) = 0.32
When x=3, P(X=x) = 0.24

I'm probably still struggling to interpret the last table. When there are three eggs, isn't the probability of an egg surviving the same as the probability of at least one surviving? Or is it only one egg surviving?
The final table is the probability of a given egg surviving. You are interested in the actual number of eggs surviving.

So, for example, if there were 3 eggs, and probability of an egg surviving is 0.8 say, then the probability of all 3 eggs surviving to become chicks is 0.8^3. Assuming eggs are independent of each other - which you have to for the question.

There is quite a bit of calculation involved in the full answer invovling a few small binomials.

Let me know if you need more.
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camfanclash
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(Original post by ghostwalker)
The final table is the probability of a given egg surviving. You are interested in the actual number of eggs surviving.

So, for example, if there were 3 eggs, and probability of an egg surviving is 0.8 say, then the probability of all 3 eggs surviving to become chicks is 0.8^3.

There is quite a bit of calculation involved in the full answer invovling a few small binomials.

Let me know if you need more.
Thanks for helping, I'll carry on giving it a go myself now.
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