Question

in a one day cricket match ,sachin played 40 balls and hit 12 sixes and saurabh played 30 balls and hit 9 fours. find the probability that sachin will hit a six in the next ball and also find the probability that saurabh will not hit four in the next ball.

Answer 1

sachin

total balls=40
number of sixes=12
probability of six=12/40=6/20=3/10

Sauabh
total balls =30
number of fours=9
so number of other balls=30-9=21
probabiity to not four= 21/30

Answer 2

• The probability that sachin will hit a six in the next ball = 3/10

• The probability that saurabh will not hit four in the next ball = 7/10

Given :

In a one day cricket match ,sachin played 40 balls and hit 12 sixes and saurabh played 30 balls and hit 9 fours.

To find :

• The probability that sachin will hit a six in the next ball

• The probability that saurabh will not hit four in the next ball

Solution :

Step 1 of 2 :

Find the probability that sachin will hit a six in the next ball

Here it is given that sachin played 40 balls and hit 12 sixes

Total number of possible outcomes = 40

Let A be the event that sachin will hit a six in the next ball

So total number of possible outcomes for the event A = 12

The probability that sachin will hit a six in the next ball

= P(A)

$\displaystyle \sf{ = \frac{12}{40} }$

$\displaystyle \sf{ = \frac{ 3}{10} }$

Step 2 of 2 :

Find the probability that saurabh will not hit four in the next ball.

Here it is given that saurabh played 30 balls and hit 9 fours

Total number of possible outcomes = 30

Let B be the event that saurabh will not hit four in the next ball.

So total number of possible outcomes for the event B = 9

So the probability that saurabh will hit four in the next ball

= P(B)

$\displaystyle \sf{ = \frac{9}{30} }$

$\displaystyle \sf{ = \frac{3}{10} }$

Hence probability that saurabh will not hit four in the next ball

= P(B')

= 1 - P(B)

$\displaystyle \sf{ =1 - \frac{3}{10} }$

$\displaystyle \sf{ = \frac{10 - 3}{10} }$

$\displaystyle \sf{ = \frac{7}{10} }$