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 complementary probability ($1 - P(E)$) is often much quicker and less prone to enumeration errors, especially in problems where the favorable event has many outcomes and the complementary event has very few.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The topic is Probability.
We are examining a single throw of a standard fair six-sided die.
We need to calculate the probability of the event where the outcome is any number other than 3.

Step 2: Key Formula or Approach:
The probability of an event $E$, denoted as $P(E)$, is defined as the ratio of the number of favorable outcomes to the total number of possible outcomes in the sample space $S$:
\[ P(E) = \frac{n(E)}{n(S)} \]
Alternatively, we can use the concept of complementary events:
\[ P(\text{not } A) = 1 - P(A) \]
where $P(A)$ is the probability of getting the number 3.

Step 3: Detailed Explanation:

• Write down the sample space $S$ for a single throw of a die:
\[ S = \{1, 2, 3, 4, 5, 6\} \]
The total number of possible outcomes is:
\[ n(S) = 6 \]

• Define the event $E$ as "getting a number other than 3".
The outcomes that are favorable to event $E$ are:
\[ E = \{1, 2, 4, 5, 6\} \]
The number of favorable outcomes is:
\[ n(E) = 5 \]

• Calculate the probability of the event $E$:
\[ P(E) = \frac{n(E)}{n(S)} = \frac{5}{6} \]

• Let us verify this using the complementary event method:
Let $A$ be the event of getting exactly the number 3.
The only favorable outcome is $\{3\}$, so $n(A) = 1$.
The probability of getting 3 is:
\[ P(A) = \frac{1}{6} \]
The probability of getting a number other than 3 (which is the complement $\bar{A}$) is:
\[ P(\bar{A}) = 1 - P(A) = 1 - \frac{1}{6} = \frac{5}{6} \]
Both methods yield the identical result.


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