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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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