Question

An unbiased dice \[1,2,3,4,5,6\] is thrown n times and the list of numbers showing up is noted. Then the probability that among the numbers \[1,2,3,4,5,6\] only three numbers appear in the list
A. \[\dfrac{[\mathop{3}^{n}-3(\mathop{2}^{n})+3]}{\mathop{6}^{n}}\]
B. \[\dfrac{6\mathop{c}_{3}[\mathop{3}^{n}-3(\mathop{2}^{n})+3]}{\mathop{6}^{n}}\]
C. \[\dfrac{6\mathop{c}_{3}[\mathop{3}^{n}-3(\mathop{2}^{n})]}{\mathop{6}^{n}}\]
D. None of these

Answer 1

Hint: Since the dice is unbiased we can randomly choose any three n numbers say \[1,2,3\]to only appear we only need to find probability.

Complete step-by-step answer:
Let us an onto function f from \[A:\left[ {{r}_{1}},{{r}_{2}}.......{{r}_{n}} \right]\]where \[{{r}_{2}}.....{{r}_{n}}\]are the readings of n throws\[\]
And \[1,2,3\]are the numbers that appear in the n throws
The total number of outcomes without any conditions applied \[=\mathop{6}^{n}\]
\[\therefore \]We have 6 options for each throws so for n throws. Total outcomes will be \[6.6.6.......\text{ n times}={{6}^{n}}\]
.We choose any three numbers out of 6 ways \[1,2,3\]in \[{{6}_{{{C}_{3}}}}\]ways. By using the number out of 6 we can get sequences of length n.
But these sequences of length n which use exactly 2 numbers and exactly 1 number.
The number of n series sequences which use exactly 2 numbers.
\[{{=}^{3}}{{C}_{2}}\left[ {{2}^{n}}-{{1}^{n}}-{{1}^{n}} \right]=3\left( {{2}^{n}}-2 \right)\].
The number of sequences which are exactly one number \[{{=}^{3}}{{C}_{1}}\left( {{1}^{n}} \right)=3\]
Thus, the number of sequences using exactly 3 numbers will be \[^{6}{{C}_{3}}\left[ {{3}^{n}}-3\left( {{2}^{n}}-2 \right)-3 \right]{{=}^{6}}{{C}_{3}}\left[ {{3}^{n}}-3\left( {{2}^{n}} \right)+3 \right]\]
\[\therefore \text{probability =}\dfrac{\text{total no}\text{. of favourable outcomes}}{\text{total no}\text{. of outcomes}}\]
\[\dfrac{6\mathop{c}_{3}[\mathop{3}^{n}-3(\mathop{2}^{n})+3]}{\mathop{6}^{n}}\]
Hence, option B is the correct option.

Note: Probability gets added or subtracted when we are considering different events and it gets multiplied when there is an event within an event.
Probability of choosing values from n values is given by \[^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!\left( r \right)!}\]