Question

homework: section 4.3
score: 1/15 answered: 1/15
question 2
american airlines flights from chicago to seattle are on time 60 % of the time. suppose 13 flights are randomly selected, and the number on - time flights is recorded.

1. the probability that at least 3 flights are on time is =
2. the probability that at most 12 flights are on time is =
3. the probability that exactly 9 flights are on time is =

Explanation:

Step1: Identify the binomial distribution parameters

Let \(n = 13\) (number of trials, i.e., number of flights), \(p=0.6\) (probability of success, i.e., probability of a flight being on - time), and \(q = 1 - p=0.4\). The binomial probability formula is \(P(X = k)=C(n,k)\times p^{k}\times q^{n - k}\), where \(C(n,k)=\frac{n!}{k!(n - k)!}\).

Step2: Calculate the probability that at least 3 flights are on time

\(P(X\geq3)=1 - P(X\lt3)=1-(P(X = 0)+P(X = 1)+P(X = 2))\)
\[

$$\begin{align*} P(X = 0)&=C(13,0)\times(0.6)^{0}\times(0.4)^{13}=\frac{13!}{0!(13 - 0)!}\times1\times(0.4)^{13}= (0.4)^{13}\approx0.000000671\\ P(X = 1)&=C(13,1)\times(0.6)^{1}\times(0.4)^{12}=\frac{13!}{1!(13 - 1)!}\times0.6\times(0.4)^{12}=13\times0.6\times(0.4)^{12}\approx0.0000154\\ P(X = 2)&=C(13,2)\times(0.6)^{2}\times(0.4)^{11}=\frac{13!}{2!(13 - 2)!}\times0.36\times(0.4)^{11}=78\times0.36\times(0.4)^{11}\approx0.000137\\ P(X\lt3)&\approx0.000000671 + 0.0000154+0.000137\approx0.000153\\ P(X\geq3)&\approx1 - 0.000153 = 0.999847 \end{align*}$$

\]

Step3: Calculate the probability that at most 12 flights are on time

\(P(X\leq12)=1 - P(X = 13)\)
\[

$$\begin{align*} P(X = 13)&=C(13,13)\times(0.6)^{13}\times(0.4)^{0}=\frac{13!}{13!(13 - 13)!}\times(0.6)^{13}\times1=(0.6)^{13}\approx0.001317 \end{align*}$$

\]
\(P(X\leq12)=1 - 0.001317 = 0.998683\)

Step4: Calculate the probability that exactly 9 flights are on time

\[

$$\begin{align*} P(X = 9)&=C(13,9)\times(0.6)^{9}\times(0.4)^{4}\\ &=\frac{13!}{9!(13 - 9)!}\times(0.6)^{9}\times(0.4)^{4}\\ &=\frac{13\times12\times11\times10}{4\times3\times2\times1}\times(0.6)^{9}\times(0.4)^{4}\\ & = 715\times0.010077696\times0.0256\\ &\approx0.184 \end{align*}$$

\]

Answer:

1. \(0.999847\)
2. \(0.998683\)
3. \(0.184\)