In a cricket match, a batter hits a boundary 9 times out of 45 balls he/she plays. The probability that in a given throw he/she does not hit any boundary is 4 5 .Is the above statement true?

Given, the batsman hits the boundary

9

times.
The probability of an event

E

is given as-

P (E) = N u m b e r o f o u t c o m e s f a v o u r a b l e t o E N u m b e r o f a l l p o s s i b l e o u t c o m e s o f t h e e x p e r i m e n t

Therefore, number of favorable outcomes is

9

.
Given, total number of balls faced by batsman is

45

.
Therefore, total number of outcomes is

45

.
Substituting the values, we get the probability of hitting the boundary-

P (E) = 945 = 15

Now, for any event

E

P E + P E ¯ = 1

where

E ¯

stands for ‘not

E

’.
Therefore, the probability of not hitting the boundary is-

P E ¯ = 1 − 15 = 45

Hence, the given statement is true and the correct option is 1.

In a cricket match, a batter hits a boundary $9$ times out of $45$ balls he/she plays.

The probability that in a given throw he/she does not hit any boundary is $45$ .

Is the above statement true?