Basic question on increased probability

[kurvmax]

Hi, this is just a curiosity question which occurred to me when I was reading the Wikipedia page on condoms. I'm sorry to start a topic on something so basic, but there isn't a category on the homework section for "other" aside from calculus/physics ect.

Let's say condoms are 98% effective over a year of use in preventing pregnancy. Let's pretend that double-bagging actually increases the effectiveness by adding another 98% chance of effectiveness (in reality, this appears to be untrue). How is that formally done? I was thinking that 98/100 of the times, pregnancy would not happen. 2/100 times it would. Since it is double-bagged, 98% of those 2/100 would not happen. That means .98 * 2 = .0196 or 1.96%. In other words, 98.04% of those using double-bagged condoms would have no pregnancies.

Is that right -- would it increase the chance of no pregnancies .04%? It just seems like such a small difference.

 

[mathman]

Assuming the condom failures are independent, then if the probability of one failing is .02, the probability of both failing is .0004 (.02²). Thus the probability of not failing is .9996.

 

[NoMoreExams]

Note that's the same as .98 + .98 - .98^2 = P(A or B) = P(A) + P(B) - P(A and B)

Which should make sense since you are defining failure as both having to fail therefore success is either one has to succeed.

Mathematically we have .98 + .98 - .98^2 = .98(1 + 1 - .98) = (1 - .02)(1 + .02) = 1 - .02^2 as mathman stated.

 

[CRGreathouse]

The above two posts have the math correct. Note that in actual use this would be more likely to increase rather than decrease pregnancies...

 

[NoMoreExams]

The OP does note that in his original post though he seems skeptical.