# Thread: Probability of monogamous pairs having at least 1 female, given fertility rates

1. ## Probability of monogamous pairs having at least 1 female, given fertility rates

Here is an interesting problem that I have come up with that involves multiple probabilities.

A species is monogamous and any pair will have at least 1 child unless genetics say otherwise.

Adolescents make up 12% of the population and pre-adolescents make up 18% of the population

There is a probability of .5% that 1 member of a pair(either the female or the male) is completely sterile due to genetics. But there is a 20% chance that 1 member of the pair is infertile(has not had a child in 1 year). Infertile couples have a 5% monthly chance of pregnancy if just 1 member is infertile and a 2% monthly chance of pregnancy if both members of a pair are infertile.

Only members above 26 years old(adult age for the species) have children without danger. Adolescent pregnancy is dangerous and so if an adolescent happens to get pregnant, that adolescent has to do 1 of 2 things.

Either:

The adolescent isolates herself from everybody else by going into the wilderness and staying pregnant

Or

The adolescent gets an induced miscarriage.

Having a miscarriage makes adolescents 50% less likely to become pregnant, by accident or trying. 95% of adolescents would prefer a miscarriage over the wilderness.

There is a 40% chance that an adolescent will survive alone in the wilderness.

Nondisjunction during meiosis is rare, it only happens 2% of the time. Number of X chromosomes does not affect fertility

Average age of death is 100 years.

Miscarriage that is not induced happens about 5% of the time. Twin births are rare, only .01%. But out of those 50% are fraternal and 50% are identical. Twin pregnancies are more common but are still rare at 2%. 20% of these will have 1 natural miscarriage, 5% will have both twins with a natural miscarriage, and the other 74.9% are mostly stillbirths with only 5% of females pregnant with twins having an induced miscarriage. Triplets and higher are so rare that they aren't taken into consideration.

Preterm labor has a 10% chance of occuring overall. Here are the chances of infant survival at different times for females in preterm labor:

• First third of Second Trimester: .1%
• Second third of Second Trimester: .5%
• Third Third of Second Trimester: 1%
• First Third of Third Trimester: 10%
• Second Third of Third Trimester: 40%
• Third Third of Third Trimester: 48.4%

There is a 1% chance that the Y chromosome does not have male genes due to meiosis and thus a 1% chance that the X chromosome has male genes.

Out of the possibilities that don't have maleless Y chromosomes or X chromosomes with male genes, there is an exact 50/50 split between males and females for singletons and twins.

Stillbirth rates are at 4%

So given all these events, what is the probability that a given member of the species who has become pregnant will have at least 1 female that survives after birth(probability I think will be the same for having a male)?

So the twin probabilities get added to the singleton probabilities.

Now, I am wanting to solve this myself but there are lots of percents here. I know I will have to add and multiply and subtract but how can I solve this complex problem for the probability of having at least 1 female?

And the stillbirths with twin pregnancies, 95% of those are just with 1 baby due to position and 5% of them are both babies.

2. It seems like there's a lot of irrelevant info here (e.g. information is given about the likelihood of pregnancy, but then in the stated problem, pregnancy is already a given).

I would draw a flow chart and look at all the events that need to occur to have at least one surviving female offspring. Remember that probabilities of independent events multiply together.

E.g. you're pregnant ---> Are you an adolescent (Y/N)? ----> if yes, do you have a miscarriage?

etc.

If you get to some intermediate stage and are stuck, post it here and we can look at it. That'll be a lot easier than people here putting in the up front effort of tracing through all these potential outcomes.

3. ## Is my solution correct

NOTE: In previous post I said .01% when I meant .1% for both twins surviving and I did the math with this .1%

I calculated all the probabilities to a thousandth of a percent, added the singleton to 1 surviving twin for adolescents(2 surviving twins gave me a millionth of a percent in magnitude so I didn't consider it)

So I then divided 1.733%(which was half of the probability of having at least 1 baby survive) by 100 and got .01733.

I then multiplied that by .12 to get .0020796 and multiplied by 100 to get the percent and I got:

.208% for the probability that a member of the species is both an adolescent and had at least 1 female survive

44.5% for singleton female
.712% for 1 surviving twin female
.002% for 2 surviving twins female

45.214% for at least 1 female

31.65% for adult and at least 1 female

31.858% for pregnancy leading to at least 1 female surviving.

Did I get it right? Is that the solution to the thousandth of a percent?

#### Posting Permissions

• You may not post new threads
• You may not post replies
• You may not post attachments
• You may not edit your posts
•