In a single throw of two dice, find P (a total greater than 8)

1.	In a single throw of two dice, find P (a total greater than 8)
Answer» We know that, Probability of occurrence of an event = \(\frac{Total\,no.of\,Desired\,outcomes}{Total\,no.of\,outcomes}\) 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) Desired outcomes are (3, 6), (4, 5), (4, 6), (5, 4), (5, 5), (5, 6), (6, 3), (6, 4), (6, 5), (6, 6) Total no. of outcomes are 36 and desired outcomes are 10 Therefore, probability of getting total greater than 8 = \(\frac{10}{36}\) = \(\frac{5}{18}\) Conclusion: Probability of getting total greater than 8, when two dice are rolled is \(\frac{5}{18}\)

Answer»

We know that,

Probability of occurrence of an event

= \(\frac{Total\,no.of\,Desired\,outcomes}{Total\,no.of\,outcomes}\)

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)

Desired outcomes are (3, 6), (4, 5), (4, 6), (5, 4), (5, 5), (5, 6), (6, 3), (6, 4), (6, 5), (6, 6)

Total no. of outcomes are 36 and desired outcomes are 10

Therefore, probability of getting total greater than 8 = \(\frac{10}{36}\) = \(\frac{5}{18}\)

Conclusion: Probability of getting total greater than 8, when two dice are rolled is \(\frac{5}{18}\)

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