Thread: Percentage Calculation Problem

Originally Posted by j198669

This is a real life word problem scenario. I need help figuring out the percent chance based on the statistics given.

Person A acts as the insertive partner in intercourse. Person B acts as a receptive partner in intercourse. Person C acts as another receptive partner in intercourse. Person B has HIV.

Every 10,000 times Person A had sex with Person B, 6.5 of the 10,000 intercourse episodes would result in Person A contracting HIV. Person A engaged in intercourse with Person B 5 times.

6.5/10000= 0.000065 is the probability person A contracts HIV any one time. 1- 0.000065= 0.999935 is the probability person A does NOT contract HIV each time. (0.999935)⁵= 0.999675 is the probability person A did NOT contract HIV after 5 times so 1- 0.999675= 0.00032496 is the probabilty that person A contracted HIV from person B at least once in the 5 times.

Every 10,000 times Person A had sex with Person C, 50 of the 10,000 intercourse episodes would result in Person C contracting HIV (if person B already gave Person A HIV).

So, given that person A has HIV, the probability that person C contracts it on any one time is 50/10000= 0.005.

What is the percent chance that Person B gave Person A HIV *AND* Person A subsequently gave Person C HIV? It is understood that Person A and Person B engaged in intercourse five times and then after that Person A and Person C engaged in intercourse twice.

Given that person A has HIV, the probability person C does NOT contract HIV each time is 1- 0.005= 0.995. The probability person C does NOT contract HIV in two times is (0.995)²= 0.990025 so the probability person C does contract HIV is 1- .990025= 0.009975.

The probability that person A contracts HIV and that person C then contracts it from person A is (0.00032496)(0.009975)= 0.0000032415.