KarunMendiratta wrote:
GeminiHeat wrote:
A fair die is rolled once and a fair coin is flipped once. What is the probability that either the die will land on 3 or that the coin will land on heads?
A. 1/2
B. 5/12
C. 7/12
D. 2/3
E. 3/4
P (getting a \(3\)) = \(\frac{1}{6}\)
P (NOT getting a \(3\)) = \(\frac{5}{6}\)
P (getting a Head) = \(\frac{1}{2}\)
P (NOT getting a Head) = \(\frac{1}{2}\)
P (getting a 3 OR a Head) = P (getting a \(3\)) + P (getting a Head) - P (both occur together) = \(\frac{1}{6} + \frac{1}{2} - (\frac{1}{6})(\frac{1}{2}) = \frac{{2 + 6 - 1}}{12} = \frac{7}{12}\)
Hence, option C
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.