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Question: An unbiased dice \[1,2,3,4,5,6\] is thrown n times and the list of numbers showing up is noted. Then...

An unbiased dice 1,2,3,4,5,61,2,3,4,5,6 is thrown n times and the list of numbers showing up is noted. Then the probability that among the numbers 1,2,3,4,5,61,2,3,4,5,6 only three numbers appear in the list
A. [3n−3(2n)+3]6n\dfrac{[\mathop{3}^{n}-3(\mathop{2}^{n})+3]}{\mathop{6}^{n}}
B. 6c3[3n−3(2n)+3]6n\dfrac{6\mathop{c}_{3}[\mathop{3}^{n}-3(\mathop{2}^{n})+3]}{\mathop{6}^{n}}
C. 6c3[3n−3(2n)]6n\dfrac{6\mathop{c}_{3}[\mathop{3}^{n}-3(\mathop{2}^{n})]}{\mathop{6}^{n}}
D. None of these

Explanation

Solution

Hint: Since the dice is unbiased we can randomly choose any three n numbers say 1,2,31,2,3to only appear we only need to find probability.

Complete step-by-step answer:
Let us an onto function f from A:[r1,r2.......rn]A:\left[ {{r}_{1}},{{r}_{2}}.......{{r}_{n}} \right]where r2.....rn{{r}_{2}}.....{{r}_{n}}are the readings of n throws$$$$
And 1,2,31,2,3are the numbers that appear in the n throws
The total number of outcomes without any conditions applied =6n=\mathop{6}^{n}
∴\therefore We have 6 options for each throws so for n throws. Total outcomes will be 6.6.6....... n times=6n6.6.6.......\text{ n times}={{6}^{n}}
.We choose any three numbers out of 6 ways 1,2,31,2,3in 6C3{{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.
=3C2[2n−1n−1n]=3(2n−2){{=}^{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 =3C1(1n)=3{{=}^{3}}{{C}_{1}}\left( {{1}^{n}} \right)=3
Thus, the number of sequences using exactly 3 numbers will be 6C3[3n−3(2n−2)−3]=6C3[3n−3(2n)+3]^{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]
∴probability =total no. of favourable outcomestotal no. of outcomes\therefore \text{probability =}\dfrac{\text{total no}\text{. of favourable outcomes}}{\text{total no}\text{. of outcomes}}
6c3[3n−3(2n)+3]6n\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 nCr=n!(n−r)!(r)!^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!\left( r \right)!}