In a single throw of two dice, what is the probability of getting odd number on one die and 6 on the other?

Please explain!

1/12

Dear student,

Favourable cases are 3

Total=36

Required probabilty=3/36=1/12

S= {(1,1)(1.2)(1,3)(1,4)(1,5)(1,6).......(6,6)} = 36.

Possible outcomes are 1,3,5 on 1st dice and 6 on other.

= {(1,6)(3,6)(5,6)} = 3.

Probability= 3 /36. = 1/12