# There are three events A, B and C, one of which must, and only one can happen. The odds against the event A are 7:4 and odds against event B are 5:3. Find the odds against event C - Mathematics and Statistics

Sum

There are three events A, B and C, one of which must, and only one can happen. The odds against the event A are 7:4 and odds against event B are 5:3. Find the odds against event C

#### Solution

Since odds against A are 7:4,

P(A) = 4/(7 + 4) = 4/11

Since odds against B are 5:3,

P(B) = 3/(5 + 3) = 3/8

Since only one of the events A, B and C can happen,

P(A) + P(B) + P(C) = 1

∴ 4/11 + 3/8 + "P"("C") = 1

∴ P(C) = 1 - (4/11 + 3/8)

= 1 - ((32 + 33)/88)

= 23/88

∴ P(C') = 1 – P(C)

= 1 - 23/88

= 65/88

∴ Odds against the event C are P(C'):P(C)

= 65/88:23/88

= 65:23

Concept: Odds (Ratio of Two Complementary Probabilities)
Is there an error in this question or solution?