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?

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Solution

(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. ∴ Odds in favour of the sum 4=333=111=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=432=18 (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. ∴ Odds against getting the sum 6=315=31:5

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