Question
A tetrahedral fair dice whose faces are numbered $1,2,3$ and 4 . The die is thrown and the face on which the pyramid lands is considered the 'winning' number. If the dice is thrown four times and the scores noted. The probability of obtaining a score of 10 is $\frac{m}{n}$ where m , n are relatively prime natural numbers, then $m + n$ is equal to
Solution
Sol. [Key Concept: Enumerate the tetrahedral-die outcomes for the event.]
$$x_{1} + x_{2} + x_{3} + x_{4} = 10,x_{i} \in \{ 1,2,3,4\}$$
$$P = \frac{\ ^{10 - 4 + 3}C_{3} - \ ^{4}C_{1}\ ^{10 - 8 + 3}C_{3}}{(4)^{4}} = \frac{\ ^{9}C_{3} - 4\ ^{5}C_{3}}{(4)^{4}} = \frac{11}{64}$$
Alternate: Favorable case $= (1,1,4,4);(1,2,3,4),(1,3,3,3),(2,2,2,4)$, (2, 2, 3, 3)
$$P = \frac{\frac{4!}{2!2!} + 4! + \frac{4!}{3!} + \frac{4!}{3!} + \frac{4!}{2!2!}}{4^{4}} = \frac{44}{256} = \frac{11}{64}$$
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