Total number of all possible outcomes when two dice thrown together T[E] = 36
Product of the numbers on both dice is less than 9 so favourable outcomes are [1, 1], [1, 2], [1, 3], [1, 4], [1, 5], [1, 6], [2, 1], [2, 2], [2, 3], [2, 4], [3, 1], [3, 2], [4, 1], [4, 2], [5, 1], [6, 1]
\begin{array}{ll} \therefore & \mathrm{F}[\mathrm{E}]=16 \\ \Rightarrow & \mathrm{P}[\mathrm{E}]=\frac{\mathrm{F}[\mathrm{E}]}{\mathrm{T}[\mathrm{E}]}=\frac{16}{36}=\frac{4}{9} \end{array}
Solution : Number of total outcomes=36
When product the number appearing on them is less than 9 then possible ways are [1,6],[15],[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]
Number of possible ways=16
`therefore` Required probability =`[16]/[36]=[4]/[9]`
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Solution
SOLUTION:
The outcomes when two dice are thrown together are
[1,1], [1,2], [1,3], [1,4], [1,5], [1,6]
[2,1], [2,2], [2,3], [2,4], [2,5], [2,6]
[3,1], [3,2], [3,3], [3,4], [3,5], [3,6]
[4,1], [4,2], [4,3], [4,4], [4,5], [4,6]
[5,1], [5,2], [5,3], [5,4], [5,5], [5,6]
[6,1], [6,2], [6,3], [6,4], [6,5], [6,6]
Total number of outcomes = 36
[i] Let A be the event of getting the numbers whose sum is less than 7.
The outcomes in favour of event A are [1, 1], [1,2], [1,3], [1,4], [1,5], [2,1], [2,2], [2,3], [2,4], [3,1], [3,2], [3,3], [4,1], [4,2] and [5,1].
Number of favourable outcomes = 15
∴ P[A ] = Number of favourable outcomesTotal number of outcomes=1536=512
[ii] Let B be the event of getting the numbers whose product is less than 16.
The outcomes in favour of event B are [1,1], [1,2], [1,3], [1,4], [1,5], [1,6], [2,1], [2,2], [2,3], [2,4], [2,5], [2,6], [3,1], [3,2], [3,3], [3,4], [3,5], [4,1], [4,2], [4,3], [5,1], [5,2], [5,3], [6,1] and [6,2].
Number of favourable outcomes = 25
∴ P[B ] = Number of favourable outcomes/Total number of outcomes=2536
[iii] Let C be the event of getting the numbers which are doublets of odd numbers.
The outcomes in favour of event C are [1,1], [3,3] and [5,5].
Number of favourable outcomes = 3
∴ P[C ] = Number of favourable outcomes/Total number of outcomes=336=112
Number of total outcomes = 36
When product of numbers appearing on them is less than 9, then possible ways are [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].
Number of possible ways = 16
∴ Required probability = 16/36
= 4/9