A lottery game has balls numbered 1 through 19. If a ball is selected at random, what is the probability of obtaining an even numbered ball or the number 10 ball?

9/19

Step-by-step explanation:

Total number of ball=19

even ball=2,4,6,8,10,12,14,16,18

total number of even ball=9

total number of ball 10=1

Now, in probability, the probability of event A OR event B happening is the probability of event A plus the probability of event B minus the probability of both A AND B.

Let A be the event of drawing an even number.

Let B be the event of drawing a 10.

In this case then, we are drawing 1 ball only.

P(A)=9/19

P(B)=1/19

P(A and B)=1/19 (because 10 is both 10 AND an even number)

Then P(A or B)=P(A)+P(B)-P(A and B)

P(A or B)=9/19+1/19-1/21

P(A or B)=9/19

0.85 or 85%

step-by-step explanation:

1) divide 23 by 27

2) round the decimal to the nearest 10

3) so your answer would be 0.85 or 85%

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