I have a probability question, I am not sure how to make of it. Hope you can give me some guidance, math has never been my strong subject.

Assuming the lady is 36 years old and is considered part of the "black" line which is 10% chance of pregnancy per cycle, and the lady gets a 6 cycle for that year, what is the probability for pregnancy for that year?

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My answer is .469?

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And if possible, can you point out the name of the concept behind the equation so I can check it out in youtube. Thanks a lot.

  • $\begingroup$ There are several concepts involved. One is independence: it is unlikely that pregnancy is independent from cycle to cycle. Another is realism: if a woman has just six cycles in a year, then she likely is either pregnant or dead by the end of that year! $\endgroup$ – whuber Apr 24 at 2:34
  • $\begingroup$ @whuber, I'm not exactly sure what the problem is. I'm just here to compute on behalf of a family member. It's really 6, or to be specific 6-8 times a year. $\endgroup$ – Earl Apr 24 at 3:57
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    $\begingroup$ @whuber is making a good point here. After various faillures, it is fair to assume that pregnancy probability is lower than first expected $\endgroup$ – David Apr 24 at 11:22

0.469 is the right answer. The Math is quite simple if we "cheat" a little bit and simplify the problem by calculating the probability of NOT getting pregnant instead:

On each cycle, there is a 90% chance of getting no pregnancy. If we are on the remaining 10%, we're done. In other case, on the next cycle, there will be another 90% of non-pregnancy chance (with 90% of 90% being 81%). Then, after three cycles, we are left with 90% of 81%, or 72,9%

In conclusion, at the $n$th, the non-pregnancy chance will be $0.9^n$ (times 100 if you want a percentage) For example, after six cycles, 0.9^6 is about 0.531, so 53.1% faillure chance (or, if you prefer, 46,9% success chance)

For more info on why this holds, take a look at conditional probalbilities!

  • $\begingroup$ Thanks @David for confirming. I will check on conditional probabilities. $\endgroup$ – Earl Apr 24 at 3:58

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