The Student Room Group

S1 discrete random variables

Hi, the question is:

An ornithologist carries out a study of the numbers of eggs laid per pair by a species of rare birds in its annual breeding season. He concludes that it may be considered as a discrete random variable X with probability distribuion given by

P(X = 0) = 0.2
P(X = r) = k(4rr2) k(4r - r^2) for r = 1, 2, 3, 4
P(X = r) = 0 otherwise

i) find the value of k and write the probability distribution as a table
I was able to do this part of the question

k = 0.08

r 0 1 2 3 4
P(X=r) 0.2 0.24 0.32 0.24 0


ii) The ornithologist observes that the probability of survival (that is of an egg hatching and of the chick living to the stage of leaving the nest) is dependent on the number of eggs in the nest. He estimates the probabilities to be as follows.

r 1 2 3
Probability of 0.8 0.6 0.4
Survival

Find, in the form of a table, the probability distribution of the number of chicks surviving per pair of adults.


pls help with this part of the question - i have tried multiplication and sum other methods but i still can't get the right answer. Pls give me sum help to put me in the right direction! Thank you :smile:

Oh, and also, can someone please tell me how to work out this:

P(Xμ<2σ)P (\mid X - \mu\mid < 2\sigma)

Just tell me what the lines are, and how i can solve the thing in brackets, if possible.

Thank you :smile:
|X-u|<2s

|X|<2s+|u|

There will be 2 values for x if it isn't 0.

You multiply the value of the probability of survival by the values you worked out earlier and then set another probability which would be 0 chicks surviving per adult and this is 1- the sum of your other values.
Reply 2
thank you!!!!
I'm not getting the answer. Please help me
Original post by Aleesha_J
thank you!!!!

How did you find the answer and was it the same as the answer at the back of the textbook?
Original post by Veeroshan97
How did you find the answer and was it the same as the answer at the back of the textbook?


This is rather an old thread, and the person you're quoting hasn't logged on for nearly 12 years! Not quite a record.

What's the answer given in the book and what working have you tried.
Original post by ghostwalker
This is rather an old thread, and the person you're quoting hasn't logged on for nearly 12 years! Not quite a record.

What's the answer given in the book and what working have you tried.

There's a table which has r 0,1,2,3 and below there's P(Y=r) 0.35104, 0.44928, 0.18432, 0.01536. I've been trying to find a way to get these answers but I don't know the correct approach. Please help
Original post by Veeroshan97
There's a table which has r 0,1,2,3 and below there's P(Y=r) 0.35104, 0.44928, 0.18432, 0.01536. I've been trying to find a way to get these answers but I don't know the correct approach. Please help



For a given number of eggs the probability of any given chick surviving is independent of any other. So, if there are 3 eggs, say, then the number of chicks surviving would follow the binomial B(3,0.4) - 0.4 as there are three eggs.

We need to use the law of total probability here - i.e. split the sum up into the different cases of the number of eggs layed.


E.g P(2 survive) = sum over n P(2 survive and n eggs to start).

= sum over n P(2 survive | n at start)P(n at start)

Edit: There's a fair bit of arithmetic involved. If it's not coming out, I suggest posting your working for one of the values.
(edited 3 years ago)
Original post by ghostwalker
For a given number of eggs the probability of any given chick surviving is independent of any other. So, if there are 3 eggs, say, then the number of chicks surviving would follow the binomial B(3,0.4) - 0.4 as there are three eggs.

We need to use the law of total probability here - i.e. split the sum up into the different cases of the number of eggs layed.


E.g P(2 survive) = sum over n P(2 survive and n eggs to start).

= sum over n P(2 survive | n at start)P(n at start)

Edit: There's a fair bit of arithmetic involved. If it's not coming out, I suggest posting your working for one of the values.


😂😂. The thing is I'm not sure how to get the zero eggs surviving value. I'm doing self study here so I may have missed out on understanding a bit theory that would lead me to the answer. If you could show me how to get to just one of them, it would be great. 🙏🏾
Original post by Veeroshan97
😂😂. The thing is I'm not sure how to get the zero eggs surviving value. I'm doing self study here so I may have missed out on understanding a bit theory that would lead me to the answer. If you could show me how to get to just one of them, it would be great. 🙏🏾


P(0 survive) = P(0 survive | 0 laid)P(0 laid) + P(0 survive | 1 laid)P(1 laid) + ... + P(0 survive | 4 laid) P(4 laid)

=1×0.2+(10.8)×0.24+(10.6)2×0.32+(10.4)3×0.24+0=1\times 0.2 + (1-0.8)\times 0.24 + (1-0.6)^2\times 0.32 +(1-0.4)^3\times 0.24+ 0

Don't really need to include the case of 4 laid, since the probability of it happening is 0.
(edited 3 years ago)
Yesss this is perfect 🙏🏾. Thank you so so much for this.
Can you please do it for p( 1 survival)
Original post by Roselin 23
Can you please do it for p( 1 survival)


All the elements of how to go about that have been presented in the thread already.

Need to see some work on your part of what you've tried already.

(Note: It's against forum rules to provide full solutions, amongst other things. See forum guidelines here
(edited 2 years ago)