# True/False Probability Problems

• Apr 19th 2009, 12:02 PM
krzyrice
True/False Probability Problems
http://i40.tinypic.com/mw5v6c.jpg

Can you confirm that these answers are correct? Thank you.
6. True
7. False
8. False
• Apr 19th 2009, 12:29 PM
krzyrice
http://www.mathhelpforum.com/math-he...a98286c0-1.gif = 1
2/3 + 2/3 - 4/9 =
4/3 - 4/9 =
12/9 - 4/9 =
9/9 = 1
• Apr 19th 2009, 02:02 PM
Soroban
Hello, krzyrice!

Here are the first two . . . I'm working on the third one.

Quote:

Let $A$ and $B$ be two events in a finite sample space.
True or false?

6. If $P(A) + P(B) \:=\:1$, then $A$ and $B$ must be complementary. . True

We have: . $P(B) \:=\:1 - P(A)$

Hence: . $B \:=\:A'$ . . . They are complementary.

Quote:

7. For all $A$ and $B, \;P(A \cap B) + P(A \cup B) \:=\:P(A) + P(B)$ . True
This is a just rearrangement of the well-known formula:

. . . . $P(A \cup B) \:=\:P(A) + P(B) - P(A \cap B)$

• Apr 19th 2009, 04:12 PM
Soroban
Sorry, Plato, I must disagree . . .

The problem gave: . ${\color{blue}P(A) + P(B) \:=\:1}$

There was no mention of an "or", as in $A \cup B$

You have: . $\begin{array}{ccccccc}P(A) &=& P(1,2,3,4) &=& \frac{2}{3} \\ \\[-4mm]P(B) &=& P(3,4,5,6) &=& \frac{2}{3} \end{array}$

. . . . and: . $P(A) + P(B) \:\neq \:1$

• Apr 19th 2009, 06:43 PM
Soroban
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.

. $P(A) \:=\:\frac{13}{52} \:=\:\frac{1}{4}$

Event B: flip two coins and get at least one Head.

. . $P(B) \:=\:\frac{3}{4}$

Hence, $P(A) + P(B) \:=\:1$ . . . but $A$ and $B$ are not complementary.

Well, duh!

• Apr 19th 2009, 09:29 PM
krzyrice
thanks for the help guys (Clapping).