Two events E and F, P(E) = 0.36, P(F) = 0.48, and ....

[Navyguy]

For two events E and F, P(E) = 0.36, P(F) = 0.48, and P(E intersec F) = 0.08, find P(FlE).

After trying working this many different ways, I came up with two answers. But if I do it the book way I think "0.167" is right:

. . .0.08 / 0.48 = 0.167

I got this by using "AREA OF E INTERSEC F / AREA OF F". Is this right? Thanks so much for helping me understand.

1. 0.222
2. 0.167
3. 0.750
4. 0.480

 

[pka]

Just learn the rule for calculating conditional probabilites:
\(\displaystyle \
P\left( {A|B} \right) = \frac{{P\left( {A \cap B} \right)}}{{P(B)}},\;P(B) \not= 0.\)

 

[Navyguy]

used the rule

so I used this
P(E intersec F) / P(E) =
0.08/0.36=.22222. Thanks for showing me that formula, it worked.[/u]