Question

Two dice are thrown. Find the odds in favour of getting the sum (i) 4 (ii) 5.
(iii) What are the odds against getting the sum 6?

Answer 1

(i)
Let A be the event of 'getting the sum 4'.
Then A= {(1, 3), (3, 1), (2, 2)}
Here, there are three favourable outcomes, while there are (36 – 3 =) 33 unfavourable outcomes.
∴ 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.
∴ Odds in favour of the sum 5 = $\frac{4}{32}=\frac{1}{8}=1:8$

(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 are five favourable outcomes, while there (36 – 5 =) 31 unfavourable outcomes.
∴ Odds against getting the sum 6 = $\frac{31}{5}=31:5$