Question

A lottery game has balls numbered 1 through 21. What is the probability of selecting an even numbered ball or an 8? Round to nearest thousandth

Answer 1

Answer: 0.476

Step-by-step explanation:

Let A = Event of choosing an even number ball.

B = Event of choosing an 8 .

Given, A lottery game has balls numbered 1 through 21.

Sample space: S= {1,2,3,4,5,6,7,8,...., 21}

n(S) = 21

Then, A= {2,4,6,8, 10,...(20)}

i.e. n(A)= 10

B= {8}

n(B) = 1

A∪B = {2,4,6,8, 10,...(20)} = A

n(A∪B)=10

Now, the probability of selecting an even numbered ball or an 8 is

$P(A\cup B)=\dfrac{n(A\cup B)}{n(S)}$

$=\dfrac{10}{21}\approx0.476$

Hence, the required probability =0.476