Question:
Two dice are thrown simultaneously. The probability of getting a multiple of \(2\) on one die and a multiple of \(3\) on the other die is:
Two dice are thrown simultaneously. The probability of getting a multiple of \(2\) on one die and a multiple of \(3\) on the other die is:
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When a probability question involves “one die” and “the other die”, count both possible arrangements and then subtract the overlap using the inclusion-exclusion principle.
Updated On: Jun 18, 2026
- \(\frac{5}{12}\)
- \(\frac{11}{36}\)
- \(\frac{13}{36}\)
- \(\frac{5}{36}\)
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The Correct Option is B
Solution and Explanation
Step 1: Find the total number of outcomes.
When two dice are thrown, \[ \text{Total outcomes} = 6\times 6 = 36. \]
Step 2: Define the required events.
Let \[ A=\text{first die shows a multiple of }2, \] \[ B=\text{second die shows a multiple of }3. \] Multiples of \(2\) are \[ \{2,4,6\} \] and multiples of \(3\) are \[ \{3,6\}. \]
Step 3: Count outcomes for the first arrangement.
First die is a multiple of \(2\) and second die is a multiple of \(3\). Number of outcomes: \[ 3\times 2=6. \]
Step 4: Count outcomes for the second arrangement.
First die is a multiple of \(3\) and second die is a multiple of \(2\). Number of outcomes: \[ 2\times 3=6. \] Thus, \[ 6+6=12. \]
Step 5: Subtract common outcomes.
The outcomes counted twice are those where both dice show multiples of both \(2\) and \(3\), i.e., multiples of \(6\). Only number \(6\) satisfies this condition.
Hence the common outcomes are \[ (6,6), \] which gives \[ 1 \] outcome.
Therefore, \[ \text{Favourable outcomes} = 12-1 = 11. \]
Step 6: Calculate the probability.
\[ P = \frac{11}{36}. \]
Step 7: Final conclusion.
Therefore, \[ \boxed{\frac{11}{36}} \]
When two dice are thrown, \[ \text{Total outcomes} = 6\times 6 = 36. \]
Step 2: Define the required events.
Let \[ A=\text{first die shows a multiple of }2, \] \[ B=\text{second die shows a multiple of }3. \] Multiples of \(2\) are \[ \{2,4,6\} \] and multiples of \(3\) are \[ \{3,6\}. \]
Step 3: Count outcomes for the first arrangement.
First die is a multiple of \(2\) and second die is a multiple of \(3\). Number of outcomes: \[ 3\times 2=6. \]
Step 4: Count outcomes for the second arrangement.
First die is a multiple of \(3\) and second die is a multiple of \(2\). Number of outcomes: \[ 2\times 3=6. \] Thus, \[ 6+6=12. \]
Step 5: Subtract common outcomes.
The outcomes counted twice are those where both dice show multiples of both \(2\) and \(3\), i.e., multiples of \(6\). Only number \(6\) satisfies this condition.
Hence the common outcomes are \[ (6,6), \] which gives \[ 1 \] outcome.
Therefore, \[ \text{Favourable outcomes} = 12-1 = 11. \]
Step 6: Calculate the probability.
\[ P = \frac{11}{36}. \]
Step 7: Final conclusion.
Therefore, \[ \boxed{\frac{11}{36}} \]
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