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:

You can also use the complement rule: \(P(\text{Not } A) = 1 - P(A)\).
The probability of getting exactly 3 is \(P(3) = \frac{1}{6}\).
So, \(P(\text{Other than 3}) = 1 - \frac{1}{6} = \frac{5}{6}\).

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The probability of an event \(E\) is the ratio of the number of favorable outcomes to the total number of possible outcomes in the sample space.
Step 2: Key Formula or Approach:
\[ P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} \]
Step 3: Detailed Explanation:
When a die is thrown once, the sample space \(S\) is:
\[ S = \{1, 2, 3, 4, 5, 6\} \]
Total number of outcomes, \(n(S) = 6\).
We need the probability of getting a number other than 3.
Let \(E\) be the event of getting a number other than 3.
The favorable outcomes are:
\[ E = \{1, 2, 4, 5, 6\} \]
Number of favorable outcomes, \(n(E) = 5\).
\[ P(E) = \frac{n(E)}{n(S)} = \frac{5}{6} \]
Step 4: Final Answer:
The probability of getting a number other than 3 is \(\frac{5}{6}\).