Three numbers are chosen at random without replacement from {1, 2, 3, ..., 8}. The probability that

Q: Why can't we just use the total sample space of 56 ways (8 choose 3)?

You absolutely could! However, conditional probability is a powerful tool precisely because it allows us to 'zoom in' on a specific scenario. By restricting the sample space to only those outcomes where the maximum is 6, we simplify the problem significantly. It's like looking at a smaller, more manageable world where the condition has already happened.

Q: Why is the third number chosen from {4, 5}?

Great question. We know the set of three numbers is {3, x, 6}. For 3 to be the minimum and 6 to be the maximum, the third number 'x' must be greater than 3 and less than 6. The only integers in that range are 4 and 5. If 'x' were 2, the minimum would be 2. If 'x' were 7, the maximum would be 7. Thus, 'x' must be 4 or 5.

Q: Does the order of selection matter here?

No, it does not. Since we are choosing a set of three numbers, the order in which we pick them doesn't change the set itself. That is why we use combinations (nCr) rather than permutations. Whether you pick {3, 4, 6} or {6, 4, 3}, it's the same set of numbers.

Question

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

Options:

(A) 3/8

(B) 1/5

(C) 1/4

(D) 2/5

Sigma Percentile · Accepted Answer

### The Elegance of Conditional Probability: A Journey Through Constraints

Welcome, fellow traveler in the world of mathematics. Today, we are going to peel back the layers of a beautiful probability problem. It might seem like a simple task of picking numbers from a hat, but beneath the surface lies a fundamental concept that defines how we navigate uncertainty: **Conditional Probability**.

Imagine you are standing before a set of eight numbers, neatly arranged on a line: $S = \{1, 2, 3, 4, 5, 6, 7, 8\}$. We are going to select three of them. We are looking for a specific outcome under a specific constraint.

### Phase 1: The Universe of Constraints

In probability, the 'sample space' is your entire universe. Usually, we calculate the probability of an event by looking at the ratio of favorable outcomes to the total possible outcomes.

However, **conditional probability** changes the rules. It tells us: "Assume this specific event has already happened." This is our Event $B$: the maximum of our three chosen numbers is $6$.

If the maximum is $6$, then any number greater than $6$ is strictly forbidden. We can instantly discard $7$ and $8$. Our new, reduced universe consists only of the set $\{1, 2, 3, 4, 5, 6\}$.

### Phase 2: Analyzing the Reduced Sample Space

Now, let's count the ways to form our set of three numbers in this new universe. We know that one of our three numbers *must* be $6$ to satisfy the condition that the maximum is $6$.

This leaves us with two spots to fill. We must choose these two numbers from the remaining set $\{1, 2, 3, 4, 5\}$.

Using the combination formula, we calculate the number of ways to choose $2$ numbers from $5$ available options:

$$n(B) = \binom{5}{2} = \frac{5 	imes 4}{2 	imes 1} = 10$$

There are exactly $10$ ways to pick three numbers such that the maximum is $6$. This is our new denominator.

### Phase 3: The Intersection of Desires

Now, we introduce the second condition: Event $A$, where the minimum of our three numbers is $3$. We need to find the intersection, $A \cap B$, where *both* conditions are true: the maximum is $6$ AND the minimum is $3$.

If the maximum is $6$ and the minimum is $3$, then two of our three numbers are already locked in stone: $3$ and $6$. We only have one spot left to fill. Let's call this unknown number $x$.

For $3$ to remain the minimum and $6$ to remain the maximum, $x$ must satisfy the inequality:

$$3 < x < 6$$

Which integers lie strictly between $3$ and $6$? Only $4$ and $5$. Therefore, our third number $x$ must be either $4$ or $5$.

### Phase 4: The Final Calculation

We have $2$ possible choices for our third number ($4$ or $5$). This means there are exactly $2$ favorable outcomes in our intersection:

1. The set $\{3, 4, 6\}$
2. The set $\{3, 5, 6\}$

So, $n(A \cap B) = 2$. Now, we apply the definition of conditional probability:

$$P(A|B) = \frac{n(A \cap B)}{n(B)} = \frac{2}{10} = \frac{1}{5}$$

The final result is **$\frac{1}{5}$**.

Animated Solution for Mathematics - Probability: 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

Select Answer:

Visualized Solution

The Sample Space $S$

Defining the Events

Analyzing Event $B$ : Maximum is $6$

Structuring Event $B$

Calculating $n (B)$

Analyzing the Intersection $A \cap B$

Structuring the Intersection

Finding the Range for $x$

Calculating $n (A \cap B)$

Applying the Conditional Probability Formula

Final Answer and Key Takeaway

The Sigma Insight: Conditional Probability

The Elegance of Conditional Probability

Phase 1

Phase 2

Phase 3

Phase 4

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