Probability distribution for discrete data

[chwala]

Homework Statement

For the attached problem below, write the probability distribution of (ii.) the number of chicks of chicks surviving per pair of adults.

Relevant Equations

probability distribution for discrete data

this is a textbook problem shared on a whattsap group by a colleague...
i have no problem in finding the value of $k=0.08$, i have a problem with part (ii) of the problem. I have attached the solution here;

how did they arrive at the probability distribution of $y$?

attached below is working from a colleague, i do not understand how he arrived at the solution. What rule is being used here?

 

[mjc123]

The probability that no chicks survive is the probability of no eggs (0.2), plus the probability of 1 egg times the probability that it does not survive (0.24 * (1-0.8)), plus the probability of two eggs times the probability that neither survives (0.32 * (1-0.6)²), etc. You get the pattern?

 

[chwala]

The probability that no chicks survive is the probability of no eggs (0.2), plus the probability of 1 egg times the probability that it does not survive (0.24 * (1-0.8)), plus the probability of two eggs times the probability that neither survives (0.32 * (1-0.6)²), etc. You get the pattern?

yes, very interesting indeed...so there is no specific rule to this? i thought i could use binomial theorem but i guess i was wrong...what about the pattern in $p(1)$, a bit confusing there on $0.32*2*0.6*0.4$

 

[chwala]

This question is a bit confusing to me, we have the probability distribution of the laying of the eggs ...i.e
$[0.2, 0.24, 0.32, 0.24, 0]$ for $[x=0, 1,2,3,4]$ , now we are given the probabilities of the survival ...ie $[0.8, 0.6,0.4]$ for $[x=1,2,3]$, my question is , is there a relationship in coming up with probability distribution for similar problems where the probability distribution , call it $x$ and probability values of another variable, call it $y$ are given...so as to find the probability distribution of $y$, call this distribution $Y$
is this not conditional probability in a way?
very interesting question, i think i now get it, one has to think in terms of combination in order to attempt this question step by step...cheers

 

[mjc123]

what about the pattern in p(1), a bit confusing there on 0.32*2*0.6*0.4

That's where your binomial theorem comes in. Probability of k successes out of n is
ⁿC_kP^k(1-P)^n-k
So P(2 eggs, 1 survives) = P(2 eggs)*2*P(survive)*P(don't survive)
Factor 2 because either egg might be the one that survives.

 

[chwala]

true, i can see now...cheers

 

[chwala]

My colleagues suggest that we cannot apply binomial distribution the way that I have done, for the simple reason that there are conditions to be followed...$p$ and $q$ has to be a fixed or rather a constant value throughout...when using binomial distribution...

 

[mjc123]

You can't use the same binomial distribution for the whole thing because the survival probabilities are not the same. But for one specific contributor to the probability (e.g. the probability of 2 eggs out of 3 surviving) you use the binomial distribution with the survival probability value for that situation.

 

[Stephen Tashi]

( The ornithologist is apparently assuming the survival of each chick in an nest with k eggs is independent of the survival of the other chicks. )

As to the general pattern of the problem. Suppose a probability space is partitioned in M+1 mutually exclusive events $C_0,C_`1,...C_{M+1}$

The probability of any given event S in the same probability space satistifes:

$Pr(S)=Pr(S∩C_0)+Pr(S∩C_1)+...Pr(S∩C_{M+1})$

$= Pr(S | C_0) Pr(C_0) + Pr(S | C_1) Pr(C_1) + ...Pr(S |C_{M+1}) Pr(C_{M+1})$In the problem, we can let $C_k=$ "k eggs are laid".

For a particular event such as S= "2 chicks survive", you can apply the above expression. A factor such as $P(S|C_3)$ is given by the probability of 2 successes of a random variable that has a binomial distribution with 3 trials and probability of success 0.4.

my question is , is there a relationship in coming up with probability distribution for similar problems where the probability distribution , call it x and probability values of another variable, call it y are given...so as to find the probability distribution of y, call this distribution Y
is this not conditional probability in a way?

As you can see from the above discussion, conditional probability is part of the general pattern. The appearance of binomial distributions is special feature of this particular problem.