<p>Person A can hit a target 4 times in 5 attempts. Person B - 3 times in four attempts. Person C - 2 times in 3 attempts. They fire a volley. The probability that the target is hit atleast two times is</p>
The probability that the target is hit at least two times is given by: \[ A B \bar{C} + A \bar{B} C + \bar{A} B C + A B C \] Substitute the given probabilities into the expression: \[ = \frac{4}{5} \times \frac{3}{4} \times \frac{1}{3} + \frac{4}{5} \times \frac{1}{4} \times \frac{2}{3} + \frac{1}{5} \times \frac{3}{4} \times \frac{2}{3} + \frac{4}{5} \times \frac{3}{4} \times \frac{2}{3} \] Simplifying: \[ = \frac{1}{5} + \frac{2}{15} + \frac{1}{10} + \frac{2}{5} \] Now, finding a common denominator: \[ = \frac{1}{5} + \frac{2}{15} + \frac{1}{10} + \frac{2}{5} = \frac{50}{60} = \frac{5}{6} \]
Reinforce your conceptual grasp with related questions from MAH MCA CET 2013:
A random variable X has the distribution law as given below:
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