Question

Three numbers are chosen at random without replacement from \left\\{ {1,2,...,8} \right\\} . The probability that their minimum is $3$, given that their maximum is $6$ is
1. $\dfrac{3}{8}$
2. $\dfrac{1}{5}$
3. $\dfrac{1}{2}$
4. $\dfrac{2}{3}$

Answer 1

Here we are given a set of numbers out of which three numbers are to be chosen at random satisfying the given conditions. For this we will first find the total number of cases and the favourable cases.
Favourable cases can be found using the conditions given in the question as the maximum of the three numbers is given to be six and minimum must be three. So, out of the three numbers, two must be equal to three and six for this condition to satisfy. Now, the third number must also lie between six and three. After finding these we will find the required probability using the formula as below:
Probability (event) $= \dfrac{{{\text{Number}}\,{\text{of}}\,{\text{favourable}}\,{\text{outcomes}}}}{{{\text{Total}}\,{\text{number}}\,{\text{of}}\,{\text{outcomes}}}}$

Complete step-by-step solution:
So, the three numbers should be chosen out of the set \left\\{ {1,2,....8} \right\\} such that their maximum is given as six and minimum is three. So, the two of the three numbers must be three and six. The third number should also lie between three and six. So, there are a total of two whole numbers between three and six out of which one should be selected. So, the number of favourable outcomes is two.
Now, we have to find the total number of outcomes. So, we are given that the maximum is six. So, the remaining two numbers should be less than six.
So, the total number of outcomes is $^5{C_2} = 10$.
We know that Probability (event) $= \dfrac{{{\text{Number}}\,{\text{of}}\,{\text{favourable}}\,{\text{outcomes}}}}{{{\text{Total}}\,{\text{number}}\,{\text{of}}\,{\text{outcomes}}}}$
So, we get,
Probability $= \dfrac{{{\text{Number}}\,{\text{of}}\,{\text{favourable}}\,{\text{outcomes}}}}{{{\text{Total}}\,{\text{number}}\,{\text{of}}\,{\text{outcomes}}}}$
$= \dfrac{{\text{2}}}{{^5{C_2}}}$
Now, we know the combination formula as $^n{C_r} = \dfrac{{n!}}{{\left( {n - r} \right)! \times r!}}$. So, we get,
$= \dfrac{{\text{2}}}{{\left( {\dfrac{{5!}}{{2! \times 3!}}} \right)}}$
$= \dfrac{{\text{2}}}{{\left( {\dfrac{{5 \times 4 \times 3!}}{{2! \times 3!}}} \right)}}$
Expanding the factorials,
$= \dfrac{{\text{2}}}{{\left( {\dfrac{{5 \times 4}}{{2!}}} \right)}}$
$= \dfrac{{\text{2}}}{{10}}$
Cancelling common factors in numerator and denominator, we get,
$= \dfrac{1}{5}$
Therefore option (2) is the correct answer.

Note: The combination is a way of selecting items from a collection, such that (unlike permutations) the order of selection does not matter. A combination is the choice of r things from a set of n things without replacement and where order doesn't matter. Probability of any event can be between zero and one only. We should always use the combination and permutation formula carefully as per the conditions stated in the question.