According to a survey, the time taken by teenagers to finish a box of donuts is uniformly distributed between [1,6] minutes. A café in New Delhi used this data for its advertisement that challenges teenagers to finish a box of donuts within 100 seconds to win a grand prize. A teenager, Ajay, accepted the café's challenge. What is the probability that he will finish the box of donuts within 4 minutes, given that he lost the challenge?

Let's break down the problem step by step.

### Step 1: Understanding the Distribution
The time taken by teenagers to finish a box of donuts is uniformly distributed between 1 and 6 minutes. This means that any time within this range is equally likely. The probability density function (PDF) for a uniform distribution \( U(a, b) \) is given by:

\[
f(x) = \frac{1}{b-a}, \quad \text{for } a \leq x \leq b
\]

Here, \( a = 1 \) minute and \( b = 6 \) minutes. So:

\[
f(x) = \frac{1}{6-1} = \frac{1}{5}, \quad \text{for } 1 \leq x \leq 6
\]

### Step 2: Event Definitions
- Let \( A \) be the event that Ajay finishes the box of donuts within 4 minutes.
- Let \( B \) be the event that Ajay loses the challenge, i.e., takes more than 100 seconds (which is 1.67 minutes).

We need to find the conditional probability \( P(A \mid B) \), which is the probability that Ajay finishes the box within 4 minutes given that he took more than 1.67 minutes.

### Step 3: Calculate Individual Probabilities
1. **Probability of Event \( A \):**
   \( A = \{1 \leq x \leq 4\} \).
   
   The probability that Ajay finishes the box within 4 minutes is:
   \[
   P(A) = \int_1^4 f(x) \, dx = \int_1^4 \frac{1}{5} \, dx = \frac{4-1}{5} = \frac{3}{5}
   \]

2. **Probability of Event \( B \):**
   \( B = \{x > 1.67\} \).
   
   The probability that Ajay took more than 1.67 minutes is:
   \[
   P(B) = \int_{1.67}^6 f(x) \, dx = \int_{1.67}^6 \frac{1}{5} \, dx = \frac{6-1.67}{5} = \frac{4.33}{5}
   \]

3. **Intersection of Events \( A \) and \( B \):**
   \( A \cap B = \{1.67 \leq x \leq 4\} \).
   
   The probability that Ajay took between 1.67 and 4 minutes is:
   \[
   P(A \cap B) = \int_{1.67}^4 f(x) \, dx = \int_{1.67}^4 \frac{1}{5} \, dx = \frac{4-1.67}{5} = \frac{2.33}{5}
   \]

### Step 4: Calculate the Conditional Probability
The conditional probability \( P(A \mid B) \) is given by:

\[
P(A \mid B) = \frac{P(A \cap B)}{P(B)} = \frac{\frac{2.33}{5}}{\frac{4.33}{5}} = \frac{2.33}{4.33} \approx 0.538
\]

### Conclusion
The probability that Ajay will finish the box of donuts within 4 minutes, given that he took more than 1.67 minutes, is approximately **0.538** or **53.8%**.

Would you like to discuss this further or have any specific questions?

### Follow-up Questions:
1. How would the probability change if the time range were different?
2. What if the challenge time was set to 3 minutes instead of 100 seconds?
3. Can we calculate the expected time for Ajay to finish the donuts?
4. What other real-life situations can be modeled using a uniform distribution?
5. How does the uniform distribution differ from other continuous distributions?

### Tip:
Conditional probability can often provide surprising insights, especially when combined with uniform distributions, where each outcome is equally likely within a given range.

Calculate the probability of finishing a box of donuts within 4 minutes, given a uniform distribution between 1 and 6 minutes. Understand the concept of conditional probability using the uniform distribution PDF. Ideal for college-level students studying probability theory.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation