There are three events E1, E2 and E3, one of which is must and only one can happen. -Maths 9th

1 Answer

Answer :

Since one and only one of the three events E1, E2 and E3 can happen, i.e, they are mutually exclusive. Therefore, P(E1) + P(E2) + P(E3) = 1        ...(i) Odds against E1 are 7 : 4 ⇒ Odds in favour of E1 are 4 : 7⇒ P(E1) = \(rac{4}{4+7}\) = \(rac{4}{11}\)                           ....(ii)Odds against E2 are 5 : 3 ⇒ Odds in favour of E2 are 3 : 5 ⇒ P(E2) = \(rac{3}{3+5}\) = \(rac{3}{8}\)                       ....(iii)∴ From (i), (ii) and (iii)\(rac{4}{11}\) + \(rac{3}{8}\) + P(E3) = 1  ⇒ P(E3) = 1 - \(\bigg(\)\(rac{4}{11}\)+\(rac{3}{8}\)\(\bigg)\)= 1 - \(\bigg(rac{32+33}{88}\bigg)\) = 1 - \(rac{65}{88}\) = \(rac{23}{88}\)∴ Odds against E3 are \(rac{1-P(E_3)}{P(E_3)}\) = \(rac{1-rac{23}{88}}{rac{23}{88}}\) = \(rac{rac{65}{88}}{rac{23}{88}}\) = 65 : 23.

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