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 dot ball.

\frac{1}{5}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

\frac{1}{10}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{1}{15}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{1}{30}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $\frac{1}{5}$
Total number of deliveries $= 30$
Number of dot balls $= 30 - (4 + 3 + 13 + 4) = 6$
$P (A) = \frac{N u m b e r o f f a v o r a b l e o u t c o m e s}{T o t a l N u m b e r o f p o s s i b l e o u t c o m e s}$
$∴ P (A) = \frac{6}{30} = \frac{1}{5}$
Hence option $A$ is correct.

Suggest Corrections

Similar questions

30

4

3

13

4

\frac{1}{4}

\frac{1}{5}

\frac{4}{5}

\frac{3}{4}

0, 1, 2, 3, 4

6

30

The last 30 balls of the Indian innings had 10 dot balls. The probability that the next ball is a dot ball is $\frac{1}{3}$

If Sachin scores 5 sixes and 15 fours in 80 deliveries, the probability of Sachin clearing the ropes on any particular delivery is 0.25.

Join BYJU'S Learning Program