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A grid of light bulbs measures x bulbs by x bulbs, where x > 2. If 4 l
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29 Mar 2021, 05:00
29 Mar 2021, 05:00
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A grid of light bulbs measures x bulbs by x bulbs, where x > 2. If 4 light bulbs are illuminated at random, what is the probability, in terms of x, that the 4 bulbs form a 2 bulb by 2 square?
A. \(\frac{4(x - 1)}{x^2}\)
B. \(\frac{24(x - 1)}{x^2(x^2 - 2)(x^2 - 3)(x + 1)}\)
C. \(\frac{24(x + 1)}{x^2(x^2 - 2)(x - 1)}\)
D. \(\frac{4(x + 1)}{x^2(x^2 - 2)(x - 1)}\)
E. \(\frac{4x - 1}{x^2(x^2 - 2)}\)
A. \(\frac{4(x - 1)}{x^2}\)
B. \(\frac{24(x - 1)}{x^2(x^2 - 2)(x^2 - 3)(x + 1)}\)
C. \(\frac{24(x + 1)}{x^2(x^2 - 2)(x - 1)}\)
D. \(\frac{4(x + 1)}{x^2(x^2 - 2)(x - 1)}\)
E. \(\frac{4x - 1}{x^2(x^2 - 2)}\)
Re: A grid of light bulbs measures x bulbs by x bulbs, where x > 2. If 4 l
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22 Feb 2024, 00:20
22 Feb 2024, 00:20
1
To find the probability that 4 randomly illuminated bulbs form a 2x2 square on an \(x\) by \(x\) grid, we need to calculate:
1. The total number of ways to choose 4 bulbs out of \(x^2\) bulbs.
2. The total number of ways to form a 2x2 square in an \(x\) by \(x\) grid.
### Total Ways to Choose 4 Bulbs
The total number of ways to choose 4 bulbs out of \(x^2\) bulbs is given by the combination formula:
\[ \binom{x^2}{4} = \frac{x^2!}{4!(x^2 - 4)!} \]
### Total Ways to Form a 2x2 Square
To form a 2x2 square, the top left bulb of the square can be any bulb except those in the last row or the last column. Therefore, there are \((x-1)\) options for the row and \((x-1)\) options for the column for the top left bulb of the square. So, the total number of 2x2 squares that can be formed is:
\[ (x-1)(x-1) = (x-1)^2 \]
### Probability
The probability \(P\) that the 4 bulbs form a 2x2 square is the ratio of the number of ways to form a 2x2 square to the total number of ways to choose 4 bulbs:
\[ P = \frac{(x-1)^2}{\binom{x^2}{4 \]
Let's calculate this probability in terms of \(x\).
The probability that 4 randomly illuminated bulbs form a 2x2 square in an \(x\) by \(x\) grid, in terms of \(x\), is \(\frac{24(x - 1)}{x^2(x + 1)(x^2 - 3)(x^2 - 2)}\).
1. The total number of ways to choose 4 bulbs out of \(x^2\) bulbs.
2. The total number of ways to form a 2x2 square in an \(x\) by \(x\) grid.
### Total Ways to Choose 4 Bulbs
The total number of ways to choose 4 bulbs out of \(x^2\) bulbs is given by the combination formula:
\[ \binom{x^2}{4} = \frac{x^2!}{4!(x^2 - 4)!} \]
### Total Ways to Form a 2x2 Square
To form a 2x2 square, the top left bulb of the square can be any bulb except those in the last row or the last column. Therefore, there are \((x-1)\) options for the row and \((x-1)\) options for the column for the top left bulb of the square. So, the total number of 2x2 squares that can be formed is:
\[ (x-1)(x-1) = (x-1)^2 \]
### Probability
The probability \(P\) that the 4 bulbs form a 2x2 square is the ratio of the number of ways to form a 2x2 square to the total number of ways to choose 4 bulbs:
\[ P = \frac{(x-1)^2}{\binom{x^2}{4 \]
Let's calculate this probability in terms of \(x\).
The probability that 4 randomly illuminated bulbs form a 2x2 square in an \(x\) by \(x\) grid, in terms of \(x\), is \(\frac{24(x - 1)}{x^2(x + 1)(x^2 - 3)(x^2 - 2)}\).
General Discussion
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A grid of light bulbs measures x bulbs by x bulbs, where x > 2. If 4 l
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29 Mar 2021, 23:56
29 Mar 2021, 23:56
4
GeminiHeat wrote:
A grid of light bulbs measures x bulbs by x bulbs, where x > 2. If 4 light bulbs are illuminated at random, what is the probability, in terms of x, that the 4 bulbs form a 2 bulb by 2 square?
A. \(\frac{4(x - 1)}{x^2}\)
B. \(\frac{24(x - 1)}{x^2(x^2 - 2)(x^2 - 3)(x + 1)}\)
C. \(\frac{24(x + 1)}{x^2(x^2 - 2)(x - 1)}\)
D. \(\frac{4(x + 1)}{x^2(x^2 - 2)(x - 1)}\)
E. \(\frac{4x - 1}{x^2(x^2 - 2)}\)
A. \(\frac{4(x - 1)}{x^2}\)
B. \(\frac{24(x - 1)}{x^2(x^2 - 2)(x^2 - 3)(x + 1)}\)
C. \(\frac{24(x + 1)}{x^2(x^2 - 2)(x - 1)}\)
D. \(\frac{4(x + 1)}{x^2(x^2 - 2)(x - 1)}\)
E. \(\frac{4x - 1}{x^2(x^2 - 2)}\)
Let us assume \(x = 3\)
Total bulbs \(= (3)(3) = 9\)
If you draw a figure, you will notice that we will have Four 2x2 square combinations out of a total of \(^9C_4\) combinations
Required Probability \(= \frac{4}{^9C_4} = \frac{2}{63}\)
Let us plug \(x= 3\) in the option choices:
A. \(\frac{4(x - 1)}{x^2}\)
\(\frac{4(3 - 1)}{3^2} = \frac{8}{9}\)
B. \(\frac{24(x - 1)}{x^2(x^2 - 2)(x^2 - 3)(x + 1)}\)
\(\frac{24(3 - 1)}{3^2(3^2 - 2)(3^2 - 3)(3 + 1)} = \frac{2}{63}\)
C. \(\frac{24(x + 1)}{x^2(x^2 - 2)(x - 1)}\)
\(\frac{24(3 + 1)}{3^2(3^2 - 2)(3 - 1)} = \frac{16}{21}\)
D. \(\frac{4(x + 1)}{x^2(x^2 - 2)(x - 1)}\)
\(\frac{4(3 + 1)}{3^2(3^2 - 2)(3 - 1)} = \frac{8}{63}\)
E. \(\frac{4x - 1}{x^2(x^2 - 2)}\)
\(\frac{4(3) - 1}{3^2(3^2 - 2)} = \frac{11}{63}\)
Hence, option B
