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in Introduction to Probability Theory by (47.8k points)
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There are three events A, B, and C of which one and only one can happen. If the odds are 7 to 4 against A and 5 to 3 against B, then odds against C is …

(a) 23 : 65 

(b) 65 : 23 

(c) 23 : 88 

(d) 88 : 23

1 Answer

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by (47.4k points)
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Best answer

(b) 65 : 23

If the probability of an event is P then the odds against its occurrence are 1 – P to P. 

Selecting 1 from the 4 number 1, 2, 3,4, can be done in 4 ways 

Here for the event A we are given that = (1 - P)/P = 7/4 

⇒ 4 – 4P = 7P

⇒ 11 P = 4 ⇒ P = 4/11 ⇒ P(A) = 4/11 for the event B 

We are given 

(1 - P)/P = 5/3 ⇒ 5P = 3 – 3P 

⇒ 8P = 3 ⇒ P = 3/8 

P(B) = 3/8 

Now we are given P(A) + P(B) + P(C) = 1 

P(C) = 1 – P(A) – P(B) 

P(C) = 1 - (4/11) - (3/8)

(i.e.,) 

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