Question

Forty percent of business travellers carry a laptop. In a sample of 15 business travelers, what is the probability that atleast three of the travelers have a laptop?

Answer 1

Given n = 5

p = ${\displaystyle \frac{40}{10}=\frac{2}{5}}$

q = 1 – p = ${\displaystyle 1-\frac{2}{5}=\frac{5-2}{5}=\frac{3}{5}}$

The binomial distribution P(X = x) = ${\displaystyle 15{\text{c}}_x{\left(\frac{2}{5}\right)}^x{\left(\frac{3}{5}\right)}^{15-x}}$

P(at least three of the travelers have a laptop)

= P(X ≥ 3)

P(X ≥ 3) = 1 – P(X < 3)

= 1 – [P(X = 0) + P(X = 1) + P(X = 2)]

P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)

= ${\displaystyle 15{\text{c}}_0{\left(\frac{2}{5}\right)}^0{\left(\frac{3}{5}\right)}^{15-0}+15{\text{c}}_1{\left(\frac{2}{5}\right)}^1{\left(\frac{3}{5}\right)}^{15-1}+15{\text{c}}_2{\left(\frac{2}{5}\right)}^2{\left(\frac{3}{5}\right)}^{15-2}}$

= ${\displaystyle \left(1\right)\left(1\right){\left(\frac{3}{5}\right)}^{15}+15\left(\frac{2}{5}\right){\left(\frac{3}{5}\right)}^{14}+105{\left(\frac{2}{5}\right)}^2{\left(\frac{3}{5}\right)}^{13}}$

= ${\displaystyle \frac{{13}^{15}}{5^{15}}+\frac{15\times 2\times 3^{14}}{5^5}+\frac{105\times 4\times 3^{13}}{5^{15}}}$

= ${\displaystyle \frac{3^{13}[3^2+(30\times 3)+(105\times 4\times 9)]}{5^{15}}}$

= ${\displaystyle \frac{1.593\times {10}^6\times (9+90+3780)}{3.055\times {10}^{10}}}$

= ${\displaystyle \frac{1.593\times 3879}{3.055\times {10}^4}}$

= ${\displaystyle \frac{6179.247}{30550}}$

= 0.20226

∴ P(X ≥ 3) = ${\displaystyle 1-020226}$

= 0.79774