Question

5.According to a local study, 80 per cent of workers in Malaysia work in an environment that could endanger their visual. If 20 Malaysia workers selected randomly as a sample, what is the probability

a) that exactly 16 workers work in such environment?

b) that more than 17 workers work in such environment?

c) that exactly 4 workers not work in such environment?

d) fewer than 3 workers the number not work in such environment?

Answer 1

$P(x) = \frac{n!}{x!(n-x)!} \times p^x \times (1-p)^{n-x}$

a) P = 80 % = 0.8

n = 20

Evaluate P(x) if x = 16

$P(x = 16) = \frac{20!}{16!(20-16)!} \times 0.8^{16} \times (1 - 0.8)^{20-16} \\ = 4845 \times 0.8^{16} \times 0.2^{4} \\ = 0.218199$

b) P(x > 17) = P(18) + P(19) + P(20)

$= \frac{20!}{18!(20-18)!} \times 0.8^{18} \times (1 – 0.8)^{20-18} + \frac{20!}{19!(20-19)!} \times 0.8^{19} \times (1 - 0.8)^{20-19} + \frac{20!}{20!(20-20)!} \times 0.8^{20} \times (1 – 0.8)^{20-20} \\ = 0.1369 + 0.0576 + 0.0115 \\ = 0.2060$

c) P = 20 % = 0.2 (not work in such environment)

n = 20

$P(x = 4) = \frac{20!}{4!(20-4)!} \times 0.2^4 \times (1-0.2)^{20-4} \\ = 4845 \times 0.0016 \times 0.02814 \\ = 0.21814$

d) P(x<3) = P(x=0) + P(x=1) +P(x=2)

$P(x=0) = \frac{20!}{0!(20-0)!} \times 0.2^0 \times (1-0.2)^{20-0} \\ = 0.011529 \\ P(x=1) = \frac{20!}{1!(20-1)!} \times 0.2^1 \times (1-0.2)^{20-1} \\ = 20 \times 0.2 \times 0.01441 \\ = 0.05764 \\ P(x=2) = \frac{20!}{2!(20-2)!} \times 0.2^2 \times (1-0.2)^{20-2} \\ = 190 \times 0.04 \times 0.01801 \\ = 0.136876 \\ P(x<3) = 0.011529 + 0.05764 + 0.136876 = 0.206045$