The odds against certain event are 5:2 and the odds in favour of anoth

1.	The odds against certain event are 5:2 and the odds in favour of another in dependent event are 6:5. The probability that at least one of the event will happen is(a) \(\frac{12}{77}\)(b) \(\frac{25}{77}\)(c) \(\frac{52}{77}\)(d) \(\frac{65}{77}\)
Answer» (c) \(\frac{52}{77}\) Given, odds against Event 1 = 5 : 2 ⇒ P(Event 1 not happening) = \(\frac{5}{5+2}\) = \(\frac{5}{7}\) Odds in favour of Event 2 = 6 : 5 ⇒ P(Event 2 happens) = \(\frac{6}{6+5}\) = \(\frac{6}{11}\) ⇒ P(Event 2 not happening) = 1 - \(\frac{6}{11}\) = \(\frac{5}{11}\) ∴ P(None of the events happen) = \(\frac{5}{7}\)x \(\frac{5}{11}\) = \(\frac{25}{77}\) (∵ Both event are independent) ⇒ P(At least one event happens) = 1 - \(\frac{25}{77}\) = \(\frac{52}{77}\).

The odds against certain event are 5:2 and the odds in favour of another in dependent event are 6:5. The probability that at least one of the event will happen is(a) \(\frac{12}{77}\)(b) \(\frac{25}{77}\)(c) \(\frac{52}{77}\)(d) \(\frac{65}{77}\)

Answer»

(c) \(\frac{52}{77}\)

Given, odds against Event 1 = 5 : 2

⇒ P(Event 1 not happening) = \(\frac{5}{5+2}\) = \(\frac{5}{7}\)

Odds in favour of Event 2 = 6 : 5

⇒ P(Event 2 happens) = \(\frac{6}{6+5}\) = \(\frac{6}{11}\)

⇒ P(Event 2 not happening) = 1 - \(\frac{6}{11}\) = \(\frac{5}{11}\)

∴ P(None of the events happen) = \(\frac{5}{7}\)x \(\frac{5}{11}\) = \(\frac{25}{77}\)

(∵ Both event are independent)

⇒ P(At least one event happens) = 1 - \(\frac{25}{77}\) = \(\frac{52}{77}\).

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The odds against certain event are 5:2 and the odds in favour of another in dependent event are 6:5. The probability that at least one of the event will happen is(a) \(\frac{12}{77}\)(b) \(\frac{25}{77}\)(c) \(\frac{52}{77}\)(d) \(\frac{65}{77}\)

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