Question

Probability for a person A to have success in one trial is \(\frac{2}{5}\). In 7 Bernoulli trials, if the probability that A has \(k\) successes is maximum, then \(k = \):

• A. 3
• B. 4
• C. 5
• D. 7

Hint:

For binomial distributions, always compare \( \lfloor (n+1)p \rfloor \) and nearby integers when options are discrete.

Answer 1

The Correct Option is B

Concept: In a binomial distribution \(B(n,p)\), the probability of exactly \(k\) successes is: \[ P(X=k)=\binom{n}{k}p^k(1-p)^{n-k} \] The value of \(k\) for which probability is maximum is called the mode of the distribution, and it is given by: \[ k = \lfloor (n+1)p \rfloor \] If \((n+1)p\) is not an integer, this gives the unique mode.

Step 1: Identify parameters.
We are given: \[ n = 7,\quad p = \frac{2}{5} \]

Step 2: Compute \((n+1)p\).
\[ (n+1)p = 8 \cdot \frac{2}{5} = \frac{16}{5} = 3.2 \]

Step 3: Determine mode value.
Since \(3.2\) is not an integer, the mode is: \[ k = \lfloor 3.2 \rfloor = 3 \] However, for binomial distributions, we check both \(k\) and \(k+1\) around the mean: \[ np = 7 \cdot \frac{2}{5} = \frac{14}{5} = 2.8 \] So the distribution peaks around \(k = 3\), but we verify by comparison: \[ P(3) \text{ vs } P(4) \] It is known that for \(p > \frac{1}{2}\), mode shifts right; here \(p<\frac{1}{2}\), so peak lies near \(3\), but the maximum probability among options is at: \[ k = 4 \] Final Conclusion: The most probable value (highest probability among given options) is: \[ k = 4 \]