Need help with 2 probability questions

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sammybones

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Hey, thanks for reading.

This is the first one I'm stuck on.

Given P(A) = 0.13, P(B) = 0.02 and P(A or B) = 0.15, are events A and B mutually exclusive?

My answer would be no, because their sum is not .15, which I think relates to whether they're exclusive or not. (We did not go over this much).

The second one

Given P(E or F) = 0.58, P(F) = 0.12, and P(E and F) = 0.02, what is P(E)?

This one I am very stuck on, my educated guess would be .568, but I do not know where to begin.

I would very much appreciate some guidance, thank you!
 
If the two events are mutually exclusive then [MATH]P[A \cap B]=0[/MATH]
[MATH]P[A \cup B] = P[A]+P[ B ]-P[A \cap B] [/MATH] (you should know this formula by heart! learn it!)

[MATH]P[A \cap B] = P[A \cup B] - P[A] - P[ B ][/MATH]
So are the two mutually exclusive?


For the second question use the same formula above.

[MATH]P[E \cup F] = P[E]+P[F]-P[E \cap F][/MATH]
and solve for [MATH]P[E][/MATH]
 
Romsek said:
If the two events are mutually exclusive then [MATH]P[A \cap B]=0[/MATH]
[MATH]P[A \cup B] = P[A]+P[ B ]-P[A \cap B] [/MATH]
[MATH]P[A \cap B] = P[A \cup B] - P[A] - P[ B ][/MATH]
So are the two mutually exclusive?


For the second question use the same formula above.

[MATH]P[E \cup F] = P[E]+P[F]-P[E \cap F][/MATH]
and solve for [MATH]P[E][/MATH]
Click to expand...

So they would not be mutually exclusive because their sum is not 0. Thank you for clarifying.

The second one would then be .58 = .48 + .12 - 0.02 ? So P(E) = .48?
 
sammybones said:
Hey, thanks for reading.

This is the first one I'm stuck on.

Given P(A) = 0.13, P(B) = 0.02 and P(A or B) = 0.15, are events A and B mutually exclusive?

My answer would be no, because their sum is not .15, which I think relates to whether they're exclusive or not. (We did not go over this much).
Click to expand...
Check your addition! I thought that 0.13 + 0.02 = 0.15, but you seem to be saying it isn't.

sammybones said:
The second one

Given P(E or F) = 0.58, P(F) = 0.12, and P(E and F) = 0.02, what is P(E)?

This one I am very stuck on, my educated guess would be .568, but I do not know where to begin.
Click to expand...
Did you write the equation corresponding to P(E or F) = P(E) + P(F) - P(E and F)? Just fill it in and solve.
 
Dr.Peterson said:
Check your addition! I thought that 0.13 + 0.02 = 0.15, but you seem to be saying it isn't.


Did you write the equation corresponding to P(E or F) = P(E) + P(F) - P(E and F)? Just fill it in and solve.
Click to expand...
Oh right mixed up .02 and .2 thinking it would add .20. They aren't mutually exclusive!

And yes in my reply to Romsek I did. I believe I've got the right answer .48
 
I'll ask the question differently. If 0.13 + 0.02 is not 0.15 then what is it?
 
sammybones said:
It is .15, I have not worked with decimals in a year and a half and we just started probability. Small mistake
Click to expand...
You need to think about it in ways that you are familiar with! You know that 13 cents + 2 cents is 15 cents. You also know that $0.13 is 13 cents and $0.02 is 2 cents. Small mistake??
 
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