Question

In a cricket math, a batsman hits a boundary 9 times out of 50 balls he plays. Find the 1)probability that he hit a boundary 2)probability that he did not hit a boundary.

Answer 1

Answer:

Total number of balls = 50

Times hit = 9

times did not it = 50-9

                         = 41

1) probability that he hit a boundary = 9/50

2)probability that he did not hit a boundary = 41/50

Answer 2

Answer:

$1)\dfrac{9}{50}$

$2)\dfrac{41}{50}$

Explanation:

Given :

• Number of times a batsman hit a boundary = 9
• Total ball he played = 50

To Find :

• Probability that he does not hit a boundary.
• Probability that he hit a boundary.

Solution :

Total number of balls played = 50

Number of times he a boundary = 9

$\sf{}P(He\ hit\ a\ boundary)=\dfrac{Number\ of\ times\ he\ hit\ a\ boundary}{Total\ number\ of\ times\ he\ played}$

$\sf{}\Rightarrow\dfrac{9}{50}$

Number of times that the batsman does not hit a boundary = 50 - 9 = 41

$\sf{}P(He\ hit\ a\ boundary)=\dfrac{Number\ of\ times\ he\ does\ not\ hit\ a\ boundary}{Total\ number\ of\ times\ he\ played}$

$\sf{}\Rightarrow\dfrac{41}{50}$