Probability help

Hey guys I'd appreciate any help on this. I determine that one team has a 30% chance of winning and I make a moneyline wager on it. My next bet, completely independent of the first bet, I determine has a 40% chance of winning and I make a ML wager on that too.

So I hold 2 separate wagers completely independent of each other. What is the probablaity that I lose both wagers? Win both? Win the 30% bet and lose the 40% bet? Lose the 30% and win the 40% bet? Thanks all. 

 On the two bets you have a 58% chance of winning 1 a 12 percent chance of winning both and a 30 % chance of losing both.

X+Y = chance of winning. X×Y chance of winning both (x+y)-(x×y) chance of winning 1

100-(x+y) chance of losing both....

A wins

30%

B wins

40%

Both A & B win

12%

At least A or B wins

58%

Exactly one wins

46%

Both A and B lose

42%

A loses

70%

B loses

60%

Therefore: 

A wins/ B loses .3 * .6 = 18%

A loses/ B wins .7 * .4 = 28%

A. This thread makes my head hurt.

B. Not a weed friendly thread.

C. One of you guys are really smart!

D. What is the correct answer?

E. Going to check some other threads.