Example — Target

Example — Target

A shooter hits the target with probability $p=0,8$. He fires $10$ times. What is the probability of hitting it at least $8$ times?

It is the sum of three binomial probabilities:

$P(k\ge 8) = P_{10}(8) + P_{10}(9) + P_{10}(10).$

$P_{10}(10) = (0,8)^{10} \approx 0,107, \qquad P_{10}(9) = 10\cdot (0,8)^9\cdot 0,2 \approx 0,268,$

$P_{10}(8) = \binom{10}{8}(0,8)^8(0,2)^2 = 45\cdot 0,168\cdot 0,04 \approx 0,302.$

Total: $P(k\ge 8) \approx 0,677$.

Links

Topics: Probability
Concepts: Binomial distribution · Complementary event
Methods: Binomial · Complementary prob
Skills: Probability calculation