Question

Question #b1268

Answer 1

Firstly we would like to solve for `p` which is the probability of success (Chance of finding an egg)... we can do this by saying:

`p = 150/15 = 0.1` or you can look at it as `10%`

And we would get `1-p = 0.9`.

This solution can be done using the Binomial Distribution

Thus giving us.

`p(k) = (nCk) p^k(1-p)^(n-k)`

`p(1) = ""_4C_1 (0.1)^1(0.9)^(3)`

`p(1) = 4(0.0729) = 0.2916`

But here is the reason as to why- the long way

So we are looking for all the different ways that a child can find `1` $10.00 bill...

We know that `P(AnnB) = P(A).P(B)` and we know that `P(AuuB) = P(A) + P(B)` for independent events.

So let us look at one option, which would be getting the egg with $10.00 of the first trial and not getting on the others.

which will give us:

(The fist egg is a success and the others are not)
`P(1st) = (p)(1-p)(1-p)(1-p) = (0.1)(0.9)(0.9)(0.9) = 0.0729`

now we look at when the second egg is a success:

`P(2nd) = (1-p)(p)(1-p)(1-p) = (0.9)(0.1)(0.9)(0.9) = 0.0729`

and the third egg:

`P(3rd) = (1-p)(1-p)(p)(1-p) = (0.9)(0.9)(0.1)(0.9) = 0.0729`

and now the fourth egg:

`P(4th) = (1-p)(1-p)(1-p)(p) = (0.9)(0.9)(0.9)(0.1) = 0.0729`

and we know that for one of four eggs to have the $10.00 bill, it is either the `1st` or `2nd` or `3rd` or `4th` egg, so we have:

`P(1st) + P(2nd) + P(3rd) + P(4th) = 0.0729 + 0.0729 + 0.0729 + 0.0729 = 4(0.0729) = 0.2916`

Thus we get that the probability of the child having one of its 4 eggs have a $10.00 bill is `= 0.2916` or `29.16%`