The probability of a man hitting a target is 1/5. If the man fires 7 times, then what is the probability that he hits the target <strong>AT LEAST</strong> twice?

  1. A. 1-(\frac{3}{5})(\frac{4}{5})^{6}
  2. B. 1-(\frac{3}{5})(\frac{4}{5})^{7}
  3. C. 1-(\frac{11}{5})(\frac{4}{5})^{6}
  4. D. 1-(\frac{11}{5})(\frac{4}{5})^{7}

Correct Answer: C. 1-(\frac{11}{5})(\frac{4}{5})^{6}

Explanation

Using the Binomial distribution with n=7, p=1/5, and q=4/5. P(X \geq 2) = 1 - [P(X=0) + P(X=1)] = 1 - [\binom{7}{0}(\frac{1}{5})^0(\frac{4}{5})^7 + \binom{7}{1}(\frac{1}{5})^1(\frac{4}{5})^6]. This simplifies to 1 - (\frac{4}{5})^6 [\frac{4}{5} + \frac{7}{5}] = 1 - (\frac{11}{5})(\frac{4}{5})^6.

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