Description : There are three events E1, E2 and E3, one of which is must and only one can happen. -Maths 9th
Last 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⇒ ... \(rac{1-rac{23}{88}}{rac{23}{88}}\) = \(rac{rac{65}{88}}{rac{23}{88}}\) = 65 : 23.