You are missing on a case where both the events can occur together i.e. getting a \(3\) and getting a Head together. Add \(\frac{1}{12}\) to it.

I think you are getting confused with OR here, OR does not mean only one event will occur. It will be the case if the question says Either.

With respect, I think this answer is arrived at with an incorrect method. Probabilities can only able be added together if they're mutually exclusive outcomes of the same action. For example, with a coin flip, if there's a 1/2 chance of heads and a 1/2 chance of tails, then I can add those up and say there is a 100% chance of either heads or tails occurring.

But since the coin and dice tosses are different events, you can't simply add probabilities together. Because this example wouldn't make sense: "If there's a 1/2 chance of 1-3 on a die toss, and 1/2 chance of heads on a coin toss, there's a 100% chance of getting 1-3 or heads".

In this situation, a 6/12 chance of heads and a 2/12 chance of 3 does not mean a 8/12 chance of heads or 3.

This is what I did to calculate the answer:

The chance of rolling a 3 or getting heads is equal to 1 - the chance of not rolling a 3 or getting heads.

The chance of not getting 3 is 5/6 and the chance of not getting heads is 1/2. So the chance of no heads and no 3 is 5/6 times 1/2 = 5/12.

1 - 5/12 = 7/12, the final answer.

But... it DOES say Either, doesn't it? So wouldn't that mean that the OR is exclusive?

Start by assuming that the die has six sides and that the coin has two sides.

(This might seem obvious, but then again, maybe it doesn't. Furthermore, the die cannot land on a corner and the coin cannot land on its edge. Not that these are impossibilities in the real world, but they are assumed not to exist on GRE.)

The odds that the die will land on "3" out of the six possibilities is: 1/6

The odds that the coin will land on Heads out of the two possibilities is: 1/2

Because we want the probability of one OR the other, we can use the standard probability equation for OR: p(a) + p(b) - p(a^b)

In other words, if you want one or the other, then it's the probability of each independently minus the crossover (both) because it's counted as part of p(a) as well as part of p(b). You double-count and then correct the mistake. Probability is silly that way, but no one has come up with a better option yet.

So we see this: 1/6 + 1/2 - 1/12 = 6/12 + 2/12 - 1/12 = 6/12 + 1/12 = 7/12. Answer C.

P(Not 3) = 5/6
P(Not heads) = 1/2

P(3 | head) = 1 - P(not 3) * P(not head) = 1 - 5/12) = 7/12

Property to recall:
P(A or B)=P(A) + P(B) - P(A and B).

P(Heads)=1/2
P(3)=1/6
P(Heads,3)=1/2*1/6=1/12

Therefore:
P(3 or Heads)=6/12+2/12-1/12=7/12

C

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