# Independent probability

• Aug 7th 2006, 09:42 PM
kwtolley
Independent probability
E and F are independent with P(E)=0.35 and P(F)=0.45. P(E union F)=___.
ok this one I have some trouble with. P(E) - P(F) = 0.10 this is my answer is it right. this looks easy but hard for me. list of possible answers are.

1. 0.80
2. 0.10
3. 0.1575
d. 0.6425
• Aug 8th 2006, 07:29 AM
galactus
Cap'n, you're much more learned than me so correct me if I am wrong, but doesn't P(E union F) mean P(E or F), not P(E and F)?.

Since E and F are independent, P(E)P(F)=(.35)(.45)=.1575

P(E or F)=P(E)+P(F)-P(E or F)=.35+.45-.1575=.6425, assuming they're not mutually exclusive.

Then again, mutually exclusive events can not be independent events.
• Aug 8th 2006, 08:32 AM
kwtolley
ok I see
Ok I see both ways, but don't a P(E union F) mean it needs to be added. I only ask this because thats what my text book shows how to do it. So please let me know hich is right, and thanks so much again for all the help.
• Aug 8th 2006, 04:16 PM
Soroban
Hello, kwtolley!

I agree with Galactus . . .

Quote:

$E$ and $F$ are independent with: $P(E)=0.35$ and $P(F)=0.45.\qquad P(E \cup F) = \_\_\_$

$1)\;0.80\qquad 2)\;0.10\qquad 3)\;0.1575\qquad 4)\;0.6425$

Formula: . $\boxed{P(E \cup F)\;=\;P(E) + P(F) - P(E \cap F)}$

Since $E$ and $F$ are independent: . $P(E \cap F) \:=\:P(E)\cdot P(F) \:=\:(0.35)(0.45)\:=\:0.1575$

Therefore: . $P(E \cup F) \;=\;0.35 + 0.45 - 0.1575 \;= \;0.6425$ . . . answer (d)

• Aug 8th 2006, 09:45 PM
kwtolley
thanks for the help
ok I got it