Key Concepts and Formulas

• Bayes' Theorem: Used to calculate the conditional probability of an event, given that another related event has occurred. If $B_1, B_2, \dots, B_n$ are mutually exclusive and collectively exhaustive events, and $E$ is an observed event, then the probability of $B_i$ given $E$ is: $P(B_i|E) = \frac{P(E|B_i)P(B_i)}{P(E)}$
• Law of Total Probability: Used to find the total probability of an event $E$ by summing the probabilities of $E$ occurring under each of the mutually exclusive and collectively exhaustive scenarios ($B_j$). $P(E) = \sum_{j=1}^{n} P(E|B_j)P(B_j)$
• Prime Numbers: A natural number greater than 1 that has no positive divisors other than 1 and itself. (e.g., 2, 3, 5, 7, ...). The number 1 is neither prime nor composite, but for this problem, it is considered a "non-prime."

Step-by-Step Solution

Step 1: Define Events and Assign Prior Probabilities

We begin by clearly defining the events involved in the problem and assigning their initial probabilities. This sets up the framework for applying Bayes' Theorem.

• Let $B_1$ be the event that Box I is selected.
• Let $B_2$ be the event that Box II is selected.
• Let $E$ be the event that the card drawn is a non-prime number.

The problem states that "A box is selected at random." Since there are two boxes, the probability of selecting either box is equal. These are our prior probabilities:

$P(B_1) = \frac{1}{2}$ $P(B_2) = \frac{1}{2}$

Reasoning: Defining events precisely prevents confusion. The prior probabilities reflect our knowledge before any card is drawn or its properties are observed.

Step 2: Calculate Conditional Probabilities (Likelihoods) of Event E

Next, we need to determine the probability of drawing a non-prime card from each specific box. These are the conditional probabilities, $P(E|B_1)$ and $P(E|B_2)$, often called likelihoods.

• For Box I (cards numbered 1 to 30):

  • Total cards: 30
  • Prime numbers in this range: {2, 3, 5, 7, 11, 13, 17, 19, 23, 29}. There are 10 prime numbers.
  • Non-prime numbers: Total cards - Number of prime cards = $30 - 10 = 20$. (This includes 1 and all composite numbers).
  • The probability of drawing a non-prime card given Box I was selected is: $P(E|B_1) = \frac{\text{Number of non-prime cards in Box I}}{\text{Total cards in Box I}} = \frac{20}{30} = \frac{2}{3}$

• For Box II (cards numbered 31 to 50):

  • Total cards: 20
  • Prime numbers in this range: {31, 37, 41, 43, 47}. There are 5 prime numbers.
  • Non-prime numbers: Total cards - Number of prime cards = $20 - 5 = 15$.
  • The probability of drawing a non-prime card given Box II was selected is: $P(E|B_2) = \frac{\text{Number of non-prime cards in Box II}}{\text{Total cards in Box II}} = \frac{15}{20} = \frac{3}{4}$

Reasoning: These likelihoods quantify how probable it is to observe the event $E$ (drawing a non-prime card) if we already know which box was chosen. This information is crucial for updating our beliefs using Bayes' Theorem.

Step 3: Calculate the Total Probability of Drawing a Non-Prime Card, P(E)

Before applying Bayes' Theorem, we need the overall probability of drawing a non-prime card, $P(E)$, considering both boxes. This is calculated using the Law of Total Probability.

$P(E) = P(E|B_1)P(B_1) + P(E|B_2)P(B_2)$

Substitute the values from Step 1 and Step 2: $P(E) = \left(\frac{2}{3}\right) \times \left(\frac{1}{2}\right) + \left(\frac{3}{4}\right) \times \left(\frac{1}{2}\right)$ $P(E) = \frac{2}{6} + \frac{3}{8}$ $P(E) = \frac{1}{3} + \frac{3}{8}$ To add these fractions, we find a common denominator, which is 24: $P(E) = \frac{1 \times 8}{3 \times 8} + \frac{3 \times 3}{8 \times 3}$ $P(E) = \frac{8}{24} + \frac{9}{24}$ $P(E) = \frac{17}{24}$

Reasoning: $P(E)$ represents the overall probability of observing the event (drawing a non-prime card) across all possible scenarios (selecting Box I or Box II), weighted by their initial probabilities. This value serves as the normalizing denominator in Bayes' Theorem.

Step 4: Apply Bayes' Theorem to Find the Required Probability

Finally, we use Bayes' Theorem to calculate the posterior probability, $P(B_1|E)$, which is the probability that the card was drawn from Box I, given that it was a non-prime number.

Using the formula: $P(B_1|E) = \frac{P(E|B_1)P(B_1)}{P(E)}$

Substitute all the values we've calculated in the previous steps: $P(B_1|E) = \frac{\left(\frac{2}{3}\right) \times \left(\frac{1}{2}\right)}{\frac{17}{24}}$ $P(B_1|E) = \frac{\frac{1}{3}}{\frac{17}{24}}$

To simplify, we multiply the numerator by the reciprocal of the denominator: $P(B_1|E) = \frac{1}{3} \times \frac{24}{17}$ $P(B_1|E) = \frac{24}{3 \times 17}$ $P(B_1|E) = \frac{8}{17}$

Reasoning: This step updates our initial belief about which box was chosen, based on the new evidence that the card drawn was non-prime. Bayes' Theorem allows us to reverse the conditioning and find the probability of the cause ($B_1$) given the effect ($E$).

Common Mistakes & Tips

• Prime Number Identification: Carefully list prime numbers. Remember that 1 is neither prime nor composite, but it is a "non-prime" for the purpose of this problem. Missing or misidentifying primes will lead to incorrect counts of non-primes.
• Confusing $P(A|B)$ and $P(B|A)$: Ensure you correctly identify what the question is asking for. Here, it's $P(\text{Box I} | \text{Non-prime})$, not $P(\text{Non-prime} | \text{Box I})$. Bayes' Theorem is specifically designed to "reverse" this conditional probability.
• Arithmetic with Fractions: Double-check all fraction additions, multiplications, and divisions. A common error is incorrect common denominators or mistakes in simplifying compound fractions.

Summary

This problem is a classic application of Bayes' Theorem. We first defined the events and their prior probabilities. Then, we calculated the likelihood of drawing a non-prime card from each box. Using these values, we determined the overall probability of drawing a non-prime card using the Law of Total Probability. Finally, we applied Bayes' Theorem to find the probability that the card came from Box I, given that it was non-prime, arriving at $\frac{8}{17}$.

The final answer is \boxed{\text{8 \over 17}}, which corresponds to option (A).