http://i40.tinypic.com/mw5v6c.jpg

Can you confirm that these answers are correct? Thank you.

6. True

7. False

8. False

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- Apr 19th 2009, 12:02 PMkrzyriceTrue/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 PMkrzyrice
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 PMSoroban
Hello, krzyrice!

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

Quote:

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.

Quote:

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) $

- Apr 19th 2009, 04:12 PMSoroban
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: .$\displaystyle P(A) + P(B) \:\neq \:1$

- Apr 19th 2009, 06:43 PMSoroban
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!*

- Apr 19th 2009, 09:29 PMkrzyrice
thanks for the help guys (Clapping).