Mathematics Asked by sirrahe73 on December 6, 2021
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
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
Answered by tommik on December 6, 2021
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