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: \(P(\text{not } E) = 1 - P(E)\). Here, \(1 - P(3) = 1 - 1/6 = 5/6\).

Answer 1

The Correct Option is C

To determine the probability of getting a number other than 3 when a die is thrown, we can use the fundamental concept of probability.

The probability \( P \) of an event happening is given by the formula:

\(P(\text{event}) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}\)

A standard die has 6 faces, numbered from 1 to 6. So, the total number of possible outcomes when throwing the die once is 6.

The favorable outcomes for getting a number other than 3 are: 1, 2, 4, 5, and 6. This gives us a total of 5 favorable outcomes.

Using the formula mentioned above, the probability of getting a number other than 3 is:

\(P(\text{not 3}) = \frac{5}{6}\)

Therefore, the correct answer is \(\frac{5}{6}\).

Conclusion: The probability of getting a number other than 3 when a die is thrown once is \(\frac{5}{6}\).