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 = 3Total bulbs
= (3)(3) = 9If 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