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Blababa [14]
2 years ago
8

each unit square of a 3-by-3 unit-square grid is to be colored either blue or red. for each square, either color is equally like

ly to be used. what is the probability of obtaining a grid that does not have a 2-by-2 red square. (hint 14)
Mathematics
1 answer:
makkiz [27]2 years ago
3 0

The probability of obtaining a grid that does not have a 2-by-2 red square 16 2/3%

probability : The probability of an event can be calculated by means of probability formula via honestly dividing the favorable quantity of outcomes by way of the total variety of viable consequences.

<u>Step-by using-step clarification:</u>

<u>given that:</u>

form are both square or circle with both equally likely :

for this reason,

P(square) = half of ; P(circle) = half of

color are either blue, pink or inexperienced with the 3 equally likely ;

P(purple) = 1/3 ; P(green) = 1/three ; P(blue) = 1/three

chance of creating a inexperienced circle :

independent probability

P(green n circle) = p(inexperienced) * p(circle)

P(green n circle) = 1/3 * half of

P(green n circle) = 1/6

= 0.16666

(0.16666 * 100%) = 16.6666%

= 16 2/3%

To learn more  about probability Visit : brainly.com/question/18920335

#SPJ4

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Let the three gp be a, ar and ar^2
a + ar + ar^2 = 21 => a(1 + r + r^2) = 21 . . . (1)
a^2 + a^2r^2 + a^2r^4 = 189 => a^2(1 + r^2 + r^4) = 189 . . . (2)
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(3) ÷ (2) => (1 + r + r^2)^2 / (1 + r^2 + r^4) = 441/189 = 7/3
3(1 + r + r^2)^2 = 7(1 + r^2 + r^4)
3(r^4 + 2r^3 + 3r^2 + 2r + 1) = 7(1 + r^2 + r^4)
3r^4 + 6r^3 + 9r^2 + 6r + 3 = 7 + 7r^2 + 7r^4
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From (1), a = 21/(1 + r + r^2)
When r = 2:
a = 21/(1 + 2 + 4) = 21/7 = 3
Therefore, the numbers are 3, 6 and 12.
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