Question

A problem in Statistics is given to three students A, B and C whose chances of solving it are 1/2, 3/4 and 1/4 respectively. What is the probability that the problem will be solved if all of them tried independently?

• A. $\frac{3}{32}$
• B. $\frac{29}{32}$
• C. $\frac{1}{32}$
• D. $\frac{1}{4}$

Hint:

For "at least one" probability questions, always calculate the product of the failure probabilities first, then subtract the result from 1.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The problem is solved if at least one of the students solves it.
Using the complement rule, it is easier to compute the probability of the opposite event (no student solves the problem) and subtract it from 1.
Key Formula or Approach:
For independent events $A, B,$ and $C$:
\[ P(\text{At least one occurs}) = 1 - P(A^c \cap B^c \cap C^c) \]
\[ P(A^c \cap B^c \cap C^c) = P(A^c) \times P(B^c) \times P(C^c) \]

Step 2: Detailed Explanation:
We are given individual probabilities of success:
$P(A) = \frac{1}{2} \implies P(A^c) = 1 - \frac{1}{2} = \frac{1}{2}$
$P(B) = \frac{3}{4} \implies P(B^c) = 1 - \frac{3}{4} = \frac{1}{4}$
$P(C) = \frac{1}{4} \implies P(C^c) = 1 - \frac{1}{4} = \frac{3}{4}$
Compute the probability that none of the students can solve the problem:
\[ P(\text{None solves}) = P(A^c) \times P(B^c) \times P(C^c) \]
\[ P(\text{None solves}) = \frac{1}{2} \times \frac{1}{4} \times \frac{3}{4} = \frac{3}{32} \]
Calculate the probability that the problem is solved:
\[ P(\text{Solved}) = 1 - P(\text{None solves}) = 1 - \frac{3}{32} = \frac{29}{32} \]
Therefore, the probability that the problem is solved is $\frac{29}{32}$.

Step 3: Final Answer
The correct option is (B).