Question

A die is thrown once. Probability of getting a number other than 3 is :

• A. $\frac{1}{6}$
• B. $\frac{3}{6}$
• C. $\frac{5}{6}$
• D. 1

Hint:

Using the complement rule ($1 - P(A)$) is extremely efficient for "other than" or "at least" questions in probability.
It simplifies counting when the number of favorable outcomes is large.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We are considering a single roll of a standard fair six-sided die.
We need to find the probability of the event where the rolled number is any number other than 3.

Step 2: Key Formula or Approach:
The probability of an event $E$, denoted as $P(E)$, is calculated as:
\[ P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}} \]
Alternatively, we can use the complement rule:
\[ P(\text{other than 3}) = 1 - P(3) \]

Step 3: Detailed Explanation:
Let us use the direct counting method:

• Determine the sample space $S$ when a die is thrown once:
The possible outcomes are the numbers on the six faces:
\[ S = \{1, 2, 3, 4, 5, 6\} \]
Thus, the total number of possible outcomes is:
\[ n(S) = 6 \]

• Identify the favorable outcomes for the event $E$ ("getting a number other than 3"):
The numbers on a die that are not equal to 3 are 1, 2, 4, 5, and 6.
\[ E = \{1, 2, 4, 5, 6\} \]
Thus, the number of favorable outcomes is:
\[ n(E) = 5 \]

• Apply the probability formula:
\[ P(E) = \frac{n(E)}{n(S)} = \frac{5}{6} \]

Let us verify using the complement method:

• The probability of getting exactly 3 is:
\[ P(3) = \frac{1}{6} \]

• The probability of getting a number other than 3 is the complement:
\[ P(\text{not 3}) = 1 - P(3) \]
\[ P(\text{not 3}) = 1 - \frac{1}{6} = \frac{5}{6} \]

Both approaches confirm the same probability.

Step 4: Final Answer:
The probability of getting a number other than 3 is $\frac{5}{6}$.
Hence, option (C) is correct.