When two dice are thrown simultaneously What is the probability that the sum of two numbers that turns up is less than 11?

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Concept used:

P(E) = Number of events/Number of possible outcomes

Calculations:

Two dice are thrown simultaneously 

So, the total possible outcomes = 6 × 6 = 36 

Total number of events of getting a sum of 2 or 8 or 12

⇒ (1,1), (2,6), (3,5), (4,4), (5,3), (6,2), (6,6) = 7

So, the probability of getting a sum of 2 or 8 or 12 = 7/36 = 7/36

∴ The probability of getting a sum of 2 or 8 or 12 is 7/36.

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When two dice are thrown simultaneously, the probability is n(S) = 6x6 = 36

Required, the sum of the two numbers that turn up is less than 12

That can be done as n(E)

= { (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) }

= 35

Hence, required probability = n(E)/n(S) = 35/36.

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