Question:
A, B, C are three events, one of which must and only one can happen. The odds in favor of A are $4:6$, the odds against B are $7:3$. Thus, odds against C are
A, B, C are three events, one of which must and only one can happen. The odds in favor of A are $4:6$, the odds against B are $7:3$. Thus, odds against C are
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Logic Tip: "Odds in favor" translates to "Successes : Failures", while "Odds against" translates to "Failures : Successes". Always remember the denominator for the probability is the sum of both parts of the ratio!
Updated On: Apr 28, 2026
- $7:3$
- $4:6$
- $6:4$
- $3:7$
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The Correct Option is A
Solution and Explanation
Concept:
If the odds in favor of an event $E$ are $a:b$, the probability of the event is $P(E) = \frac{a}{a+b}$. If the odds against an event $E$ are $a:b$, the probability of the event is $P(E) = \frac{b}{a+b}$. When exactly one of a set of mutually exclusive and exhaustive events must occur, the sum of their probabilities equals 1.
Step 1: Calculate the probabilities of events A and B.
The odds in favor of A are given as $4:6$. $$P(A) = \frac{4}{4 + 6} = \frac{4}{10}$$ The odds against B are given as $7:3$. $$P(B) = \frac{3}{7 + 3} = \frac{3}{10}$$
Step 2: Find the probability of event C.
Since exactly one of the events A, B, or C must happen, they are mutually exclusive and exhaustive. Therefore, the sum of their probabilities must be 1. $$P(A) + P(B) + P(C) = 1$$ Substituting the known values: $$\frac{4}{10} + \frac{3}{10} + P(C) = 1$$ $$\frac{7}{10} + P(C) = 1$$ $$P(C) = 1 - \frac{7}{10} = \frac{3}{10}$$
Step 3: Determine the odds against C.
To find the odds against C, we first need the probability of C not happening, which is $P(C')$. $$P(C') = 1 - P(C) = 1 - \frac{3}{10} = \frac{7}{10}$$ The odds against C are the ratio of the probability of it not happening to the probability of it happening: $$\text{Odds against } C = P(C') : P(C)$$ $$\text{Odds against } C = \frac{7}{10} : \frac{3}{10}$$ $$\text{Odds against } C = 7 : 3$$
If the odds in favor of an event $E$ are $a:b$, the probability of the event is $P(E) = \frac{a}{a+b}$. If the odds against an event $E$ are $a:b$, the probability of the event is $P(E) = \frac{b}{a+b}$. When exactly one of a set of mutually exclusive and exhaustive events must occur, the sum of their probabilities equals 1.
Step 1: Calculate the probabilities of events A and B.
The odds in favor of A are given as $4:6$. $$P(A) = \frac{4}{4 + 6} = \frac{4}{10}$$ The odds against B are given as $7:3$. $$P(B) = \frac{3}{7 + 3} = \frac{3}{10}$$
Step 2: Find the probability of event C.
Since exactly one of the events A, B, or C must happen, they are mutually exclusive and exhaustive. Therefore, the sum of their probabilities must be 1. $$P(A) + P(B) + P(C) = 1$$ Substituting the known values: $$\frac{4}{10} + \frac{3}{10} + P(C) = 1$$ $$\frac{7}{10} + P(C) = 1$$ $$P(C) = 1 - \frac{7}{10} = \frac{3}{10}$$
Step 3: Determine the odds against C.
To find the odds against C, we first need the probability of C not happening, which is $P(C')$. $$P(C') = 1 - P(C) = 1 - \frac{3}{10} = \frac{7}{10}$$ The odds against C are the ratio of the probability of it not happening to the probability of it happening: $$\text{Odds against } C = P(C') : P(C)$$ $$\text{Odds against } C = \frac{7}{10} : \frac{3}{10}$$ $$\text{Odds against } C = 7 : 3$$
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