Q. 2.7 - A First Course in Probability

Question

Use Venn diagrams

$\left(a\right)$ to simplify the expressions $(E \cup F)(E \cup F^c)$;

$\left(b\right)$to prove DeMorgan’s laws for events$E$ and$F$. [That is, prove${(E \cup F)}^c = E^c{F}^c$, and ${\left(EF\right)}^c = E^c \cup F^c]$

Step-by-Step Solution

Answer

By using the commutative, distributive, and associative properties of$(\cup ,\cap )$

$a)E$

$b) E F$

$c)$$F\cup EG$

Step 1 Given Information.

$\left(a\right) expressions (E \cup F)(E \cup Fc);$

$\left(b\right) (E \cup F)c = EcFc, and \left(EF\right)c = Ec \cup Fc.$

Step 2 Part (a) Explanation.

a) Simplification of $(E\cup F)\left(E\cup F^c\right)$

$(E\cup F)\left(E\cup F^c\right)$$=(E\cup F)E\cup (E\cup F)F^c$

$=E\cup (E\cup F)F^c$

$=E\cup \left[E\left(F^c\right)\cup F\left(F^c\right)\right]$

$=E\cup E\left(F^c\right)\cup F\left(F^c\right)$ 

$=E\cup E\left(F^c\right)\cup \varnothing$

$=E\cup E\left(F^c\right)$

$=E$

Step 3 part (b) Explanation.

b) Simplification of $(E\cup F)\left(E^c\cup F\right)\left(E\cup F^c\right)$

$(E\cup F)\left(E^c\cup F\right)\left(E\cup F^c\right)$$=(E\cup F)\left(E\cup F^c\right)\left(E^c\cup F\right)$

$=\left[(E\cup F)\left(E\cup F^c\right)\right]\left(E^c\cup F\right)$

$=E\left(E^c\cup F\right)$

$=E\left(E^c\right)\cup EF$

$=\varnothing \cup EF$

$=E F$

Step 4 part (c) Explanation.

c) Simplification of $(E\cup F)(F\cup G)$

$(E\cup F)(F\cup G)=(E\cup F)F\cup (E\cup F)G$

$=F\cup (E\cup F)G$

$=F\cup EG\cup FG$

$=F\cup FG\cup EG$

$=F\cup EG$