Question

A month is selected at random in a year. The probability of it being June or September is

• A. \(\frac{3}{4}\)
• B. \(\frac{1}{12}\)
• C. \(\frac{1}{6}\)
• D. \(\frac{1}{4}\)

Hint:

In "or" probability problems where the events are mutually exclusive (like selecting June or September - you can't select both at once), you can find the probability of each and add them: P(June) + P(September) = 1/12 + 1/12 = 2/12 = 1/6.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
We need to find the probability of selecting one of two specific months from a total of 12 months in a year.

Step 2: Key Formula or Approach:
\[ \text{Probability} = \frac{\text{Number of Favorable Outcomes}}{\text{Total Number of Possible Outcomes}} \]

Step 3: Detailed Explanation:
The total number of possible outcomes is the total number of months in a year, which is 12.
The favorable outcomes are the months "June" or "September".
The number of favorable outcomes is 2.
Now, calculate the probability:
\[ P(\text{June or September}) = \frac{2}{12} = \frac{1}{6} \]

Step 4: Final Answer:
The probability is \(\frac{1}{6}\).