A box contains 100 tickets numbered 1, 2, ..., 100. Two tickets are chosen at random. It is given th

Q: Wait, why do we restrict the sample space to 10? Shouldn't it be 100?

That is a classic trap! The phrase 'it is given that' acts as a filter. By telling us the maximum number is at most 10, the problem effectively removes all tickets numbered 11 through 100 from consideration. We are now in a new, smaller world where only 10 tickets exist.

Q: Why can't the second ticket also be 5? If I pick {5, 5}, is the minimum 5?

Great question! The problem states we are choosing 'two tickets' from a box of 100 numbered tickets. This implies we are picking two distinct tickets. If we were picking with replacement, {5, 5} might be possible, but in standard probability problems involving picking tickets from a set, we assume distinct selection unless stated otherwise. Thus, the second ticket must be different from the first.

Q: Why do we use combinations $\binom{n}{r}$ instead of permutations?

In this problem, the order in which we pick the two tickets does not matter. Whether you pick ticket 5 then ticket 6, or ticket 6 then ticket 5, the set of tickets in your hand is the same: {5, 6}. Since the outcome is defined by the set of tickets, combinations are the correct tool.

Question

A box contains 100 tickets numbered $1, 2, \dots, 100$. Two tickets are chosen at random. It is given that the maximum number on the two chosen tickets is not more than 10. The minimum number on them is 5 with probability .........

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in front of a box containing 100 tickets, each uniquely numbered from 1 to 100. You are asked to reach in and select two tickets at random.

In a standard scenario, the total number of ways to pick these tickets would be $\binom{100}{2}$. However, the problem introduces a powerful constraint: "It is given that the maximum number on the two chosen tickets is not more than 10."

In the world of probability, "given" is a magical word. It acts as a filter, a lens that narrows your vision. Because we know the maximum is at most 10, any pair containing a number like 11, 50, or 99 is now impossible.

They have been wiped from existence in our current reality. Our new universe is restricted to the set $S' = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$.

### The Anatomy of the Sample Space

Now that we have defined our new universe, we must determine the size of this restricted sample space. We are choosing 2 tickets from these 10 available numbers.

The number of ways to do this is given by the combination formula:
$$n(A) = \binom{10}{2}$$

Calculating this, we get:
$$n(A) = \frac{10 	imes 9}{2 	imes 1} = 45$$

This number, 45, is the denominator of our probability. It represents every possible pair of tickets that could exist under the given condition.

### The Favorable Event: The Hunt for the Minimum

Now, we turn our attention to the specific event we are interested in: the minimum number on the two tickets must be exactly 5. Let this be Event $B$.

We are looking for the intersection of our two conditions: $A \cap B$. This means we need pairs where the maximum is $\le 10$ AND the minimum is 5.

If the minimum is 5, one of your tickets MUST be 5. The other ticket must be greater than 5 to ensure that 5 remains the minimum. If the other ticket were, say, 4, then 4 would be the minimum, not 5.

So, the second ticket must be in the set $\{6, 7, 8, 9, 10\}$.

### The Final Calculation

We have 5 options for that second ticket: 6, 7, 8, 9, or 10. Each of these, when paired with 5, creates a valid set where the minimum is 5 and the maximum is $\le 10$.

Thus, there are exactly 5 favorable outcomes:
$$n(A \cap B) = \binom{5}{1} = 5$$

Finally, we apply the definition of conditional probability:
$$P(B|A) = \frac{n(A \cap B)}{n(A)}$$

Substituting our values, we get:
$$P(B|A) = \frac{5}{45}$$

Simplifying this fraction, we arrive at the elegant result: **$\frac{1}{9}$**.

This problem teaches us that probability is not just about counting; it is about understanding the boundaries of the space you are working in. When you encounter a "given" condition, embrace the restriction—it makes the math simpler, not harder.

Animated Solution for Mathematics - Probability: A box contains 100 tickets numbered $1, 2, \dots, 100$ . Two tickets are chosen at random. It is given that the maximum number on the two chosen tickets is not more than 10. The minimum number on them is 5 with probability .........

Visualized Solution

The Original Sample Space

The Given Condition (Event $A$ )

Reducing the Sample Space

Calculating $n (A)$

Evaluating $n (A)$

The Favorable Event (Event $B$ )

Analyzing the Favorable Pair

Options for the Second Ticket

Calculating $n (A \cap B)$

Applying Conditional Probability

Final Calculation

The Sigma Insight: Conditional Probability

Analyzing the Setup

The Anatomy of the Sample Space

The Favorable Event

The Final Calculation

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