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Question: In a cricket, a batswoman hits a boundary \(6\) times, out of \(30\) balls she plays. Find the proba...

In a cricket, a batswoman hits a boundary 66 times, out of 3030 balls she plays. Find the probability that she did not hit a boundary.

Explanation

Solution

Number of balls the batswoman does not hit a boundary=306=24=30-6=24.
PP(She does not hit the boundary)=2430=45=\dfrac{24}{30}=\dfrac{4}{5}.
Let us understand this problem by finding the terms needed for finding the probability that she did not hit a boundary. For this, we need an event and a sample space. So the event will be that the batswoman did not hit a boundary and sample space will be the balls she played.

Complete step by step answer:
Let us consider EE to be an event that the batswoman does hit a six and SS be the sample space of the balls she played.
Here, the total number balls she played is 3030 out of which the batswoman hit a boundary 66 times. It means she hit a boundary 66 times in 66 balls, and in the remaining balls she did not hit a boundary.
So we can write it as,
Total number of balls the batswoman played=30=30.
Number of times the batswoman hit a boundary=6=6.

We will use these terms to find the number of balls the batswoman does not hit a boundary.
So, the number of times the batswoman does not hit a boundary =306=24=30-6=24.
As is EE the event that the batswoman does not hit a six, therefore n(E)=24n\left( E \right)=24.
Also, SSis the sample space of balls the batswoman played, therefore n(S)=30n\left( S \right)=30.

Now we have n(E)n\left( E \right) and n(S)n\left( S \right) , so we can find P(E)P\left( E \right), i.e., PP(She does not hit the boundary).
\therefore PP(She does not hit the boundary)
=P(E)=P\left( E \right)
== number of balls when she did not hit a boundarytotal number of balls she played\dfrac{\text{number of balls when she did not hit a boundary}}{\text{total number of balls she played}}
=n(E)n(S)=\dfrac{n\left( E \right)}{n\left( S \right)}
=2430=\dfrac{24}{30}
=45=\dfrac{4}{5}.

Hence, the probability that she did not hit a boundary is 45\dfrac{4}{5}.

Note: We can also solve this problem by this method:
Total number of balls played=30=30.
Number of times the batswoman hit a boundary=6=6.

PP(She does not hit the boundary) =1P=1-P(She hit the boundary).
PP (She hit the boundary) == number of balls when she hits a boundarytotal number of balls she played\dfrac{\text{number of balls when she hits a boundary}}{\text{total number of balls she played}}
=630=\dfrac{6}{30}
=15=\dfrac{1}{5}.
\therefore PP(She did not hit the boundary)
=1P=1-P(She hit the boundary)
=115=1-\dfrac{1}{5}
=515=\dfrac{5-1}{5}
=45=\dfrac{4}{5}.