Respuesta :
Answer:
21 drawings
Step-by-step explanation:
The number labelings are 0 - 49
The minimum set that could be the same number is the labeling 0, 10 , 20
It could also be 1, 11, 21 or even 2,12,22
In all the cases, we can be sure that we might have three labelings ending with the same number when we pull our ball labeled 20
That makes a total of 21 drawings
Following are the response to the given question:
Pigeon Hole's Principle can also be used to tackle this problem. Assuming that we already have labels ranging from [tex]0\ to\ 49.[/tex]
- We possess five balls which conclude with the number 0 [tex]\text{(ball \ 0, ball\ 10, ball\ 20, ball\ 30, and\ ball\ 40)}.[/tex]
- We possess five balls which conclude with the digit 1[tex]\bold{(ball\ 1, ball\ 11, ball\ 21, ball\ 31, ball\ 41)}.[/tex]
- .....
- I have five balls that end in the number 9 [tex]\text{(ball\ 9, ball\ 19, ball\ 29, ball\ 39, and ball\ 49)}.[/tex]
Each digit ending is represented by five balls.
In the pigeonhole principle, it [tex]kn+1[/tex] objects [tex](k \geq 1)[/tex] are distributed among n boxes, one of the boxes will contain at least [tex]k+1[/tex] objects. Assume that you draw a ball and place it in a box labeled with its last digit. So there will be n = 10 boxes.
- Box 0: we place balls that end in the digit 0.
- Box 1: we place balls that end in the digit 1.
- .....
- Box 9 is where we put balls that end in the number 9.
According to the pigeon hole concept, you need to draw [tex]kn+1[/tex] balls to assure that at least one of the boxes includes [tex]k+1[/tex] objects. Because [tex]k+1 = 3,[/tex] we require three objects in at least one box (we need three balls ending with the same digit).
[tex]k + 1 = 3 \\\\k =2[/tex]
Therefore, we have to draw [tex]kn+1[/tex] balls, [tex]k = 2, n = 10[/tex], at least[tex]21\ balls[/tex] must be drawn to ensure that at least three of them have the same 1 digit.
Learn more:
brainly.com/question/10898731
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