1. Help Now Please hard problem!!!

Suppose 30% of the individuals in a community carry a dreaded virus in their blood. Of those with the virus, 5% test negative for the disease and 95% of those who do not have the virus test negative for the disease. Find the probability that a person has the virus even though they test negative for it.

2. Hello, JP101!

We need Bayes' Theorem: .$\displaystyle P(A|B) \:=\:\frac{P(A \cap B)}{P(B)}$

Suppose 30% of the individuals in a community are infected with a dreaded virus.
Of those with the virus, 5% test negative for the disease
and 95% of those who do not have the virus test negative for the disease.
Find the probability that a person has the virus, given they tested negative for it.

We are given: .$\displaystyle \begin{array}{cc}P(\text{inf}) = 0.3 & {\color{blue}[1]} \\ P(\text{not}) = 0.7 & {\color{blue}[2]}\end{array}$

Also: .$\displaystyle \begin{array}{ccc}P(\text{neg}|\text{inf}) = 0.05 & {\color{blue}[3]} \\ P(\text{neg}|\text{not}) = 0.95 & {\color{blue}[4]}\end{array}$

We want: .$\displaystyle P(\text{inf}|\text{neg}) \;=\;\frac{P(\text{inf} \cap \text{neg})}{P(\text{neg})}$ .(a)

Using Bayes' Theorem, [3] becomes: .$\displaystyle P(\text{neg}|\text{inf}) \:=\:\frac{P(\text{neg} \cap \text{inf})}{P(\text{inf})}$
Substituting [1], we have: .$\displaystyle \frac{P(\text{neg}\cap\text{inf})}{0.3} \:=\:0.05\quad\Rightarrow\quad\boxed{ P(\text{neg} \cap \text{inf}) \:=\:0.015}$ .[5]

Using Bayes' Theorem, [4] becomes: .$\displaystyle P(\text{neg}|\text{not}) \;=\;\frac{P(\text{neg}\cap\text{not})}{P(\text{no t})}$
Substituting [2], we have: .$\displaystyle \frac{P(\text{neg} \cap \text{not})}{0.7} \:=\:0.95\quad\Rightarrow\quad P(\text{neg}\cap\text{not}) \:=\:0.665$ .[6]

Combining [5] and [6]: .$\displaystyle P(\text{neg}) \;=\;0.0.015 + 0.665\quad\Rightarrow\quad \boxed{P(\text{neg})\;=\;0.68}$

Substitute into (a): .$\displaystyle P(\text{inf}|\text{neg}) \;=\;\frac{0.015}{0.68} \;\approx\;{\bf{\color{red}0.022}}$