Question

An unbased dice is rolled twice find the
probability of getting a sum equal to 10 on
their top faces​

Answer 1

Answer:

A dice is rolled twice.

Sample Space (S)= {(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)}

n(S)= 6^2= 36.

Let A be the event of getting the sum of 10.

A={(4,6)(5,5)(6,4)}.

n(A)= 3.

So required probability assigned to event A will be n(A)/n(S)= 3/36= 1/12.

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