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. \(\frac{4x - 1}{x^2(x^2 - 2)}\)
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