mutual exclusivity
 

p(A)= .5 p(B) = .35

p(A' and B') = .2

Find the probability that both A and B occur.

Re: mutual exclusivity
 

0.05?

Is this a trick question?

Re: mutual exclusivity
 

How did you get to 0.05?

Re: mutual exclusivity
 

Perhaps I should be more clear.

p(A') is where A does NOT happen so .5 for A' and .65 for B'

Re: mutual exclusivity
 

[ QUOTE ]
How did you get to 0.05?

[/ QUOTE ]

Of course he is right:

<font class="small">Code:</font><hr /><pre>
| A = 1 | A = 0 |
-------------------------------------------
B = 1 | | | .35
-------------------------------------------
B = 0 | | .2 |
--------------------------------------------
| .5 | | 1.00
</pre><hr />
Now you go ahead and fill in the missing values:
<font class="small">Code:</font><hr /><pre>
| A = 1 | A = 0 |
-------------------------------------------
B = 1 | (3) .05 | | .35
-------------------------------------------
B = 0 | (2) .45 | .20 |(1) .65
--------------------------------------------
| .50 | | 1.00
</pre><hr />

Re: mutual exclusivity
 

[ QUOTE ]
[ QUOTE ]
How did you get to 0.05?

[/ QUOTE ]

Of course he is right:

<font class="small">Code:</font><hr /><pre>
| A = 1 | A = 0 |
-------------------------------------------
B = 1 | | | .35
-------------------------------------------
B = 0 | | .2 |
--------------------------------------------
| .5 | | 1.00
</pre><hr />
Now you go ahead and fill in the missing values:
<font class="small">Code:</font><hr /><pre>
| A = 1 | A = 0 |
-------------------------------------------
B = 1 | (3) .05 | | .35
-------------------------------------------
B = 0 | (2) .45 | .20 |(1) .65
--------------------------------------------
| .50 | | 1.00
</pre><hr />

[/ QUOTE ]

Ok now now explain your calculations, I'm quite lost on this one.

Re: mutual exclusivity
 

I've managed to figure out that you've .35 * .5 - .2

But I don't understand why you've done it this way, if we take .5 and .65 and * them together we get .325 so there is a .125 gap which means they are not mutually exclusive, why do we just - the .2?

Re: mutual exclusivity
 

Let's try this a different way.

Prob ( A or B ) = 1 - Prob ( A' and B') = 0.8

Prob ( A or B ) = Prob (A) + Prob (B) - Prob (A and B)

Prob (A and B) = 0.5 + 0.35 - 0.8 = 0.05

Basically, for A or B to happen, add up the probabilities of A and B, but the subtract the "A and B" case so that you don't double count.

Re: mutual exclusivity
 

.35 + .5 does not equal the probability of A or B happening, they are not two sides of a coin, they are independent events each with differing probabilities.

Re: mutual exclusivity
 

There are four possibilities. AB, A'B, AB' and A'B'. You have four equations:

AB+AB' = 0.35
AB+A'B = 0.70
A'B' = 0.2
AB+A'B+AB'+A'B' = 1

If we add the first three and subtract the fourth, we get:

AB = 0.35 + 0.70 + 0.20 - 1 = 0.25.

To check, that means:

AB = 0.25
AB' = 0.10
A'B = 0.45
A'B' = 0.20

Those add up to 1 and meet the stated conditions.