HELP PLEASE!! Calculate probability of the independent event
P(A or B) when P(a)=.3 and P(B)=.7

- anonymous

HELP PLEASE!! Calculate probability of the independent event
P(A or B) when P(a)=.3 and P(B)=.7

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- schrodinger

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- anonymous

You just have to multiply .3 by .7

- anonymous

oh ok well what about conditional probability with the same variables?

- ybarrap

$$
P(A\cup B)=P(A)+P(B)-P(A\cap B)
$$
Since A and B are independent, \(P(A\cap B)=0\)
So just SUM the independent probabilities.

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## More answers

- anonymous

so just 1?

- ybarrap

Yes.
See - http://en.wikipedia.org/wiki/Inclusion%E2%80%93exclusion_principle

- ybarrap

Specifically, this section - http://en.wikipedia.org/wiki/Inclusion%E2%80%93exclusion_principle#In_probability

- anonymous

ok what about if the variables are P(a)=.5 P(b)=.4 and then P(a and b)=.1?????

- anonymous

I'm supposed to calculate the probability

- ybarrap

You've got everything you need, P(a), P(b) and P(a and b) -- see my formula above! Just plug in for P(a or b).

- ybarrap

$$
P(A\cup B)=P(A)+P(B)-P(A\cap B)=0.5+0.4-0.1
$$
See how I did that?

- anonymous

yes i just thought a and b were supposed to = P(A and B) but the answer is .8

- ybarrap

Are you asking me?

- anonymous

yes(:

- ybarrap

What are you asking, not sure

- anonymous

the answer is .8 right?

- ybarrap

Yes for P(a or b).

- anonymous

ok but theres nothing else i need to find? it just says calculate the probability

- ybarrap

Tell me the whole question to be sure

- anonymous

its 27

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- ybarrap

Oh, those are different problems that P(A or B)
Ok

- ybarrap

Do you know what P(A|B) means?

- anonymous

i know its a conditional probability problem

- ybarrap

$$
P(A|B)=\cfrac{PA\cap B}{P(B)}=\cfrac{0.1}{0.4}
$$
You can do the next.
BRB

- anonymous

i got .25

- ybarrap

Do #27 now

- anonymous

umm isn't that what i did?

- anonymous

oh okay for 27 i got 4

- anonymous

if you can check the other ones i got .21 on 23 and on 25 i got 1 but if you can't then thank you for helping!

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- ybarrap

Probabilities can not be greater than 1 so #27 is not 4

- ybarrap

For #27 what do you have for P(B|A)=P(B and A)/P(A) ?

- ybarrap

We already did #23

- anonymous

oh well you had .1/.4 so in 27 i just did .4/.1

- ybarrap

No. The "|" does NOT mean division. It means "given" P(A|B) means Probability of event A given event B.

- anonymous

you did .1/.4

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- ybarrap

$$
P(B|A)=\cfrac{P(A\cap B)}{P(A)}\\
P(A|B)=\cfrac{P(A\cap B)}{P(B)}
$$
Do you see the difference above between the two?

- ybarrap

Yes and that one is correct

- anonymous

i guess I'm not understanding this then

- ybarrap

So we are dealing with probabilities of events. When we are given that an event occurs, we are asked what is the probability that the other will occur. If they are independent, then there is no effect. But if they are not independent, then conditional probability tells us that
Probability of A given B is the Probability of A and B divided by the probability of B.
Here is a visual
|dw:1433295585568:dw|

- anonymous

ok

- ybarrap

That fraction is
$$
P(A|B)=\cfrac{P(A\cap B)}{P(B)}
$$

- anonymous

that looks like alien writing to me

- ybarrap

$$
P(A|B)=P(A \text{ given }B)=\cfrac{P(A\cap B)}{P(B)}=\cfrac{P(A\text{ and }B)}{P(B)}
$$
Is this better?

- anonymous

haha no I'm sorry... what does it look like with the values plugged in

- ybarrap

|dw:1433296024303:dw|

- ybarrap

$$
P(A|B)=P(A \text{ given }B)=\cfrac{P(A\cap B)}{P(B)}=\cfrac{P(A\text{ and }B)}{P(B)}=.1/.4
$$
For #26

- ybarrap

For #27
$$
P(B|A)=P(B \text{ given }A)=\cfrac{P(A\cap B)}{P(A)}=\cfrac{P(A\text{ and }B)}{P(A)}=.1/.5
$$

- anonymous

ok so its .2

- anonymous

it just made a lot more sense

- anonymous

so because (a and b) the value is .1 and P(a)=.5

- ybarrap

#24
$$
P(A|B)=P(A \text{ given }B)=\cfrac{P(A\cap B)}{P(B)}=\cfrac{P(A\text{ and }B)}{P(B)}=(0.3\times0.7)/0.7
$$
Notice that the result is independent of B because P(A|B) = P(A)

- ybarrap

This happened because A and B are Independent

- ybarrap

If you understood that, then you can do #25 without any calculations

- ybarrap

Correct on your last comment

- anonymous

ok thank you i got it now

- ybarrap

You're welcome

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