In an innings of $30$ deliveries, a batsman scored $4$ boundaries, $3$ sixes, $13$ singles, $4$ twos and the remaining were dot balls. Find the probability that batsman would score on the next delivery: A six

\frac{1}{5}

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\frac{1}{30}

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\frac{1}{20}

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\frac{1}{10}

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Solution

The correct option is D $\frac{1}{10}$
Total number of deliveries $= 30$
Number of sixes $= 3$
$P (A) = \frac{N u m b e r o f f a v o r a b l e o u t c o m e s}{T o t a l N u m b e r o f p o s s i b l e o u t c o m e s}$
$∴ P (A) = \frac{3}{30} = \frac{1}{10}$
Hence option $D$ is correct.

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