Question:
A die is rolled once. If the die shows odd number, then the probability of getting other than \( 5 \) is
A die is rolled once. If the die shows odd number, then the probability of getting other than \( 5 \) is
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In conditional probability, always first reduce the sample space based on the given condition, then compute probability within that reduced space.
Updated On: Apr 28, 2026
- \( \dfrac{1}{6} \)
- \( \dfrac{5}{6} \)
- \( \dfrac{1}{3} \)
- \( \dfrac{2}{3} \)
- \( \dfrac{1}{5} \)
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The Correct Option is D
Solution and Explanation
Step 1: Understand the given condition.
A die is rolled, and it is given that the outcome is an odd number.
So we are working under the condition that the sample space is restricted to odd numbers only.
Step 2: List the odd outcomes.
The possible outcomes when a die is rolled are: \[ \{1,2,3,4,5,6\} \] The odd outcomes are: \[ \{1,3,5\} \] So the reduced sample space contains \( 3 \) outcomes.
Step 3: Identify the required event.
We need the probability of getting a number other than \( 5 \), given that the number is odd.
From the set \( \{1,3,5\} \), the outcomes other than \( 5 \) are: \[ \{1,3\} \]
Step 4: Count favourable outcomes.
Number of favourable outcomes: \[ 2 \quad (\text{namely } 1 \text{ and } 3) \]
Step 5: Count total possible outcomes under condition.
Total outcomes in the reduced sample space: \[ 3 \quad (\text{namely } 1,3,5) \]
Step 6: Apply conditional probability formula.
Thus, \[ P(\text{other than }5 \mid \text{odd}) = \frac{\text{favourable outcomes}}{\text{total outcomes}} = \frac{2}{3} \]
Step 7: Final conclusion.
Hence, the required probability is \[ \boxed{\frac{2}{3}} \] Therefore, the correct option is \[ \boxed{(4)\ \dfrac{2}{3}} \]
A die is rolled, and it is given that the outcome is an odd number.
So we are working under the condition that the sample space is restricted to odd numbers only.
Step 2: List the odd outcomes.
The possible outcomes when a die is rolled are: \[ \{1,2,3,4,5,6\} \] The odd outcomes are: \[ \{1,3,5\} \] So the reduced sample space contains \( 3 \) outcomes.
Step 3: Identify the required event.
We need the probability of getting a number other than \( 5 \), given that the number is odd.
From the set \( \{1,3,5\} \), the outcomes other than \( 5 \) are: \[ \{1,3\} \]
Step 4: Count favourable outcomes.
Number of favourable outcomes: \[ 2 \quad (\text{namely } 1 \text{ and } 3) \]
Step 5: Count total possible outcomes under condition.
Total outcomes in the reduced sample space: \[ 3 \quad (\text{namely } 1,3,5) \]
Step 6: Apply conditional probability formula.
Thus, \[ P(\text{other than }5 \mid \text{odd}) = \frac{\text{favourable outcomes}}{\text{total outcomes}} = \frac{2}{3} \]
Step 7: Final conclusion.
Hence, the required probability is \[ \boxed{\frac{2}{3}} \] Therefore, the correct option is \[ \boxed{(4)\ \dfrac{2}{3}} \]
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