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The question

Assume the rate of transmission of HIV is between $\frac{1}{100}$ and $\frac{1}{1000}$. Taking the chance of transmission to be 1%, find the risk of becoming infected if you are exposed 100 times.

Here's what I did:

I made two assumptions

  1. person "A" has sexual intercourse with the same person "B" who is infected for each and every encounter
  2. the likelihood of exposure does not change with each successive encounter.

The problem states that the disease is contracted with a probability of 1% which I will call "p". So (1-p) is the probability of not obtaining the disease. I believe this is a process that repeats itself until the first "success" which I will define as person "A" contracting the disease. The problem also states that person "A" was exposed 100 times.

So,

1st exposure--success occurs with probability $p$

2nd exposure--success occurs with probability $(1-p)p$

3rd exposure--success occurs with probability $(1-p)(1-p)p$

nth exposure--success occurs with probability $(1-p)^{n-1}p$

$S_n= \frac{a(1-r^n)}{1-r}$ , $r=(1-p)$ , $a=p$

So the sum of the first 100 terms is

$S_{100} = \frac{.01(1-0.99^{100})}{1-0.99} \approx 0.634$

Did I approach this problem correctly?

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  • $\begingroup$ You can add probabilities of events only if they are mutually exclusive. $\endgroup$ – Somos Jul 21 at 16:54
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simply

$$1-(\frac{99}{100})^{100}\approx 63.4\%$$

that is the complementary probability of never being infected during 100 exposures

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    $\begingroup$ Will you add a few remarks as to why my approach is incorrect? $\endgroup$ – sirrahe73 Jul 21 at 15:33
  • $\begingroup$ @sirrahe73 : your sum is wrong. The denominator is $1-r=a$ $\endgroup$ – tommik Jul 21 at 17:12

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