JEST 2019

Question

Three dice are rolled independently. Probability of obtaining the difference from largest and smallest number as exactly 4 :

Answer 1

Total case possible for throwing 3 dice independently $= 6*6*6 = 216.$

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Favorable cases :

 $(6,2)$ and $(1,5)$ are only possible pairs that give 4 as the difference.

Cases with (6,2) possible are

$(6,6,2)$ and these can be arranged in $\frac{3!}{2!}$ ways = 3 ways

$(6,5,2)$ and these can be arranged in $3!$ ways = 6 ways

$(6,4,2)$ and these can be arranged in $3!$ ways = 6 ways

$(6,3,2)$ and these can be arranged in $3!$ ways = 6 ways

$(6,2,2)$ and these can be arranged in $\frac{3!}{2!}$ ways = 3 ways

$\therefore$ 3+6+6+6+3 = $24$ cases possible

Cases with (5,1) possible are

$(5,5,1)$ and these can be arranged in $\frac{3!}{2!}$ ways = 3 ways

$(5,4,1)$ and these can be arranged in $3!$ ways = 6 ways

$(5,3,1)$ and these can be arranged in $3!$ ways = 6 ways

$(5,2,1)$ and these can be arranged in $3!$ ways = 6 ways

$(5,1,1)$ and these can be arranged in $\frac{3!}{2!}$ ways = 3 ways

$\therefore$ 3+6+6+6+3 = $24$ cases possible

$\therefore$ Total number of favorable cases = 24 + 24 = $48$

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$\therefore$ Required Probability = $\frac{Total number of favorable cases}{Total number of cases}$ = $\frac{48}{216}$ = $0.22$

Answer 2

Sample space be $S = \{(d_1, d_2, d_3) \ | \ 1 \le d_1,d_2,d_3 \le 6\}$ where $d_i$ correspond to $i^{th}$ dice. 

$\underline{\textbf{Case 1:}} \text{ }$ 

$\forall_{1 \le j,k,z \le 3} \ j\neq k \land  d_j=1 \land d_k=5 \land z \neq j \land z \neq k\Longrightarrow \forall_{2\le m\le 4} \ d_z = m$

In other way, we write it as $\underbrace{^3C_2}_{\text{selecting j, k}} \times \underbrace{2!}_{\text{arranging j,k}} \times \underbrace{^3C_1}_{\text{choosing }d_z} = 18$

And,

$\forall_{1 \le j,k,z \le 3} \ j\neq k \land  d_j=2 \land d_k=6 \land z \neq j \land z \neq k\Longrightarrow \forall_{3\le m\le 5} \ d_z = m$

In other way, we write it as $\underbrace{^3C_2}_{\text{selecting j, k}} \times \underbrace{2!}_{\text{arranging j,k}} \times \underbrace{^3C_1}_{\text{choosing }d_z} = 18$

$\underline{\textbf{Case 2:}}$

$\forall_{1 \le j,k,z \le 3} \ j\neq k \land  d_j=1 \land d_k=5 \land z \neq j \land z \neq k\Longrightarrow \forall_{m \in \{1,5\}} \ d_z = m$

In other way, we write it as $\dfrac{\underbrace{^3C_2}_{\text{selecting j, k}} \times \underbrace{2!}_{\text{arranging j,k}} \times \underbrace{^2C_1}_{\text{choosing }d_z}}{\underbrace{2!}_{\text{overcounting; e.g. (1,1,5) counted twice}}} = 6$

And, 

$\forall_{1 \le j,k,z \le 3} \ j\neq k \land  d_j=2 \land d_k=6 \land z \neq j \land z \neq k\Longrightarrow \forall_{m \in \{2,6\}} \ d_z = m$

In other way, we write it as $\dfrac{\underbrace{^3C_2}_{\text{selecting j, k}} \times \underbrace{2!}_{\text{arranging j,k}} \times \underbrace{^2C_1}_{\text{choosing }d_z}}{\underbrace{2!}_{\text{overcounting; e.g. (2,2,6) counted twice}}} = 6$

$\underline{\textbf{Total}}$

$\text{Favourable Cases} = 18+18+6+6 = 48$

$|S| = 6^3 = 216$

$\text{Probability} = \dfrac{48}{216} = 0.\overline{2}$