Question

A fair die is tossed twice. What are the odds in favour of getting a 4, 5 or 6 on the first toss and a 1, 2, 3 or 4 on the second toss?

Answer 1

As we know that sample space of Tossing two dice contains = 36 numbers

Favourable outcome (4,1) (4,2) (4,3) (4,4) (5,1) (5,2) (5,3) (5,4) ,(6,1) (6,2) (6,3) (6,4) =12

Probability of getting given condition E is

$p(E) = \frac{12}{36} = \frac{1}{3}$
odds in favour of the event
$= \frac{p(E)}{1 - p(E)} \\ \\ = \frac{ \frac{1}{3} }{1 - \frac{1}{3} } \\ \\ = \frac{ \frac{1}{3} }{ \frac{3 - 1}{3} } \\ \\ = \frac{1}{2}$
odds in favour of the event is 1:2

Hope it helps you