Question

It is observed that 25% of the cases related to child labour reported to the police station are solved. If 6 new cases are reported, then the probability that at least 5 of them will be solved is

• A. $\frac{19}{1024}$
• B. $\frac{19}{4096}$
• C. $\left( \frac{1}{4} \right)^6$
• D. $\frac{19}{2048}$

Hint:

When the options have large common denominators like $4^6 = 4096$, do not simplify intermediate fractions like $\binom{6}{5}$. Keep the denominators identical throughout the calculation to make the final addition trivial.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The problem presents a scenario where a specific event (solving a child labour case) has a constant probability of success.
We are given a fixed number of independent trials ($n = 6$) and need to find the probability of obtaining at least 5 successful outcomes.
Because each case is independent and has exactly two outcomes (solved or unsolved), this follows a Binomial Distribution.

Step 2: Key Formula or Approach:
The Probability Mass Function for a Binomial Distribution is given by:
$$P(X = x) = \binom{n}{x} p^x q^{n-x}$$ Where $n$ is the total number of trials, $x$ is the number of successes, $p$ is the probability of success, and $q = 1 - p$ is the probability of failure.
To find the probability of "at least 5", we calculate:
$$P(X \geq 5) = P(X = 5) + P(X = 6)$$

Step 3: Detailed Explanation:
From the problem description, identify the given parameters:
Probability of a case being solved, $p = 25\% = \frac{25}{100} = \frac{1}{4}$.
Probability of a case remaining unsolved, $q = 1 - \frac{1}{4} = \frac{3}{4}$.
Total number of cases reported, $n = 6$.
Now, calculate the individual probabilities for $x = 5$ and $x = 6$:
For $x = 5$:
$$P(X = 5) = \binom{6}{5} \left(\frac{1}{4}\right)^5 \left(\frac{3}{4}\right)^1 = 6 \times \frac{1}{1024} \times \frac{3}{4} = \frac{18}{4096}$$ For $x = 6$:
$$P(X = 6) = \binom{6}{6} \left(\frac{1}{4}\right)^6 \left(\frac{3}{4}\right)^0 = 1 \times \frac{1}{4096} \times 1 = \frac{1}{4096}$$ Sum the two values to obtain the total cumulative probability:
$$P(X \geq 5) = \frac{18}{4096} + \frac{1}{4096} = \frac{19}{4096}$$

Step 4: Final Answer:
The probability that at least 5 cases will be solved is $\frac{19}{4096}$, which perfectly matches option (B).