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I got in a Facebook argument and did some math
Posted: Posted January 25th by Ghowilo
 Could someone check this for me? I'm rather curious if I did it correctly. The argument occurred in a group specifically made for people to be shit-heads to each other, so it really doesn't matter if I did it right the first time. Honestly wouldn't have even bothered if I knew I was posting in that group... Condoms are said to be effective 98% of the time, that means they have a 2% failure rate. Birth control has an 8% failure rate measured to the same standard (100% - 92% effectiveness = 8% failure rate) P(Condom∪Pill) = .02 * .08 P(C∪Pill) = 0.0016 P(C∪Pill) = 0.16% = P(P) (P = Pregnant) There are 157,000,000 women in the United States. If we assume that 2/3 of them are sexually active, and that all of those who are sexually active used both forms of birth control we have. 104,666,667 sexually active women in the United States P(P) = 104,666,667 * 0.0016 P(P) = 167,466 = P(Accidental Pregnancies / year)
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First, I think the symbol you want is $$\cap$$, not $$\cup$$. The first means and, while the second is or. The probability of A AND B is $$P(A) \cdot P(B)$$, assuming that A and B are independent events (maybe that can be assumed here?).

You also have to consider how the failure rate is calculated. Usually they are over a period of time, assuming correct usage (sometime correct usage isn't assumed, which gives different statistics). To accurately combine these values, the failure rates should be over the same amount of time (say, a year).

2/3 is a large percentage. I have no better guess at the moment, but that sounds generous.

Not all of those women are having sex with men.

You're multiplying number correctly, but I don't know how valid a lot of the assumptions that you're making are.

Posted January 27th by EN
EN
Reply to: I got in a Facebook argument and did some math