Can you confirm that these answers are correct? Thank you.
6. True
7. False
8. False
Hello, krzyrice!
Here are the first two . . . I'm working on the third one.
Let $\displaystyle A$ and $\displaystyle B$ be two events in a finite sample space.
True or false?
6. If $\displaystyle P(A) + P(B) \:=\:1$, then $\displaystyle A$ and $\displaystyle B$ must be complementary. . True
We have: .$\displaystyle P(B) \:=\:1 - P(A)$
Hence: .$\displaystyle B \:=\:A'$ . . . They are complementary.
This is a just rearrangement of the well-known formula:7. For all $\displaystyle A$ and $\displaystyle B, \;P(A \cap B) + P(A \cup B) \:=\:P(A) + P(B)$ . True
. . . . $\displaystyle P(A \cup B) \:=\:P(A) + P(B) - P(A \cap B) $
Sorry, Plato, I must disagree . . .
The problem gave: .$\displaystyle {\color{blue}P(A) + P(B) \:=\:1}$
There was no mention of an "or", as in $\displaystyle A \cup B$
You have: .$\displaystyle \begin{array}{ccccccc}P(A) &=& P(1,2,3,4) &=& \frac{2}{3} \\ \\[-4mm]P(B) &=& P(3,4,5,6) &=& \frac{2}{3} \end{array}$
. . . . and: .$\displaystyle P(A) + P(B) \:\neq \:1$
Okay, Plato, you win!
I had to come up with a truly embarrassing example to convince me.
Event A: draw card from a deck and get a Heart.
.$\displaystyle P(A) \:=\:\frac{13}{52} \:=\:\frac{1}{4}$
Event B: flip two coins and get at least one Head.
. . $\displaystyle P(B) \:=\:\frac{3}{4}$
Hence, $\displaystyle P(A) + P(B) \:=\:1$ . . . but $\displaystyle A$ and $\displaystyle B$ are not complementary.
Well, duh!