There are three events X, Y and Z, out of which only one must happen. If the odds are 8 to 3 against X, 5 to 2 against Y, the odds against Z must be: Option 1) 34 to 43 Option 2) 43 to 34 Option 3) 8 to 11 Option 4) 7 to 5 Option 5) None of these
Question
There are three events X, Y and Z, out of which only one must happen. If the odds are 8:3 against X, 5:2 against Y, the odds against Z must be:
Option 1) 34:43
Option 2) 43:34
Option 3) 8:11
Option 4) 7:5
Option 5) None of these
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Answer
:
The correct answer is (B) 43:34.
Given:
 The odds are 8:3 against event ( X ).
 The odds are 5:2 against event ( Y ).
To find the odds against event ( Z ):

According to the given odds, we can express the probabilities as follows: $$ \frac{P(X')}{P(X)} = \frac{8}{3} $$
 Thus, ( P(X') = \frac{8}{11} ) and ( P(X) = \frac{3}{11} ).
Similarly, for event ( Y ): $$ \frac{P(Y')}{P(Y)} = \frac{5}{2} $$
 Thus, ( P(Y') = \frac{5}{7} ) and ( P(Y) = \frac{2}{7} ).

Since only one of the events ( X ), ( Y ), and ( Z ) must happen, the sum of their probabilities is 1: $$ P(X) + P(Y) + P(Z) = 1 $$

Substitute the known probabilities: $$ \frac{3}{11} + \frac{2}{7} + P(Z) = 1 $$

To solve for ( P(Z) ), convert the fractions to a common denominator, which is 77: $$ \frac{3}{11} = \frac{21}{77}, \quad \frac{2}{7} = \frac{22}{77} $$ Thus, $$ \frac{21}{77} + \frac{22}{77} + P(Z) = 1 $$

Combine the fractions: $$ \frac{43}{77} + P(Z) = 1 $$
Therefore: $$ P(Z) = 1  \frac{43}{77} = \frac{34}{77} $$

The probability of event ( Z' ): $$ P(Z') = 1  P(Z) = \frac{43}{77} $$

Hence, the odds against ( Z ) are: $$ \text{Odds against } Z = \frac{P(Z')}{P(Z)} = \frac{43}{34} $$
Thus, the odds against ( Z ) are 43:34.
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