Box I contains 30 cards numbered 1 to 30 and Box II contains 20 cards numbered 31 to 50. A box is se

Q: Wait, is 1 a prime number? I always get confused about this.

This is a classic trap! A prime number is defined as a natural number greater than 1 that has exactly two distinct positive divisors: 1 and itself. Since 1 only has one divisor (itself), it does not meet the criteria. Therefore, 1 is neither prime nor composite. It is a non-prime. Always list your primes carefully to avoid this common pitfall.

Q: Why do we use Bayes' Theorem here instead of just simple probability?

Simple probability is used when you know the 'cause' (the box) and want to predict the 'effect' (the card). Here, the situation is reversed: you already know the 'effect' (the card is non-prime) and you are trying to infer the 'cause' (which box it came from). Bayes' Theorem is the mathematical bridge that allows us to travel backward in time from the effect to the cause.

Q: Can I just ignore the $\frac{1}{2}$ probability of picking the box since it's the same for both?

Mathematically, yes, because it cancels out in the numerator and denominator. However, as a student, you should always write it down in your initial setup. It demonstrates that you understand the full structure of the problem. Only after you have written the complete Bayes' formula should you perform the cancellation. It keeps your logic rigorous and prevents errors in more complex problems where the box probabilities might not be equal.

Question

Box I contains 30 cards numbered 1 to 30 and Box II 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 1 is :

Options:

(A) 4/17

(B) 2/3

(C) 2/5

(D) 8/17

Sigma Percentile · Accepted Answer

### The Detective's Dilemma: Unraveling the Mystery of the Boxes

Imagine you are standing in a room with two mysterious boxes. Box I holds a collection of cards numbered 1 through 30, and Box II holds cards numbered 31 through 50. You are blindfolded, you pick a box at random, and you draw a card.

You look at the card, and it is a non-prime number. The question is: can you deduce which box you picked? This is not just a math problem; it is a detective story. We are using the language of probability to reverse-engineer the past.

### Phase 1: The Anatomy of the Boxes

Before we can solve the mystery, we must understand our environment. We have two boxes, $B_1$ and $B_2$. Since the problem states the box is selected at random, we assign an equal probability to each: $P(B_1) = P(B_2) = \frac{1}{2}$.

In Box I, we have numbers from 1 to 30. The prime numbers are $\{2, 3, 5, 7, 11, 13, 17, 19, 23, 29\}$, totaling 10 primes. Since there are 30 cards in total, the number of non-primes is $30 - 10 = 20$.

Thus, the conditional probability of drawing a non-prime given that we are in Box I is:
$$P(NP | B_1) = \frac{20}{30} = \frac{2}{3}$$

Now, let us turn our attention to Box II. This box contains numbers from 31 to 50, meaning the total count is $50 - 31 + 1 = 20$ cards. The primes in this range are $\{31, 37, 41, 43, 47\}$, which is 5 prime numbers.

Consequently, the number of non-primes is $20 - 5 = 15$. The conditional probability of drawing a non-prime given that we are in Box II is:
$$P(NP | B_2) = \frac{15}{20} = \frac{3}{4}$$

### Phase 2: The Power of Bayes' Theorem

We have our data. We want to know the probability that the card came from Box I, given that we know it is non-prime. This is the classic application of Bayes' Theorem, where we seek $P(B_1 | NP)$.

The formula is defined as:
$$P(B_1 | NP) = \frac{P(NP | B_1)P(B_1)}{P(NP)}$$

The denominator $P(NP)$ is the total probability of drawing a non-prime, calculated as:
$$P(NP) = P(NP | B_1)P(B_1) + P(NP | B_2)P(B_2)$$

### Phase 3: The Elegant Cancellation

Now, let us substitute our values into this framework. We have $P(B_1) = \frac{1}{2}$, $P(NP | B_1) = \frac{2}{3}$, and $P(NP | B_2) = \frac{3}{4}$.

$$P(B_1 | NP) = \frac{\frac{1}{2} 	imes \frac{2}{3}}{\frac{1}{2} 	imes \frac{2}{3} + \frac{1}{2} 	imes \frac{3}{4}}$$

We can factor out the $\frac{1}{2}$ from the numerator and the denominator and cancel them out entirely. This leaves us with a much cleaner expression:
$$P(B_1 | NP) = \frac{\frac{2}{3}}{\frac{2}{3} + \frac{3}{4}}$$

### Phase 4: The Final Calculation

To solve the denominator, we find a common denominator for $\frac{2}{3}$ and $\frac{3}{4}$, which is 12. We convert them:
$$\frac{2}{3} = \frac{8}{12}, \quad \frac{3}{4} = \frac{9}{12}$$

Adding them together, we get $\frac{8}{12} + \frac{9}{12} = \frac{17}{12}$. Now, we substitute this back into our main equation:
$$P(B_1 | NP) = \frac{\frac{2}{3}}{\frac{17}{12}}$$

To divide by a fraction, we multiply by its reciprocal:
$$P(B_1 | NP) = \frac{2}{3} 	imes \frac{12}{17} = \frac{8}{17}$$

The final probability is **$\frac{8}{17}$**.

Select Answer:

Visualized Solution

The Random Experiment Setup

Analyzing Box I

Primes and Non-Primes in Box I

Analyzing Box II

Primes and Non-Primes in Box II

The Core Question & Bayes' Theorem

Substituting the Values

Simplifying the Expression

Calculating the Denominator

Final Probability

The Sigma Insight: Bayes' Theorem

The Detective's Dilemma

Phase 1

Phase 2

Phase 3

Phase 4

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Select Answer:

Visualized Solution

The Random Experiment Setup

Analyzing Box I

Primes and Non-Primes in Box I

Analyzing Box II

Primes and Non-Primes in Box II

The Core Question & Bayes' Theorem

Substituting the Values

Simplifying the Expression

Calculating the Denominator

Final Probability

The Sigma Insight: Bayes' Theorem

The Detective's Dilemma

Phase 1

Phase 2

Phase 3

Phase 4

Similar Questions

Comprehension Passage

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