Probability Ques 17
- An unbiased die, with faces numbered 1, 2, 3, 4, 5 and 6 is thrown $n$ times and the list of $n$ numbers showing up is noted. What is the probability that among the numbers 1, 2, 3, 4, 5 and 6 only three numbers appear in this list?
(2001, 5M)
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Answer:
Correct Answer: 17.$\frac{\left(3^{n}-3 \cdot 2^{n}+3\right) \times{ }^{6} C _3}{6^{n}}$
Solution:
- Let us define a onto function $F$ from $A:\left[r _1, r _2, \ldots, r _n\right]$ to $B: \quad[1,2,3]$, where $r _1, r _2, \ldots, r _n$ are the readings of $n$ throws and 1,2,3 are the numbers that appear in the $n$ throws.
Number of such functions, $M=N-[n(1)-n(2)+n(3)]$ where, $N=$ total number of functions and $n(t)=$ number of function having exactly $t$ elements in the range.
$$ \begin{array}{ll} \text { Now, } & N=3^{n}, n(1)=3 \cdot 2^{n}, n(2)=3, n(3)=0 \\ \Rightarrow & M=3^{n}-3 \cdot 2^{n}+3 \end{array} $$
Hence, the total number of favourable cases
$$ \begin{aligned} & =\left(3^{n}-3 \cdot 2^{n}+3\right) \cdot{ }^{6} C _3 \\ \therefore \text { Required probability } & =\frac{\left(3^{n}-3 \cdot 2^{n}+3\right) \times{ }^{6} C _3}{6^{n}} \end{aligned} $$
BETA
