Question

2 dice are thrown together ,find the probability of (i) getting a doublet ,(ii) a sum 10 ot the 2 numbers on the dice

Answer 1

I) all doublet digit are (1,1),(2,2),(3,3),(4,4),(5,5),(6,6) so the total doublet no. are 6 and total probable is 36 so the probability is
6/36 = 1/6
ii) sum of 10 digits are (4,6),(5,5),(6,4) so the probability is
3/36= 1/12

Answer 2

(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)

total number of outcomes = 36
(i) P(getting a doublet) = 6/36
=1/6

(ii)P(getting some 10) = 3/36
= 1/12

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