The odds against a certain events are $5 : 2$ and the odds in favour of another events are $6 : 5$ . The probability that at least one of the events will happens is:

\frac{52}{77}

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\frac{13}{36}

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\frac{27}{36}

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\frac{26}{36}

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Solution

The correct option is C $\frac{52}{77}$
$O d d s a g a i n s t f i r s t e v e n t i s 5 : 2 \Rightarrow P r o b a b i l i t y o f a g a i n s t i n f i r s t e v e n t = 5 / 7 O d d s i n f a v o u r o f s e c o n d e v e n t i s 6 : 5 \Rightarrow P r o b a b i l i t y o f f a v o u r i n s e c o n d e v e n t i s 6 / 11 \Rightarrow P r o b a b i l i t y o f a g a i n s t i n s e c o n d e v e n t i s 5 / 11 P r o b a b i l i t y t h a t a t l e a s t o n e e v e m t h a p p e n s = 1 - p r o b a b i l i t y t h a t n o n e h a p p e n s = 1 - \frac{5}{7} * \frac{5}{11} = \frac{52}{77}$

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The odds against a certain events are $5 : 2$ and the odds in favour of another events are $6 : 5$ . The probability that at least one of the events will happens is: