Multiple choice

Let a red die, a blue die, a green die and a white die are rolled once, the dice being fair. The outcomes on the red, blue, green and white denote the numbers, $a,b,c$ and $d$ respectively. Let $E$ denotes the event that absolute value of $(a-1)(b-2)(c-3)(d-6)=1$, then $P(E)$ is:

  1. $\dfrac{1}{324}$
  2. $\dfrac{1}{648}$
  3. $\dfrac{2}{324}$
  4. $\dfrac{1}{162}$
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A Correct answer
Explanation

The condition |(a-1)(b-2)(c-3)(d-6)| = 1 means each factor must be 1 or -1. For each die, count the values that satisfy this. For (a-1), a=2 or 0 (impossible). So a=2. For (b-2), b=3 or 1. For (c-3), c=4 or 2. For (d-6), d=5 (d=7 impossible). Total favorable outcomes = 1 * 2 * 2 * 1 = 4. Total outcomes = 6^4 = 1296. P(E) = 4 / 1296 = 1 / 324.

AI explanation

For the absolute value of (a-1)(b-2)(c-3)(d-6) to equal 1, each factor must be either 1 or -1. This requires a to be 2, b to be 1 or 3, c to be 2 or 4, and d to be 5 or 6. There are exactly 4 favorable outcomes out of the 1296 total possible outcomes, making the probability 4/1296, which simplifies to 1/324.

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