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Basic question on increased probability

  1. Dec 22, 2008 #1
    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.
  2. jcsd
  3. Dec 22, 2008 #2


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    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.
  4. Dec 22, 2008 #3
    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.
  5. Dec 22, 2008 #4


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    The above two posts have the math correct. Note that in actual use this would be more likely to increase rather than decrease pregnancies...
  6. Dec 23, 2008 #5
    The OP does note that in his original post though he seems skeptical.
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