# Probability question?

## A sample space $S$ consists of the events ${E}_{1} , {E}_{2} , {E}_{3} , {E}_{4} , {E}_{5.}$ We know $\text{P"(E_1)="P"(E_2)=0.3," P} \left({E}_{3}\right) = 0.1 ,$ and ${E}_{4}$ is twice as likely as ${E}_{5}$. (a) What is the probability of event ${E}_{4}$? of ${E}_{5}$? (b) Let $A = \left\{{E}_{1} , {E}_{3} , {E}_{4}\right\}$ and $B = \left\{{E}_{2} , {E}_{3}\right\} .$ What are $\text{P} \left(A\right)$ and $\text{P} \left(B\right)$? (c) What is $A \cup B$? (d) What is $A \cap B$?

Oct 19, 2017

a) P(E_4)=0.2; " "P(E_5)=0.1.
b) P(A)=0.6;" " P(B)=0.4.
c) ${E}_{1} , {E}_{2} , {E}_{3} , {E}_{4}$
d) ${E}_{3}$

#### Explanation:

We assume the events ${E}_{1}$ through ${E}_{5}$ are mutually exclusive and cover all of $S$. (Like how when you roll a die, you must land on one and only one number.)

a) We know that the sum of all 5 single probabilities must be 1:

$\textcolor{red}{\left[1\right]} \text{ } P \left({E}_{1}\right) + P \left({E}_{2}\right) + P \left({E}_{3}\right) + P \left({E}_{4}\right) + P \left({E}_{5}\right) = 1$

Given

$P \left({E}_{1}\right) = P \left({E}_{2}\right) = 0.3$
$P \left({E}_{3}\right) = 0.1$
$P \left({E}_{4}\right) = 2 P \left({E}_{5}\right)$

we substitute these into $\textcolor{red}{\left[1\right]}$:

$0.3 + 0.3 + 0.1 + 2 P \left({E}_{5}\right) + P \left({E}_{5}\right) = 1$

which simplifies to

$0.7 + 3 P \left({E}_{5}\right) = 1$
$\text{ } 3 P \left({E}_{5}\right) = 0.3$
$\text{ } P \left({E}_{5}\right) = 0.1$

And since $P \left({E}_{4}\right) = 2 P \left({E}_{5}\right)$, we get $P \left({E}_{4}\right) = 2 \left(0.1\right) = 0.2 .$

b) If $A = \left\{{E}_{1} , {E}_{3} , {E}_{4}\right\} ,$ then

$P \left(A\right) = P \left({E}_{1} \cup {E}_{3} \cup {E}_{4}\right)$

and since ${E}_{1} , {E}_{3} ,$ and ${E}_{4}$ are mutually exclusive, we get

$P \left(A\right) = P \left({E}_{1}\right) + P \left({E}_{3}\right) + P \left({E}_{4}\right)$
$\textcolor{w h i t e}{P \left(A\right)} = 0.3 + 0.1 + 0.2$
$\textcolor{w h i t e}{P \left(A\right)} = 0.6$

Similarly, $P \left(B\right) = P \left({E}_{2}\right) + P \left({E}_{3}\right) = 0.4 .$

c) We are asked to describe the union of events $A$ and $B .$ This consists of all the events that show up in at least one of $A$ or $B$ (and potentially both). Observing the elements of both $A$ and $B$, we find

$A \cup B = \left\{{E}_{1} , {E}_{2} , {E}_{3} , {E}_{4}\right\}$.

d) Similarly, here we are asked to find the intersection of $A$ and $B ,$ which consists of all the events that appear in both $A$ and $B .$ Since the only event to appear in both $A$ and $B$ is ${E}_{3} ,$ we can say

$A \cap B = \left\{{E}_{3}\right\}$.

## Note:

It is impossible for any event to have negative probability. By definition, a probability must be a number between 0 and 1 (inclusive). For example, it would be ridiculous if your local weather forecast said there was a –15% chance of rain today.