Step 1: Write down the information clearly.
- Die I is a normal die with 6 faces: 1, 2, 3, 4, 5, 6.
- Die II is not normal. It has 6 faces too, but the numbers are: 1, 1, 2, 2, 3, 3.
Step 2: Find the total number of outcomes.
- When two dice are thrown, the number of total outcomes = (number of faces on Die I) × (number of faces on Die II).
- That is, \(6 \times 6 = 36\).
- So, there are 36 equally likely outcomes.
Step 3: Understand the sums we need.
- We want the probabilities of sums = 2, 3, 4, 5, 6, 7, 8, 9.
- For each sum, we count how many outcomes give that sum.
Step 4: Count outcomes for each sum.
- Sum = 2: Only possible if Die I = 1 and Die II = 1.
But Die II has two faces with "1".
So number of outcomes = 2.
- Sum = 3: (Die I=1, Die II=2) or (Die I=2, Die II=1).
Each case repeats twice because Die II has two "1"s and two "2"s.
So total outcomes = 4.
- Sum = 4: Possible by (1,3), (2,2), (3,1).
- (1,3): 2 outcomes (because 3 has 2 faces).
- (2,2): 2 outcomes (because 2 has 2 faces).
- (3,1): 2 outcomes (because 1 has 2 faces).
Total = 6.
- Sum = 5: (2,3), (3,2), (4,1).
Each gives 2 outcomes. Total = 6.
- Sum = 6: (3,3), (4,2), (5,1).
Each gives 2 outcomes. Total = 6.
- Sum = 7: (4,3), (5,2), (6,1).
Each gives 2 outcomes. Total = 6.
- Sum = 8: (5,3), (6,2).
Each gives 2 outcomes. Total = 4.
- Sum = 9: (6,3).
Gives 2 outcomes. Total = 2.
Step 5: Convert counts into probabilities.
- Formula: \( P(E) = \dfrac{\text{Favourable outcomes}}{36} \).
- Sum = 2: \(\tfrac{2}{36} = \tfrac{1}{18}\).
- Sum = 3: \(\tfrac{4}{36} = \tfrac{1}{9}\).
- Sum = 4: \(\tfrac{6}{36} = \tfrac{1}{6}\).
- Sum = 5: \(\tfrac{6}{36} = \tfrac{1}{6}\).
- Sum = 6: \(\tfrac{6}{36} = \tfrac{1}{6}\).
- Sum = 7: \(\tfrac{6}{36} = \tfrac{1}{6}\).
- Sum = 8: \(\tfrac{4}{36} = \tfrac{1}{9}\).
- Sum = 9: \(\tfrac{2}{36} = \tfrac{1}{18}\).
Final Note: Always check that the sum of probabilities is 1.
\(\tfrac{2+4+6+6+6+6+4+2}{36} = \tfrac{36}{36} = 1\). ✔ Correct!