Question

The probability of setting an easy exam paper by three setters are \( \frac{1}{2}, \frac{1}{3}, \) and \( \frac{1}{4} \). If all three are setting one paper each, then the probability that at least one of the papers will be easy is ................... (round off to 2 decimal places).

Hint:

Whenever a probability question uses the phrase "at least one," your first thought should be to calculate "1 minus the probability of none." It almost always simplifies the problem and reduces the chance of calculation errors.

Answer 1

Step 1: Understanding the Concept:
This problem asks for the probability of "at least one" of several independent events occurring. The most efficient way to solve this type of problem is to calculate the probability of the complementary event (i.e., that none of the events occur) and subtract this from 1.
Step 2: Key Formula or Approach:
Let A, B, and C be the events that setters 1, 2, and 3 set an easy paper, respectively.
We want to find \( P(\text{at least one easy paper}) \).
The complementary approach formula is: \[ P(\text{at least one easy}) = 1 - P(\text{no easy papers}) \] \[ P(\text{no easy papers}) = P(\text{A is not easy}) \times P(\text{B is not easy}) \times P(\text{C is not easy}) \] Step 3: Detailed Calculation:
First, let's find the probability that each setter does not set an easy paper.
- Probability that Setter 1 does not set an easy paper: \[ P(A') = 1 - P(A) = 1 - \frac{1}{2} = \frac{1}{2} \] - Probability that Setter 2 does not set an easy paper: \[ P(B') = 1 - P(B) = 1 - \frac{1}{3} = \frac{2}{3} \] - Probability that Setter 3 does not set an easy paper: \[ P(C') = 1 - P(C) = 1 - \frac{1}{4} = \frac{3}{4} \] Next, calculate the probability that none of them set an easy paper. Since the events are independent, we multiply their probabilities: \[ P(\text{all are not easy}) = P(A') \times P(B') \times P(C') = \frac{1}{2} \times \frac{2}{3} \times \frac{3}{4} \] \[ P(\text{all are not easy}) = \frac{1 \times 2 \times 3}{2 \times 3 \times 4} = \frac{6}{24} = \frac{1}{4} \] Finally, calculate the probability that at least one paper is easy: \[ P(\text{at least one easy}) = 1 - P(\text{all are not easy}) = 1 - \frac{1}{4} = \frac{3}{4} \] Step 4: Final Answer:
To express the answer as a decimal rounded to two places: \[ \frac{3}{4} = 0.75 \] The probability is 0.75.
Step 5: Why This is Correct:
The solution correctly uses the principle of complementary probability for independent events. The calculations for the individual and combined probabilities are accurate, leading to the final correct answer.