Problem 4 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

Box I contains 30 cards numbered 1 to 30 and Box Il contains 20 cards numbered 31 to 50 . A box is selected at random and a card is drawn from it. The number on the card is found to be a non-prime number. The probability that the card was drawn from Box I is: \([\) Sep. 02, 2020 (I)] (a) \(\frac{2}{3}\) (b) \(\frac{8}{17}\) (c) \(\frac{4}{17}\) (d) \(\frac{2}{5}\)

Step-by-Step Solution

Answer

\(\frac{8}{17}\)

Step 1: Define Event

Let's define the events as follows: let \(A\) be the event that a card is drawn from Box I, and \(B\) be the event that the drawn card is a non-prime number.

Step 2: Find Probability of Selecting Box I

The probability of selecting Box I is \(P(A) = \frac{1}{2}\) since either Box I or Box II can be chosen with equal probability.

Step 3: Count Non-Prime Numbers in Box I

Box I contains cards numbered 1 to 30. The non-prime numbers are: 1, 4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21, 22, 24, 25, 26, 27, 28, and 30. Total non-prime numbers = 20.

Step 4: Calculate Conditional Probability for Box I

The probability of drawing a non-prime number from Box I is \(P(B|A) = \frac{20}{30} = \frac{2}{3}\).

Step 5: Count Non-Prime Numbers in Box II

Box II contains cards numbered 31 to 50. The non-prime numbers are: 32, 33, 34, 35, 36, 38, 39, 40, 42, 44, 45, 46, 48, 49, 50. Total non-prime numbers = 15.

Step 6: Calculate Conditional Probability for Box II

The probability of drawing a non-prime number from Box II is \(P(B|A^c) = \frac{15}{20} = \frac{3}{4}\).

Step 7: Calculate Total Probability of Non-Prime Number

Using the law of total probability, \(P(B) = P(B|A) \cdot P(A) + P(B|A^c) \cdot P(A^c) = \frac{2}{3} \cdot \frac{1}{2} + \frac{3}{4} \cdot \frac{1}{2} = \frac{5}{12} + \frac{3}{8} = \frac{17}{24}\).

Step 8: Find Probability Card is from Box I (Bayes Theorem)

Using Bayes' theorem, calculate \(P(A|B) = \frac{P(B|A) \cdot P(A)}{P(B)} = \frac{\frac{2}{3} \cdot \frac{1}{2}}{\frac{17}{24}} = \frac{8}{17}\).

Step 9: Final Answer

Thus, the probability that the card was drawn from Box I given that it is a non-prime number is \(\frac{8}{17}\).

Key Concepts

Bayes Theorem

Bayes Theorem is a fundamental concept in probability that enables us to update the probability estimate for an event based on new evidence. It essentially allows us to flip conditional probabilities. In the context of our problem, we wanted to find the probability that a card was drawn from Box I, given that the card showed a non-prime number. Bayes Theorem helps us calculate this by using known probabilities and the total probability of drawing a non-prime number.
For this problem, we denote two events:

• Event A: Drawing a card from Box I.
• Event B: Drawing a non-prime number.

Given Bayes Theorem, the formula is: \[ P(A|B) = \frac{P(B|A) \cdot P(A)}{P(B)} \] Where:

• \( P(A|B) \) is the probability that the card came from Box I given it's non-prime.
• \( P(B|A) \) is the probability of getting a non-prime if the card is from Box I.
• \( P(A) \) is the initial probability of selecting Box I.
• \( P(B) \) is the total probability of drawing a non-prime number, calculated previously with the law of total probability.

Using Bayes Theorem, we determined that the probability a non-prime card was from Box I is \( \frac{8}{17} \). It is a powerful tool for decision-making based on evidence.

Non-prime Numbers

Understanding non-prime numbers is essential for tackling probability problems involving them. A non-prime number is a positive integer that has more than two distinct positive divisors. This means it's not a prime number, as prime numbers have exactly two distinct positive divisors: 1 and the number itself.
In the given problem, Box I contains cards numbered 1 through 30, and Box II cards 31 through 50. Non-prime numbers for each box were counted as follows:

• Box I: 1, 4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21, 22, 24, 25, 26, 27, 28, and 30. There are 20 non-prime numbers in Box I.
• Box II: 32, 33, 34, 35, 36, 38, 39, 40, 42, 44, 45, 46, 48, 49, and 50. There are 15 non-prime numbers in Box II.

Counting these non-primes was crucial as it directly impacted our probability calculations, including conditional probabilities and total probabilities, leading to finding the likelihood of Box I being chosen.

Law of Total Probability

The Law of Total Probability is an important aspect in probability theory used to find the overall probability of a specific event happening given multiple possible scenarios or conditions. It states that the total probability of an event is the sum of the probabilities of that event given different mutually exclusive conditions, each multiplied by the probability of each condition occurring.
In our exercise, we calculated the total probability of drawing a non-prime number (denoted as Event B) from either of two boxes. This was crucial for the subsequent use of Bayes Theorem because it helped establish the denominator of our probability equation.
The Law of Total Probability formula used was: \[ P(B) = P(B|A) \cdot P(A) + P(B|A^c) \cdot P(A^c) \] Where:

• \( P(B|A) \) is the probability of a non-prime from Box I.
• \( P(B|A^c) \) is the probability of a non-prime from Box II.
• \( P(A) \) and \( P(A^c) \) are the probabilities of selecting Box I and Box II respectively, both equal to \( \frac{1}{2} \).

After using this law, we found that the total probability of drawing a non-prime card was \( \frac{17}{24} \). This highlighted the essential role of this principle in combining different scenarios to obtain meaningful results.