A couple will stop given birth immediately a

female child is born. If a male is 5 times more
likely to be born as a female. Find
a) The expected number of delivery of that
family.
b)the probability that they ll be 5 children in the family.
3) Probability that they ll be at most 4 children
in the family.

Help show work

let prob(female) = p

prob(male) = 5p
??? very unrealistic ???

p + 5p=1
p = 1/6 <---- prob(female)

so they could have a female as a first child, as a second child, as a third child, etc
expected outcome = 1/6 + (1/6)(5/6) + (1/6)(5/6)(5/6) + ....

this becomes the sum of an infinite series, with a = 1/6 and r = 5/6
= (1/6) / (1 - 5/6)
= (1/6)(6/1) = 1 ?????
There is of course a limit to the fertility of a mother.
The numbers are weird because your prob of male vs female is totally out of wack!!!

prob(5 children)
= must have had 4 boys, then a girl
= (5/6)^4 (1/6)
= 625/7776

at most 4 children
---> girl as first child
or ---> one boy, then girl
or ---> two boys, then girl
or ---> 3 boys, then girl
= 1/6 + 5/12 + 5/24 + 5/48
= 43/48

To find the expected number of deliveries for the family, we need to consider the probabilities of having a male or female child and calculate the weighted average. Let's denote the probability of having a male child as p and the probability of having a female child as 1-p.

Given that a male is 5 times more likely to be born than a female, we can set up the equation:

p = 5(1-p)

To solve for p, we can simplify the equation:

p = 5 - 5p
6p = 5
p = 5/6

So, the probability of having a male child is 5/6, and the probability of having a female child is 1/6.

a) Expected number of deliveries:
The probability of having a child of a specific gender does not change with subsequent deliveries, as mentioned in the question. So, we can use the concept of expected value to calculate the expected number of deliveries.

Let X be the number of deliveries until a female child is born. X follows a geometric distribution with parameter p = 1/6, as the probability of having a female child is 1/6.

The expected value of a geometric distribution is given by E(X) = 1/p, where p is the probability of success.

Therefore, E(X) = 1 / (1/6) = 6.

So, the expected number of deliveries for the family is 6.

b) Probability of having 5 children in the family:
To calculate the probability of having 5 children in the family, we need to have 4 male children and 1 female child in any order.

The probability of having a male child is 5/6, and the probability of having a female child is 1/6.

Using the binomial probability formula, the probability of having k successes (female children) in n trials (total children) is given by:

P(X = k) = (nCk) * p^k * (1-p)^(n-k)

Here, n = 5 (total children) and k = 1 (female child).

P(X = 1) = (5C1) * (1/6)^1 * (5/6)^(5-1)
= 5 * (1/6) * (5/6)^4
= 5 * (1/6) * (625/1296)
= 3125 / 7776

So, the probability of having 5 children in the family is 3125/7776.

3) Probability of having at most 4 children in the family:
To find the probability of having at most 4 children, we need to calculate the sum of probabilities of having 0, 1, 2, 3, or 4 female children.

P(X ≤ 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)

Using the same binomial probability formula, we can calculate each individual probability and add them up.

P(X = 0) = (5C0) * (1/6)^0 * (5/6)^5
= (1) * (1) * (3125/7776)
= 3125 / 7776

P(X = 1) = (5C1) * (1/6)^1 * (5/6)^4
= 5 * (1/6) * (625/1296)
= 3125 / 7776

P(X = 2) = (5C2) * (1/6)^2 * (5/6)^3
= 10 * (1/36) * (125/216)
= 1250 / 7776

P(X = 3) = (5C3) * (1/6)^3 * (5/6)^2
= 10 * (1/216) * (25/36)
= 250 / 7776

P(X = 4) = (5C4) * (1/6)^4 * (5/6)^1
= 5 * (1/1296) * (5/6)
= 25 / 7776

P(X ≤ 4) = 3125/7776 + 3125/7776 + 1250/7776 + 250/7776 + 25/7776
= 8750 / 7776

So, the probability of having at most 4 children in the family is 8750/7776.