Question about pregnancy condition probability

gary

New member
Joined
Apr 11, 2019
Messages
2
Hi,

Strange post, I hope someone can help. My girlfriend has 2 unlinked genetic conditions which affect her ability to get pregnant, hold the baby to full term and give birth to a natural baby. She is 40 years old and there is a 50% chance she will pass one of these conditions on and she is wanting to go 'forward' with IVF but I am hoping to convince her otherwise but I wanted to give her some real stats before we decide.

So. how do I calculate the overall chance of a healthy baby.

e.g.
a) there is a 50% chance she will pass on one genetic condition
2) She has a 30% chance of having a down syndrome baby
3) She has a 25% chance pass of passing on the other genetic condition
4) she has a 20% chance of stillborn

So how do I calculate the overal probability of a giving birth to a healthy child, assuming she has how problems in pregnancy.


Is it 100% a) =50% , b)15%(2 conditions so odds /2, 3) 8.33 (3rd condition so odds/3) = 73.3% = 26.7% ?

Please help, I'm exhausted. Maths only please. I'm too tired for opinions..Her condition also dictates that there are multiple reasons why she could miscarry so in laymens terms if someone could explain the logic I would really appreciatte it. Please assume I am dumb.
 
Last edited:

MarkFL

Super Moderator
Staff member
Joined
Nov 24, 2012
Messages
1,608
Hello, and welcome to FMH! :)

Assuming the 4 conditions you stated are not interrelated in any way, the probability of none of those things happening is:

\(\displaystyle P(X)=(1-0.5)(1-0.3)[(1-0.25)(1-0.2)=0.21=21\%\)

I know you asked only for math, but I feel I should recommend that you speak to a medical professional who will know much more about how all the stated issues may be related and can give you a more accurate figure/advice. :)
 

gary

New member
Joined
Apr 11, 2019
Messages
2
Thanks Mark, I really appreciate ur reply. We have got referrals lined up, but we have neurological, then cardiologist, then genetics, then back to neuro and a private visit to a specialist for pregnancy advice, but I just wanted to prepare her for what lies ahead. A long wait for bad news....

Could you explain the formula above. I haven't done any math studies since 1990 and I don't understand the format of your calculations.

Regards
Gary
 

MarkFL

Super Moderator
Staff member
Joined
Nov 24, 2012
Messages
1,608
If we know the probability of something happening (we'll call this event \(A\)), then we can, armed with the knowledge that it is certain (probability of 1) that either the event will happen or it will not, state:

\(\displaystyle P(A)+P(\text{not }A)=1\)

Hence:

\(\displaystyle P(\text{not }A)=1-P(A)\)

If we have multiple unrelated events that have known probabilities, and we wish to find the probability that all of them will come to pass, then we simply find the product of all the separate probabilities to compute to probability that all of them will happen.

That was the basic reasoning I used to come up with the result I posted.
 
Top