In an experiment, there are exactly three elementary events. The probability of two of them are $\frac{2}{7}$ and $\frac{1}{7}$ . What is the probability of third event?

\frac{4}{7}

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\frac{6}{7}

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\frac{3}{7}

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\frac{2}{7}

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Solution

The correct option is A $\frac{4}{7}$
$S i n c e t h e e v e n t h a s e x a c t l y t h r e e e v e n t t h a n s o m e o f p r o b a b i l i t y o f t h e s e t h r e e e v e m t = 1 l e t t h e p r o b a b i l i t y o f 3 r d e v e n t b e p t h e n \Rightarrow p + \frac{1}{7} + \frac{2}{7} = 1 \Rightarrow p = 1 - \frac{3}{7} = \frac{4}{7}$

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Similar questions

If 2/7 is the probability of an event, then the probability of the event 'not A' is:

P (A) = \frac{1}{3}, P (B) = \frac{1}{2}

P (C) = \frac{1}{4}

P (A) = \frac{1}{3}, P (B) = \frac{1}{2}, a n d P (C) = \frac{1}{4} .

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