Question

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 $P\left(E\right)$ is:

• A. $\frac{1}{324}$
• B. $\frac{1}{648}$
• C. $\frac{2}{324}$
• D. $\frac{1}{162}$

Answer 1

The correct option is A $\frac{1}{324}$

Solution- Red die $-a$

Blue die $-b$

Green die $-c$

White die $-d$

$(a-1)(b-2)(c-3)(d-4)=1$

$XYWZ=1$

Where $X=a-1,Y=b-2,Z=c-3,W=d-6$

$\{X,Y,Z,W\}$ values for this,

Total ways $\to \{1,1,1,1\}\to (d-6)$ cannot be $1\times \{-1,-1,-1,-1\}$

$\{-1,-1,1,1\}\to (d-6)$ cannot be $1\times \{-1,1,-1,1\}\to (d-6)$ cannot be $1\times \{-1,1,1,-1\}$

$\{1,-1,1,-1\}\to (d-6)$ cannot be $1\times \{1,1,-1,-1\}$.

Total ways $=4$

Probability $=\frac{4}{6\times 6\times 6\times 6}=\frac{2}{3\times 2}=\frac{1}{324}$

$A$ is correct