In a cricket match, a batsman hits a boundary 5 times every 30 balls he plays. Find the probability that he did not hit a boundary.

\frac{1}{2}

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\frac{5}{6}

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\frac{1}{6}

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\frac{1}{3}

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Solution

The correct option is B $\frac{5}{6}$
Probability = $\frac{N o . o f t r i a l s w h e r e t h e e v e n t h a p p e n e d}{T o t a l n o . o f t r i a l s}$

P (Hitting a boundary) $= \frac{5}{30}$

We know that the sum of probabilities of all possible outcomes of an event is 1.

So, P(not hitting a boundary)
$= 1 - \frac{5}{30} = \frac{25}{30} = \frac{5}{6}$

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In a cricket match, a batsman hits a boundary 5 times every 30 balls he plays. Find the probability that he did not hit a boundary.

The correct option is B 56Probability = No. of trials where the event happenedTotal no. of trials P (Hitting a boundary) =530 We know that the sum of probabilities of all possible outcomes of an event is 1. So, P(not hitting a boundary) =1−530=2530=56