Probability problem for independent events I have tried to solve a  probability  problem which is: 

In a certain game, you perform three tasks.  You flip a quarter, and
  success would be heads.  You roll a single die, and success would be a
  six.  You pick a card from a full playing-card deck, and success would
  be picking a spades card.   If any of these task are successful, then
  you win the game.  What is the probability of winning?

The way I have solved is: 
Probability of getting head +  Probability of getting six on the dice +Probability of getting spades card
= $1/2 +1/6 + 13/52 = 11/12$
Have I done it correctly? 
If that correct then I see the violation of the probability theory in my method of the calculation, because if I had opportunity to throw more dice or coins, the probability could exceed 1.  
 A: Call $A,B,C$ the successes at the different tasks. Using the inclusion-exclusion principle
$$P(\operatorname{win})=P(A)+P(B)+P(C) - P(A\cap B) - P(A\cap C) - P(B\cap C) + P(A \cap B \cap C)$$
So
$$\frac{1}{2} + \frac{1}{6} + \frac{13}{52} - \frac{1}{2}\frac{1}{6} - \frac{1}{2}\frac{13}{52} - \frac{1}{6}\frac{13}{52}  + \frac{1}{2}\frac{1}{6}\frac{13}{52} = \frac{11}{16}$$
A: The key here is that any of the tasks has to be successful, ie. you may get heads and fail at getting six and spades, but you also may get both heads and roll a six. At both chances you have won the game. 
Let $A_1$ denote the event you get heads, $A_2$ the event you get six and $A_3$ the event you get spades. 
$Pr(win) = P(A_1 A_2' A_3') + P(A_1' A_2 A_3') +P(A_1' A_2' A_3) + P(A_1 A_2 A_3) + P(A_1' A_2 A_3) + P(A_1 A_2' A_3) + P(A_1 A_2 A_3')$
So
$$Pr(win) = \frac{1}{2}\cdot \frac{5}{6}\cdot \frac{3}{4} +... + \frac{1}{2}\cdot \frac{1}{6} \cdot \frac{3}{4} = \frac{33}{48} = \frac{11}{16}$$
A: The easiest way to solve problems like this is to compute the probabily of not winning instead. In order to not win you have to fail all the tasks. And since they are independant you can just multiply the probabilities. So
$$P(\text{lose}) = \frac{1}{2}\cdot \frac{5}{6}\cdot\frac{3}{4} = \frac{5}{16}$$
And therefore $P(\text{win}) = 1-P(\text{lose}) = \frac{11}{16}$.
A: A much simpler way is to take the complement of  $P$(none of the events occur)
$ 1 - \dfrac12\cdot\dfrac56\cdot\dfrac34 = \dfrac{11}{16}$
