Question

A coin is tossed and a die is thrown. The probability that the outcome will be head or a number greater than 4 or both, is

• A. $\frac{2}{3}$
• B. $\frac{1}{6}$
• C. $\frac{1}{2}$
• D. $\frac{1}{3}$

Hint:

Whenever a probability question uses the phrasing "A or B or both," it is explicitly asking for the union $P(A \cup B)$. Don't forget to subtract the intersection $P(A \cap B)$ so you don't double-count the scenario where both happen!

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We are dealing with two independent events: tossing a coin and rolling a standard 6-sided die. We need to find the probability of getting a Head, getting a number strictly greater than 4, or both occurring simultaneously.

Step 2: Key Formula or Approach:
This requires the fundamental formula for the probability of the union of two events:
$$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$ Because the coin toss and die roll are completely independent, the probability of both occurring is their product: $P(A \cap B) = P(A) \times P(B)$.

Step 3: Detailed Explanation:
Let Event $A$ be getting a Head on the coin.
$$P(A) = \frac{1}{2}$$ Let Event $B$ be getting a number greater than 4 on the die (i.e., rolling a 5 or 6).
$$P(B) = \frac{2}{6} = \frac{1}{3}$$ Calculate the probability of both happening, $P(A \cap B)$:
$$P(A \cap B) = \frac{1}{2} \times \frac{1}{3} = \frac{1}{6}$$ Now, substitute these into the union formula:
$$P(A \cup B) = \frac{1}{2} + \frac{1}{3} - \frac{1}{6}$$ Find a common denominator (which is 6):
$$P(A \cup B) = \frac{3}{6} + \frac{2}{6} - \frac{1}{6}$$ $$P(A \cup B) = \frac{4}{6} = \frac{2}{3}$$

Step 4: Final Answer:
The required probability is $\frac{2}{3}$, which corresponds to option (A).