Hi,
It is known that in Europe, between 5% and 9% of all births of babies
occur prematurely. 5% of PBs occur at less than 28 weeks gestation.

Taking an average of 5%-9% for all premature births, is the
proabability that a child is born prematurely at 24 weeks gestation
therefore (0.07 x 0.05) = 0.0035? Is that a way of stating that "if
you have a baby, the chances that it will be born at less than 28
weeks gestation is 0.35%"?


Thanks v. much for any help.

No, you can't make any assumptions about the distribution of the gestation period from the information given. So assuming any kind of linear distribution is unjustified, since the two facts you have don't contain that information.

Ah, so there is not enough information to put the two together?

I thought it's like tossing a coin and a dice together. The probability of getting a head on the coin AND a six on the dice is:

(1/2) x (1/6) = 1/12

Why can't I do this same thing with the birth question?

Cheers for your help

You multiply to find the chance of two independent events both occurring, which isn't the case here. You have one statistic that is a subset of another one. You also have only one data point on the distribution. Was it a typo to put 24% in the question after you quoted the 28% statistic?

As to your second question, it appears you answered it in the original premise. You state: Then you ask, "if you have a baby, the chances that it will be born at less than 28 weeks gestation is 0.35%"? But you already said it was 5%. I think you may have meant to ask, "If you have a premature baby, the chances.... But you can't answer that either with the information given.

I think the questions were perhaps just poorly worded and I'm not sure I've given you what you are looking for.

There is some confusion in the question, since you use 24 weeks one time and 28 weeks two other times. If we assume you meant to say 28 weeks, what you have is the following:

P(premature) = 7% (using your midpoint value)
P(<28 weeks|premature) = 5%

Then, indeed, P(<28 weeks) = P(premature) x P(<28 weeks|premature) = 0.35%

Not defined in the problem statement is what is premature, but it would have to be some period after 28 weeks.

statmanhal is correct and I misunderstood the question, and was further confused by the "24%" in one place. I think you need to use the range in the answer instead of the midpoint however, which gives:

(5% * 5%) < P < (9% * 5%)

0.25% < P < 0.45%

That's brilliant. Yes the baby was born at 24 weeks gestation, so I used the <28 weeks statistic.

But P(<28 weeks|premature) = 5% is a subset of P(premature) = 7% ... and the two are not independent statistics, so is it correct to multiply them out in this way?

BTW, they regard any child born before 37 weeks of pregnancy as premature.

Yes, it is ALWAYS true that

P(A AND B) = P(A)*P(B|A).

When A and B are INDEPENDENT, then P(B|A) = P(B) by definition, so this becomes

P(A AND B) = P(A)*P(B).

Super.

So, now the best way to phrase this in the book I'm writing. "if
you have a baby, the chances that it will be born at less than 28
weeks gestation is just 0.35%"? I hope that reads well... if not I sure would appreciate your guidance for correcting it.

The weirdest part of this birth is also this. This child was also born on someone's birthday! And prematurely! And also at <28 weeks gestation, which is classed as the most extreme end of prematurity in order to deliver a live birth!

So now, I know that the probability of having a child on one's own birthday is (1/365) (unless it was born on Feb 29, which is not the case here.)

And it was a girl (1/2).

So the probability that it was born on one's birthday AND that it was a girl, would that be just (1/365) x (1/2) = 0.00137? Is this technically valid?

And of the ~50 million species on earth, it was human. So the chances of this particular baby actually coming into existence (being human, female, born on someone else's birthday, AND premature) are something like 1 in 100,000,000,000.

It is that baby's lucky day! She should buy a lotto ticket as soon as possible.