A Die is thrown twice.find the probabaility of getting a total 9
Anonymous:
1/9 or 9
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Answered by
3
Total outcomes, n = 36. Favourable outcomes, m = 4. Probability is m/n = 4/36 = 1/9.
Answered by
1
When A Die is thrown twice , total outcomes 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) .
Hence total outcomes = 36
Total outcomes of getting a total 9=
(3,6) , (4,5) , (5,4) , (6,3)
Hence probabaility of getting a total 9 = 4/36
= 1/9
(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) .
Hence total outcomes = 36
Total outcomes of getting a total 9=
(3,6) , (4,5) , (5,4) , (6,3)
Hence probabaility of getting a total 9 = 4/36
= 1/9
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