Possible outcomes:
(1, 6), (1, 5), (1, 4), (1, 3), (1, 2), (1, 1), (2, 2), (2, 3), (2, 4), (3, 2), (4, 2), (4, 1), (3, 1), (5, 1), (6, 1) and (2, 1)
T. ways=36
favourable outcome=16
p(getting product less than 9) =favourable outcome/T.ways=16/36=4/9

Understanding the Problem
When two dice are thrown, each die has 6 faces numbered from 1 to 6. The total possible outcomes when throwing two dice is 6 x 6 = 36.
Finding Valid Outcomes
To find the probability of the product of the two dice being less than 9, we need to identify all the pairs (a, b) such that the product a * b < 9.="" />
Possible Pairs and Their Products
Here are the pairs and their corresponding products:
- (1, 1) → Product: 1
- (1, 2) → Product: 2
- (1, 3) → Product: 3
- (1, 4) → Product: 4
- (1, 5) → Product: 5
- (1, 6) → Product: 6
- (2, 1) → Product: 2
- (2, 2) → Product: 4
- (2, 3) → Product: 6
- (2, 4) → Product: 8
- (3, 1) → Product: 3
- (3, 2) → Product: 6
- (4, 1) → Product: 4
- (4, 2) → Product: 8
- (5, 1) → Product: 5
- (6, 1) → Product: 6
Count of Valid Outcomes
The valid pairs with products less than 9 are:
- (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)
- (2, 1), (2, 2), (2, 3), (2, 4)
- (3, 1)
- (4, 1), (4, 2)
- (5, 1)
- (6, 1)
Counting these, we find 15 valid outcomes.
Calculating Probability
To find the probability:
- Total outcomes = 36
- Favorable outcomes (product < 9)="" />
Probability = Favorable Outcomes / Total Outcomes = 15 / 36 = 5 / 12.
Conclusion
The probability of finding the product of two dice being less than 9 is 5/12.