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
- person "A" has sexual intercourse with the same person "B" who is infected for each and every encounter
- 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?
simply
$$1-(\frac{99}{100})^{100}\approx 63.4\%$$
that is the complementary probability of never being infected during 100 exposures