# How to solve this?There are three events A,B,C one of which must and only one can happen.The odds are 7 to 3 against A and 6 to 4 against B,find the odds against C

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I cant understand what to take as the favourable event and total event.And how to solve this

I cant understand what to take as the favourable event and total event.And how to solve this

The odds for C are the same as A, or 7 to 3 against.

Notice that for events A and B, we're looking at 10 events in total for both (7 to 3, 6 to 4).

Also notice that we're talking about the odds against the events happening out of 10 tries. A won't happen 7 out of 10 times (which means it will happen 3 out of 10 times). And for B, it won't happen 6 out of 10 times (which means it will happen 4 out of 10 times).

And so out of 10 times, A happen 3 times and B happens 4 times. That leaves 3 times that C can happen (and, in fact, must happen since one and only one event must occur).

And so the odds for C are the same as A, or 7 to 3 against.

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To find the odds against event C, you can use the fact that the sum of the probabilities of all possible outcomes must equal 1.

First, calculate the probabilities of events A and B occurring:

- Probability of event A = 3 / (7 + 3) = 3/10
- Probability of event B = 4 / (6 + 4) = 4/10

Since only one event can occur, the probability of event C is 1 minus the sum of the probabilities of events A and B:

- Probability of event C = 1 - (3/10 + 4/10) = 1 - 7/10 = 3/10

Now, to find the odds against event C, you can use the ratio of the probability of event C not happening to the probability of it happening:

- Odds against event C = (Probability of C not happening) : (Probability of C happening) = (7/10) : (3/10)

Thus, the odds against event C are 7 to 3.

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