What is the probability of getting heads or a 4?

Question

When a coin is flipped and a die is thrown, what is the probability of getting a heads or a 4 ?

What I've tried:

P(Getting Heads) = $\frac12$

P(Getting a 4) = $\frac16$

Thus, P(Getting Heads or 4) $= \frac12 + \frac16$ $ =\frac23$

But the right answer is $\frac{7}{12}$. What am I doing wrong?

Answer 1

You're ignoring ​ P(Getting Heads and 4) .

P(A or B) ​ + ​ P(A and B) ​ ​ ​ = ​ ​ ​ P(A) + P(B)

Answer 2

Think of it as 3 different scenarios where it can be P(Heads or 4):

Scenario One: P(4 and not Heads): (1/6) * (1/2) = 1/12 Scenario Two: P(Heads and not 4): (5/6) * (1/2) = 5/12 Scenario Three: P(Heads and 4): (1/6) * (1/2) = 1/12

Add up these three scenarios and you get 5/12 + 1/12 + 1/12 which equals to 7/12.