Q. 254.5( 2 Votes )

# In a single throw

Total no.of cases will be 6 x 6 x 6 = 216(because each die can have values from 1 to 6)

Desired outcomes are those whose sum up to 5. Desired outcomes are (1, 1, 3), (1, 3, 1), ( 1, 2, 2), (2, 1, 2), (2, 2, 1), (3, 1, 1) i.e. total of 6 cases

As we know,

Probability of occurrence of an event

Therefore, the probability of outcome whose sum is 5

Conclusion: Probability of getting a total of 5 when three dice are thrown is

(ii) Total no.of cases will be 6 x 6 x 6 = 216(because each die can have values from 1 to 6)

Desired outcomes are those whose sum up to 5. Desired outcomes are (1, 1, 3), (1, 3, 1), ( 1, 2, 2), (2, 1, 2), (2, 2, 1), (3, 1, 1) (1, 1, 1), (1, 1, 2), (1, 2, 1), (2, 1, 1), i.e. total of 10 cases

As we know,

Probability of occurrence of an event

Therefore, the probability of outcome whose sum is at most 5

Conclusion: Probability of getting total of at most 5 when three dice are thrown is

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