Problem 92 - A Complete Resource Book in Mathematics for JEE Main

Question

The probablity that the birthdays of six different people will fall in exactly two calendar months is (A) \(\frac{1}{6}\) (B) \({ }^{12} C_{2} \times \frac{2^{6}}{12^{6}}\) (C) \({ }^{12} C_{2} \times \frac{2^{6}-2}{12^{6}}\) (D) \(\frac{341}{12^{5}}\)

Step-by-Step Solution

Answer

(C) \(^{12}C_{2} \times \frac{2^{6}-2}{12^{6}}\)

Step 1: Determine Total Possibilities

Consider that there are 12 calendar months for each of the 6 people to have a birthday. Thus, there are a total of \(12^6\) possible combinations for the months in which the 6 individuals' birthdays could occur.

Step 2: Choose Two Months

Select 2 out of the 12 calendar months for the birthdays to fall into. The number of ways to choose 2 months from 12 is \(^\text{12}C_2\), which is calculated as follows:\[^\text{12}C_2 = \frac{12!}{2!(12-2)!} = \frac{12 \times 11}{2 \times 1} = 66\]

Step 3: Assign Birthdays to Two Months

Since we need all 6 birthdays in exactly two chosen months, we can choose whether each birthday falls into the first or the second month. There are \(2^6\) ways to assign 6 birthdays to two months, including the cases where all are in one month.

Step 4: Subtract Invalid Assignments

The problem requires that the birthdays be in exactly two months, meaning they cannot all fall into just one month. There are 2 cases where this happens (all birthdays in the first month or all in the second month), so subtract 2 from \(2^6\):\[2^6 - 2 = 64 - 2 = 62\]

Step 5: Calculate the Probability

Finally, calculate the probability by dividing the number of valid assignments (from Step 4) by the total possibilities (from Step 1):\[\text{Probability} = \frac{{}^\text{12}C_2 \times (2^6 - 2)}{12^6} = \frac{66 \times 62}{12^6}\]

Step 6: Choose the Correct Option

Among the options provided, option (C) matches our calculation:(C) \(^{12}C_{2} \times \frac{2^{6}-2}{12^{6}}\)

Key Concepts

Combinatorics

Combinatorics is a branch of mathematics that deals with counting, arranging, and analyzing discrete structures. It is a crucial tool in probability theory because it helps us determine how many different ways an event can occur. By understanding these counts, we can calculate probabilities more effectively.

When faced with problems like the one involving the six individuals and their birthdays, combinatorics is used to calculate the total number of possible outcomes. In our exercise, the counting involved selecting 2 months out of 12 for the birthdays to occur, which is calculated using combinations.

• The formula for combinations, denoted as \(^{n}C_{k}\), represents the number of ways to choose \(k\) items from \(n\) items without considering the order. This is given by \(^{n}C_{k} = \frac{n!}{k!(n-k)!}\), where \(!\) denotes factorial, the product of all positive integers up to that number.
• In our case, choosing 2 months from 12, the calculation is \(^{12}C_{2} = 66\), reflecting the different ways to allocate the months.

These foundational combinatorial principles make it manageable to break down complex probability scenarios into understandable and solvable problems.

Birthday Problem

The "Birthday Problem" is a famous problem in probability theory that examines the likelihood that, in a group of people, some birthdays will coincide. The classic problem often seeks to find the probability that at least two people in a group share a birthday. However, in this specific exercise, we are exploring a variation – the probability that all birthdays fall within exactly two months.

To understand this better, consider the simple scenario where each person's birthday is equally likely to fall in any of the 12 months. Given this simplicity:

• We must choose two specific months where birthdays should occur, and there are \(^{12}C_{2}\) ways to choose these months.
• Next, distribute all six birthdays between these two months. Since there are no restrictions in a typical birthday problem except the month constraint, arrangements can be made using binary choices, leading to \(2^6\) potential arrangements.
• However, we reject cases where all six birthdays are within a single month, hence subtracting those scenarios.

This adjusted birthday problem underscores the importance of carefully considering constraints while calculating probability.

Conditional Probability

Conditional probability is the likelihood of an event happening given that another event has already occurred. It's a crucial concept when dealing with probabilities involving multiple dependent scenarios, like our current exercise where we have imposed criteria regarding months for birthdays.

In the problem, conditional probability isn't directly calculated, but understanding it can help refine solving methods. If one were to actually assess the probability of these outcomes happening,

• we'd first consider the total possibilities: allocating any birthday in any month, resulting in \(12^6\) combinations.
• Then, one might compare different event combinations – applying conditional frameworks to refine probable outcomes and eliminate invalid scenarios.

Conditional probability explicates why probabilities hinge on prior events' outcomes. Here, understanding why we subtract the invalid scenario (birthdays all in one month) is enhanced by considering conditional probabilities, ensuring valid outcomes alone are considered for crafting the final solution.