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

To solve this problem, we need to calculate the probability of getting a number on a die that is not 3.

When you throw a standard six-sided die (which has faces numbered from 1 to 6), each face has an equal probability of landing face-up.

Step 1: Determine the Total Number of Outcomes

• A standard die has 6 faces. Therefore, there are 6 possible outcomes when the die is thrown once.

Step 2: Determine the Favorable Outcomes

• We want any number other than 3. The favorable outcomes are therefore: 1, 2, 4, 5, and 6.
• Thus, the number of favorable outcomes = 5.

Step 3: Calculate the Probability

• The probability of an event is given by the formula: \(\text{Probability} = \frac{\text{Number of Favorable Outcomes}}{\text{Total Number of Possible Outcomes}}\)
• Here, the probability of getting a number other than 3 is: \(\frac{5}{6}\)

Conclusion:

• Therefore, the probability of rolling a number other than 3 is \(\frac{5}{6}\).