Two dice are thrown together find the probability that the product is less than 9

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 (adsbygoogle = window.adsbygoogle || []).push({}); (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). (adsbygoogle = window.adsbygoogle || []).push({}); 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

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