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

Hint: First find the probability of event A as getting three numbers with one number 6 as maximum value. Then find the probability of B as getting three numbers with one number 3 as minimum value. Then find the conditional probability of event A, with the prior happening of event B. Formula of conditional probability will be used.

Complete step-by-step answer:
This question will use the combinatorial formula with the rules of conditional probability. Let us suppose A is the event for getting maximum as 6
Also, suppose B is the event for getting a minimum as 3.
 Now, the probability for getting maximum 6,
$P[A] = \dfrac{{favourable\;count}}{{total\;count}}$
For A, 6 is already chosen as maximum, and 2 more has to be taken from remaining (1,2,3,4,5). Here favourable count for A will be ${}^5{C_2}$ , and total count will be 3 from 8 numbers, so \[{}^8{C_3}\] , as total numbers are 8.
Thus, $P[A] = \dfrac{{{}^5{C_2}}}{{{}^8{C_3}}}$
Now, the probability for getting minimum 3,
$P[B] = \dfrac{{favourable\;count}}{{total\;count}}$
For, B, 3 is already chosen as minimum, and 2 more has to be taken from remaining (4,5,6,7,8). Here favourable count for B will be ${}^5{C_2}$, and total count will be 3 from 8 numbers, so \[{}^8{C_3}\], as total numbers are 8.
Thus, $P[B] = \dfrac{{{}^5{C_2}}}{{{}^8{C_3}}}$
Also, probability of event$A \cap B$ will be,
$P[A \cap B] = \dfrac{{favourable\;count}}{{total\;count}}$
Here favourable count is ${}^2{C_1}$, and total count is \[{}^8{C_3}\], as total numbers are 8.
$P[A \cap B] = \dfrac{{{}^2{C_1}}}{{{}^8{C_3}}}$
Now, we will find conditional probability for getting minimum is 3, given that their maximum is 6,as
$P(\dfrac{B}{A}) = \dfrac{{P(A \cap B)}}{{P(B)}}$
Now, substituting the values in above equation, we get
$
  P(\dfrac{B}{A}) = \dfrac{{\dfrac{{{}^2{C_1}}}{{{}^8{C_3}}}}}{{\dfrac{{{}^5{C_2}}}{{{}^8{C_3}}}}} \\
   \Rightarrow P(\dfrac{B}{A}) = \dfrac{{{}^2{C_1}}}{{{}^5{C_2}}} \\
   \Rightarrow P(\dfrac{B}{A}) = \dfrac{2}{{10}} = \dfrac{1}{5} \\
 $
In the above equation we have computed ${}^2{C_1}$as$\dfrac{2}{{1 \times 1}}$= 2
Also, we have computed ${}^5{C_2}$as $\dfrac{{5 \times 4}}{2}$= 10.

$\therefore $The required probability will be $\dfrac{1}{5}$.

Additional Information: Formula for conditional probability is:
$P(\dfrac{B}{A}) = \dfrac{{P(A \cap B)}}{{P(B)}}$. Where A and B are two events may be dependent or independent. If both are independent then $P(\dfrac{B}{A}) = P(B)$.

Note: Proper combination and hence combinatorial formula will make its solution easy. Precautions must be taken while framing the group of numbers, without missing any from the set. In the probability theory, conditional probability will be applicable when we have to find the probability of an event occurring while some other event has already occurred, before it. Here events A and B will be dependent as it is happening without replacement.