# In a certain year, 87% of all Caucasians in the U.S., 77% of all African-Americans, 77% of all...

## Question:

In a certain year, 87% of all Caucasians in the U.S., 77% of all African-Americans, 77% of all Hispanics, and 77% of residents not classified into one of these groups used the Internet for e-mail. At that time, the U.S. population was 66% Caucasian, 11% African-American, and 14% Hispanic.

What percentage of U.S. residents who used the Internet for e-mail were Hispanic? (Round your answer to the nearest whole percent.)

## Application of Theorem of Total Probability and Bayes' Rule

Let {eq}A {/eq} denote an event and {eq}B_1, B_2,.... {/eq} denote a sequence of events which are mutually exclusive and union of which makes the whole sample space.
Further, let {eq}P(.) {/eq} denote the probability of an event.

Then , {eq}P(A) = \sum\limits_{i=1}^n P(A \cap B_i) = \sum\limits_{i=1}^n P(A| B_i)P(B_i). {/eq}
(where {eq}P(A|B_i) {/eq} denote the probability of {eq}A {/eq} given that {eq}B_i {/eq} has already occurred.)

This is the Theorem of Total Probability.

Bayes' Rule is defined as:
{eq}P(A|B) = \frac{P(B|A)P(A)}{P(B)}. {/eq}

Let {eq}A, B, C, D {/eq} denote the events that a resident is Caucasian, African-American, Hispanic and Others respectively.
Let {eq}E {/eq} denote the event that internet is used for e-mail.
Then, {eq}P(A)= 0.66, P(B) = 0.11, P(C) = 0.14, P(D) = 0.09. \\ P(E|A) = 0.87, P(E| B) =0.77, P(E|C)= 0.77, P(E|D) = 0.77. {/eq}
We are required to find {eq}P(C|E). {/eq}

Now, from the Theorem of Total Probability,
{eq}\begin{align*} P(E) &= P(E|A)P(A) +P(E|B)P(B) +P(E|C)P(C) +P(E|D)P(D)\\ & = (0.87\times 0.66) +(0.77\times 0.11)+(0.77\times 0.14)+(0.77\times 0.09)\\ &= 0.5742 + 0.0847 +0.1078 + 0.0693 \\ &= 0.836. \end{align*} {/eq}
Therefore, using Bayes' Theorem, we get,
{eq}P(C|E) = \frac{P(E|C)P(C)}{P(E)}= \frac{0.77 \times 0.14}{0.836}= \frac{0.1078}{0.836}= 0.1289. {/eq}
Therefore, the required percentage of residents is around {eq}13 \%. {/eq}