Question

a die is thrown thrown twice what is the probability that sum of numbers appear is 10?

Answer 1

TE =36

FE=3(4,6)(5,5)(6,4)

P=FE/TE

$p = \frac{3}{36} \\ p = \frac{1}{12}$

Hope it helps

Answer 2

Total no. of favourable outcomes = 36

no. of possible outcomes = 2 (4,6) (6,4)

the probability,P(E)= 2/36=1/18