SOLUTION: Two events, E and F, are such that P(E)=2/5, P(F)=1/6 and P(E U F)=13/30. Show that E and F are neither mutually exclusive nor independent. I'm honestly not even sure what it's

Algebra ->  Probability-and-statistics -> SOLUTION: Two events, E and F, are such that P(E)=2/5, P(F)=1/6 and P(E U F)=13/30. Show that E and F are neither mutually exclusive nor independent. I'm honestly not even sure what it's      Log On


   

Question 868492: Two events, E and F, are such that P(E)=2/5, P(F)=1/6 and P(E U F)=13/30. Show that E and F are neither mutually exclusive nor independent.
I'm honestly not even sure what it's asking me to do, does it want me to prove that it's not mutually exclusive by calculating if it is mutually exclusive? I'm quite lost with this problem..
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
If they were mutually exclusive, then
P%28E%29%2BP%28F%29=1 and P%28E_U_F%29=0
Like a coin can't land heads and tails, it can only land heads or tails.
.
.
.
Independent events follow this rule,
P%28E_U_F%29=P%28E%29%2AP%28F%29
In this case,
13%2F30=%282%2F5%29%281%2F6%29
13%2F30=2%2F30
False, so E and F are not independent either.
As an example, the probability of flipping a coin twice and getting a head and then a tail is,
P%28H%2CT%29=P%28H%29%2AP%28T%29=%281%2F2%29%281%2F2%29=1%2F4
Proves that coin flipping is independent, your second flip doesn't depend on what you got on the first flip.