There are three events E1, E2 and E3, one of which is must and only on

1.	There are three events E1, E2 and E3, one of which is must and only one can happen. The odds are 7 to 4 against E1 and 5 to 3 against E2. Find the odds against E3.?
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) = \(\frac{4}{4+7}\) = \(\frac{4}{11}\) ....(ii) Odds against E₂ are 5 : 3 ⇒ Odds in favour of E₂ are 3 : 5 ⇒ P(E₂) = \(\frac{3}{3+5}\) = \(\frac{3}{8}\) ....(iii) ∴ From (i), (ii) and (iii) \(\frac{4}{11}\) + \(\frac{3}{8}\) + P(E3) = 1 ⇒ P(E3) = 1 - \(\bigg(\)\(\frac{4}{11}\)+\(\frac{3}{8}\)\(\bigg)\) = 1 - \(\bigg(\frac{32+33}{88}\bigg)\) = 1 - \(\frac{65}{88}\) = \(\frac{23}{88}\) ∴ Odds against E₃ are \(\frac{1-P(E_3)}{P(E_3)}\) = \(\frac{1-\frac{23}{88}}{\frac{23}{88}}\) = \(\frac{\frac{65}{88}}{\frac{23}{88}}\) = 65 : 23.

There are three events E1, E2 and E3, one of which is must and only one can happen. The odds are 7 to 4 against E1 and 5 to 3 against E2. Find the odds against E3.?

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) = \(\frac{4}{4+7}\) = \(\frac{4}{11}\) ....(ii)

Odds against E₂ are 5 : 3 ⇒ Odds in favour of E₂ are 3 : 5

⇒ P(E₂) = \(\frac{3}{3+5}\) = \(\frac{3}{8}\) ....(iii)

∴ From (i), (ii) and (iii)

\(\frac{4}{11}\) + \(\frac{3}{8}\) + P(E3) = 1 ⇒ P(E3) = 1 - \(\bigg(\)\(\frac{4}{11}\)+\(\frac{3}{8}\)\(\bigg)\)

= 1 - \(\bigg(\frac{32+33}{88}\bigg)\) = 1 - \(\frac{65}{88}\) = \(\frac{23}{88}\)

∴ Odds against E₃ are \(\frac{1-P(E_3)}{P(E_3)}\) = \(\frac{1-\frac{23}{88}}{\frac{23}{88}}\) = \(\frac{\frac{65}{88}}{\frac{23}{88}}\) = 65 : 23.

There are three events E1, E2 and E3, one of which is must and only one can happen. The odds are 7 to 4 against E1 and 5 to 3 against E2. Find the odds against E3.?

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