Total possible outcomes in throwing two dice = 6×6 = 36
(i)The favourable outcomes of the event ", a total of 6" are,
(1,5), (2,4), (3,3), (4,2), (5,1)
Thus, the total number of favourable outcomes = 5
[tex]P = \frac{ Number \: of \: favourable \: outcomes }{ Total \: number \: of \: possible \: outcomes} [/tex]
[tex]
P (the \: total \: of \: 6) = \frac{5}{36} [/tex]
(ii) The favourable outcomes to the event "a total of 10" are (4,6), (5,5), (6,4)
Thus, the total number of favourable outcomes = 3
[tex]P = \frac{ Number \: of \: favourable \: outcomes }{ Total \: number \: of \: possible \: outcome } [/tex]
[tex]
Hence, P(total \: of \: 10) = \frac{3}{36} = \frac{1}{12} [/tex]
