Question

1. it is reported that 77% of workers aged 16 and over drive to work alone. choose 8 workers at random. find the probability that, n = 8 p =.77 q =.23

a) at most 2 drive
b) at least 6 drive
c) all 8 drive

Explanation:

Step1: Recall binomial probability formula

The binomial probability formula is \(P(X = k)=C(n,k)\times p^{k}\times q^{n - k}\), where \(C(n,k)=\frac{n!}{k!(n - k)!}\), \(n\) is the number of trials, \(k\) is the number of successes, \(p\) is the probability of success, and \(q = 1 - p\) is the probability of failure.

Step2: Solve part (a)

For \(P(X\leq2)=P(X = 0)+P(X = 1)+P(X = 2)\)

• \(P(X = 0)=C(8,0)\times(0.77)^{0}\times(0.23)^{8}=\frac{8!}{0!(8 - 0)!}\times1\times(0.23)^{8}=1\times1\times0.23^{8}\approx0\)
• \(P(X = 1)=C(8,1)\times(0.77)^{1}\times(0.23)^{7}=\frac{8!}{1!(8 - 1)!}\times0.77\times0.23^{7}=8\times0.77\times0.23^{7}\approx0\)
• \(P(X = 2)=C(8,2)\times(0.77)^{2}\times(0.23)^{6}=\frac{8!}{2!(8 - 2)!}\times0.77^{2}\times0.23^{6}=28\times0.5929\times0.23^{6}\approx0.0006\)

\(P(X\leq2)\approx0 + 0+0.0006 = 0.0006\)

Step3: Solve part (b)

For \(P(X\geq6)=P(X = 6)+P(X = 7)+P(X = 8)\)

• \(P(X = 6)=C(8,6)\times(0.77)^{6}\times(0.23)^{2}=\frac{8!}{6!(8 - 6)!}\times0.77^{6}\times0.23^{2}=28\times0.77^{6}\times0.0529\approx0.373\)
• \(P(X = 7)=C(8,7)\times(0.77)^{7}\times(0.23)^{1}=\frac{8!}{7!(8 - 7)!}\times0.77^{7}\times0.23 = 8\times0.77^{7}\times0.23\approx0.299\)
• \(P(X = 8)=C(8,8)\times(0.77)^{8}\times(0.23)^{0}=\frac{8!}{8!(8 - 8)!}\times0.77^{8}\times1=1\times0.77^{8}\approx0.12\)

\(P(X\geq6)\approx0.373+0.299 + 0.12=0.792\)

Step4: Solve part (c)

For \(P(X = 8)=C(8,8)\times(0.77)^{8}\times(0.23)^{0}=\frac{8!}{8!(8 - 8)!}\times0.77^{8}\times1\approx0.12\)

Answer:

a) \(0.0006\)
b) \(0.792\)
c) \(0.12\)