Problem 44 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

At a railway staition a passenger leaves his luggage in a locker which is opened by dialling a three digit code (say \(253,009,325\) etc.). The passenger chooses the code, closes the locker and leaves for the town. A strange man, who does not know the code, tries to open the locker by dialling three digits at random. The probability that the locker opens after \(k\) trials is (Here \(k<1000\) ) (A) \(\frac{k}{100}\) (B) \(\frac{k}{1000}\) (C) \(\frac{1000 \quad k}{1000}\) (D) none of these.

Step-by-Step Solution

Answer

The probability is \(\frac{k}{1000}\); option (B) is correct.

Step 1: Understand the total possible outcomes

The lock requires a three-digit code, where each digit can be from 0 to 9. Therefore, the total number of possible different three-digit codes is given by the number of permutations with repetition, which is \(10 \times 10 \times 10 = 1000\).

Step 2: Determine the probability of opening the locker in one trial

The probability of guessing the correct code in one try is the number of successful outcomes (guessing the correct code) divided by the total number of possible outcomes (all possible codes). Thus, this is \(\frac{1}{1000}\).

Step 3: Calculate the probability of successful attempt after k trials

When attempting \(k\) trials, the strange man has \(k\) chances to guess one of the 1000 possible combinations. Since each trial is independent, the probability that the locker opens after \(k\) trials is simply \(\frac{k}{1000}\), because each trial is a separate chance to enter the correct code.

Step 4: Evaluate the given options

From the calculation in Step 3, the probability of opening the locker after \(k\) trials is \(\frac{k}{1000}\). Comparing this result with the given options, the correct choice is (B) \(\frac{k}{1000}\).

Key Concepts

Permutations with Repetition

Permutations with repetition refer to a scenario where you can use the same items more than once in different orders. In our locker example, a three-digit code is created using numbers 0 through 9. Since each digit in the code can be any of these 10 numbers, and the same number can be reused, this is a case of permutation with repetition.

When calculating permutations with repetition, we use a simple formula: if there are \(n\) possible choices for each item and \(r\) items in total, the number of different permutations is \(n^r\). For the locker, \(n = 10\) (digits 0-9) and \(r = 3\) (three digits), so the calculation is:

• \(10^3 = 1000\) possible combinations

This means there are 1000 different codes that could unlock the locker. Each combination is equally likely, making this a uniform probability distribution. This foundation is crucial to understanding how random trials will interact with probability.

Three-digit Code

A three-digit code requires basic understanding of combinations in the context of probability. Each digit can be selected independently and allows for repetition. So, a code like '253' uses three independent selections.

To construct a three-digit code, follow these simple steps:

• Select a digit from 0 to 9 for the first spot.
• Repeat the digit selection for the second and third spots.

With this flexibility, the three-digit code can indeed be any number from 000 to 999, making it 1000 combinations in total. The knowledge of the total combinations helps clarify why certain probability calculations, like guessing the code, are the way they are. For the strange man attempting random guesses, understanding this total number of possibilities forms the basis for calculating the chances of success in multiple trials.

Independent Trials

Independent trials are fascinating in probability because each event does not affect others. Each guess the man makes at the code is an independent trial, having no influence on the outcomes of past or future guesses.

The probability remains constant throughout, emphasizing that each trial is a fresh start. Since there are 1000 possible combinations, at each trial, the probability of correctly guessing the code remains \(\frac{1}{1000}\).
When the man tries to guess,

• Each try is one independent trial.
• With \(k\) attempts, the probability that the code is guessed correctly at least once is the probability "added up" for each trial: \(\frac{k}{1000}\).
• This cumulative probability still depends on the constant probability of each individual event.

The independence of these trials reassures that attempts aren't wasted and adds up to increase the likelihood of success when multiple tries are allowed, just as calculated in the solution.