Question

Two dice are thrown. Find the odds in favour getting the sum

(i) 4

(ii) 5

(iii) What are the odd against getthing the sum 6?

Answer 1

(i) Let A be the event of 'getting the sum 4'.

Then A at {(1, 3), (3, 1), (2, 2)}

Here, there are three favourable outcomes, while there are (36 - 3) = 33 unfavourable outcomes.

$\therefore$ Odds in favour of the sum $4=\frac{3}{33}=\frac{1}{11}=1:11$

(ii) Let A be the event of 'getting the sum 5'.

Then A = {(1, 4), (4, 1),(2, 3), (3, 2)}
Here, there are four favourable outcomes, while there are (36-4)=32 unfavourable outcomes.

$\therefore$ Odds in favour of the sum $5=\frac{4}{32}=\frac{1}{8}$

(iii) Let A be the event of 'getting the sum 6'.

Then A = {(1, 5), (5, 1), (2, 4), (4, 2), (3, 3)}

Here, there am five favourable outcomes, while there (36-5) = 31 unfavourable outcomes.

$\therefore$ Odds against getting the sum $6=\frac{31}{5}=31:5$