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Three positive numbers are selected at random(without replacement) from first six positive numbers. If X denotes the smallest of the three numbers obtained. Find the probability distribution of X. Also find the mean and variance.

I have tried solving the question by making a sample space and calculating the probability but that method is taking too long, is there any method to solve this question faster?

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    Hint: compute $P(X≥i)$ for $i\in \{1,2,3,4,5,6\}$. – 2017-02-25
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    First six positive numbers? Is $\Bbb R_+$ countable? – 2017-02-25
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    To explain @lulu's hint (which is MUCH preferable to the suggestions in the two currently not deleted answers), note that the event $[X\geqslant i]$ corresponds to choosing three numbers in $\{i,i+1,\ldots,6\}$, a set with $7-i$ elements hence $$P(X\geqslant i)=\frac{{7-i\choose 3}}{{6\choose 3}}=\frac{(7-i)(6-i)(5-i)}{6\cdot5\cdot4}$$ Next, use $$E(X)=\sum_{i=1}^\infty P(X\geqslant i)$$ – 2017-02-25

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Hint

  • There are ${6 \choose 3}=20$ equally probable ways of choosing the three numbers

  • Of these, there are ${6-n \choose 2}$ ways of choosing the three numbers so the smallest is $n$ and the other two are in $n+1,n+2,\ldots, 6$

  • Division will give probabilities $P_n$ for $n=1,2,3,4$

  • The mean is $\sum(nP_n)$ and the variance is $\sum(n^2P_n)-(\sum(nP_n))^2$

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    I'm getting 120 ways of choosing three numbers and i did not understand what you did there. I'm only in 12th grade is there any other way to explain me what u did there – 2017-02-25
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Without knowing any stats, you can simply:

  • evaluate the 120 possible permutations of 3 numbers from $\{1,2,3,4,5,6\}$
  • then record the Min of each permutation, and
  • then Tally the number of 1's, the number of 2s, the number of 3's and the number of 4's.

Here is Mathematica code that calculates the exact pmf in less than $\frac{1}{1000}$th of a second:

Map[Min, Permutations[Range[6], {3}]] // Tally

$\left( \begin{array}{cc} 1 & 60 \\ 2 & 36 \\ 3 & 18 \\ 4 & 6 \\ \end{array} \right)$

i.e. $P(X=1) = \frac{60}{120} = \frac12$, $\quad P(X=2) = \frac{36}{120} \quad $ etc

There won't be any 5's or 6's, because in a sample of size 3, there will always be a number smaller than 5 or 6.