Question

An unbiased die, with faces numbered $1,2,3,4,5,6$ is thrown n times and the list of n numbers showing up is noted. Then the probability that, among the numbers $1,2,3,4,5,6$ only three numbers appear in this list ?

• A. $\frac{[3^n-3(2^n)+3]}{6^n}$
• B. $\frac{{}^6{C}_3[3^n-3(2^n)+3]}{6^n}$
• C. $\frac{{}^6{C}_3[3^n-3(2^n\left)\right]}{6^n}$
• D. None of these

Answer 1

The correct option is B $\frac{{}^6{C}_3[3^n-3(2^n)+3]}{6^n}$

The total no. of outcomes = $6^n$

We can choose three numbers out of 6 in ${}^6{C}_3$ ways. By using three numbers out of 6 we can get $3^n$ sequences of length n. But these sequences of length n which use exactly two numbers and exactly one number.

The number of n – sequences which use exactly two numbers

= ${}^3{C}_2[2^n–1^n–1^n]=3(2^n–2)$ and the number of n sequence which are exactly one number

= ${(}^3{C}_1\left)\right(1^n)=3$

Thus, the number of sequences, which use exactly three numbers

= ${}^6{C}_3[3^n–3(2^n–2)–3]{=}^6{C}_3[3^n–3(2^n)+3]$

∴ Probability of the required event,

= $\frac{{}^6{C}_3[3^n–3(2^n)+3]}{6^n}$