Probability of 1 survivor

Suppose that the probability of an insect's laying r eggs is given by a Poisson
distribution with parameter 5. Assume that the probability of an egg surviving to develop is
p. Assume also that the eggs are independent as far as survival is concerned.

If we know that the insect had at most three eggs, and p = 0:08, find the probability of
exactly one survivor.

I used the binomial formula and found out what the probability was for 1survivor based on 1, 2, 3 eggs being laid, then added them up, is this correct?
If I am incorrect could you point me in the right direction!

Thanks

I used the binomial formula and found out what the probability was for 1survivor based on 1, 2, 3 eggs being laid, then added them up, is this correct?

Click to expand...

That is hard to say. It would help immensely if you would show hor you implemented this binomial probability. At first glance, you may have missed something. Suppose the question were "at most 10,000 eggs". What do you suppose you might get if you added up 10,000 values not all that close to zero?

Let's see your work and you can be directed.

For one egglaid 1C1 x o.o8 exp1(1-0.08)exp0=0.08

For 2 laid 2C1 X 0.08exp1(1-0.08)exp1=0.1472


For 3 laid 3C1 X 0.08exp1(1-0.08)exp2=0.2031

I then added them up and came up with 0.4303?????

This is where my 10,000-question comes in. If you keep on the way you are going, your sum is very likely to exceed unity. What's the chance the correct answer shouldbe 112%?

You have not used the Poisson Distribution.

I'l use Bin(a,n,p) to represent the probability of getting 'a' from a binomial distribution with parameters n (number fo experiments) and p (probability of success).

You should have P(1|1 laid) + P(1|2 laid) + P(1|3 laid) = Bin(1,1,0.08)*P(Laying 1) + Bin(1,2,0.08)*P(Laying 2) + Bin(1,3,0.08)*P(Laying 3)

You have the Bin()-parts (I didn't check your arithmetic), but not the P(Laying) parts.

While thinking about the Poisson Distribution, you must also adjust for the truncation.