In a single throw of two dice, determine the probability of not gettin

1.	In a single throw of two dice, determine the probability of not getting the same number on the two dice.
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 all outcomes except (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6) Total no. of outcomes are 36 and desired outcomes are 30 probability of not getting same number = \(\frac{30}{36}\) = \(\frac{5}{6}\) Conclusion: Probability of not getting the same number on the two dice is \(\frac{5}{6}\)

In a single throw of two dice, determine the probability of not getting the same number on the two dice.

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 all outcomes except (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)

Total no. of outcomes are 36 and desired outcomes are 30 probability of not getting same number = \(\frac{30}{36}\) = \(\frac{5}{6}\)

Conclusion: Probability of not getting the same number on the two dice is \(\frac{5}{6}\)

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