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

Let the sample space be all ways of choosing $$3$$ distinct numbers from $$\{1,2,3,\dots ,8\}$$. We work with combinations because order does not matter.

Define the events
  $$A$$ : the minimum of the three chosen numbers is $$3$$,
  $$B$$ : the maximum of the three chosen numbers is $$6$$.

We need $$P(A\mid B) = \dfrac{P(A\cap B)}{P(B)}$$.

Step 1: Count the selections for event $$B$$ (maximum = 6).
If the maximum is exactly $$6$$, then
  • the number $$6$$ must be in the set, and
  • the other two numbers must be chosen from $$\{1,2,3,4,5\}$$ (all < 6).
Thus the required selections are obtained by fixing $$6$$ and choosing any $$2$$ numbers from the $$5$$ numbers $$\{1,2,3,4,5\}$$.

Number of such selections: $$\binom{5}{2}=10$$.

Step 2: Count the selections for event $$A\cap B$$ (minimum = 3 and maximum = 6).
Now the set must contain both $$3$$ (to make the minimum 3) and $$6$$ (to make the maximum 6).
The third number must lie between $$3$$ and $$6$$ so that the minimum stays $$3$$ and the maximum stays $$6$$. Hence the third number can be $$4$$ or $$5$$.

Number of such selections: $$2$$ (namely $$\{3,4,6\}$$ and $$\{3,5,6\}$$).

Step 3: Evaluate the conditional probability.
$$P(A\mid B)=\dfrac{|A\cap B|}{|B|}=\dfrac{2}{10}=\dfrac{1}{5}$$.

Therefore, the required probability is $$\frac{1}{5}$$.

Option B which is: $$\frac{1}{5}$$.