Question

The probability that a man failing to hit a target is \(\frac{1}{3}\). If he fires 4 times, then the probability that he hits the target at least thrice is

• A. \(\frac{16}{27}\)
• B. \(\frac{11}{27}\)
• C. \(\frac{8}{81}\)
• D. \(\frac{32}{81}\)

Hint:

When calculating probabilities of multiple independent events, always remember to consider all possible successful outcomes that meet the criteria.

Answer 1

The Correct Option is A

Step 1: Calculate the probability of hitting the target, which is \(1 - \frac{1}{3} = \frac{2}{3}\). 
Step 2: To find the probability of hitting the target at least 3 times in 4 attempts, consider the cases where he hits exactly 3 times or exactly 4 times. \[ P(\text{3 hits}) = \binom{4}{3} \left(\frac{2}{3}\right)^3 \left(\frac{1}{3}\right)^1 = 4 \times \frac{8}{27} \times \frac{1}{3} = \frac{32}{81} \] \[ P(\text{4 hits}) = \binom{4}{4} \left(\frac{2}{3}\right)^4 = \frac{16}{81} \] Step 3: Add the probabilities of these two events. \[ P(\text{at least 3 hits}) = P(\text{3 hits}) + P(\text{4 hits}) = \frac{32}{81} + \frac{16}{81} = \frac{48}{81} = \frac{16}{27} \]