Fill in the Blank: Let E and F be events such that P(E ∩ F ) = .34, P(E|F ) = .53. Find P(F).

8.
Fill in the Blank: Let E and F be events in a sample space S such that P(E ∪ F) = .64, P(F) = .34 and P(E) = .43. Find P(E|F ).

9.
Fill in the Blank: A box contains 4 yellow balls, 5 red balls and 6 green balls. Two balls are drawn without replacement. What is the probability that one yellow and one green ball were drawn?

Question

Fill in the Blank: Let E and F be events such that P(E ∩ F ) = .34, P(E|F ) = .53. Find P(F).
  
8. 
Fill in the Blank: Let E and F be events in a sample space S such that P(E ∪ F) = .64, P(F) = .34 and P(E) = .43. Find P(E|F ).
  
9. 
Fill in the Blank: A box contains 4 yellow balls, 5 red balls and 6 green balls. Two balls are drawn without replacement. What is the probability that one yellow and one green ball were drawn?

anonymous · Answer

PLease Help...I'm desperate!!

anonymous · Answer

P(E ∩ F ) =  P(E|F ) x P(F)

anonymous · Answer

9. (4)(6)/(15)(14)

anonymous · Answer

ok so I got .114 for number 9

anonymous · Answer

P(F)+P(E)-P(E ∪ F)= P(E ∩ F ) 
And then use the same formula in my first comment

anonymous · Answer

9. 4/35

anonymous · Answer

is this what the first one would look like .34=.53xP(F)

anonymous · Answer

(4)(6)/(15)(14) = 24/210 = 4/35

anonymous · Answer

yes you are right about the first one

anonymous · Answer

then would i subtract .53 from both sides?

anonymous · Answer

No, you have to divide by 53 on both sides

anonymous · Answer

i'm getting .642

anonymous · Answer

for question 9 : it will be |dw:1349531938277:dw|