Question

a shipping company claims that 90% of its packages are delivered on time. jenny noticed that out of the last 10 packages shipped, 2 were late. what is the probability that 2 out of 10 randomly selected shipments would be late?
0.04
0.10
0.19
0.43

Explanation:

Step1: Determine the probability of success and failure

The probability of a package being on - time \(p = 0.9\), so the probability of a package being late \(q=1 - p=1 - 0.9 = 0.1\). We use the binomial probability formula \(P(X = k)=C(n,k)\times p^{n - k}\times q^{k}\), where \(n = 10\) (the number of trials, i.e., the number of packages), \(k = 2\) (the number of "successes" where success is a late package), and \(C(n,k)=\frac{n!}{k!(n - k)!}\)

Step2: Calculate the binomial coefficient

Calculate \(C(10,2)=\frac{10!}{2!(10 - 2)!}=\frac{10\times9\times8!}{2\times1\times8!}=\frac{10\times9}{2\times1}=45\)

Step3: Calculate the probability using the binomial formula

Substitute \(n = 10\), \(k = 2\), \(p = 0.9\), and \(q = 0.1\) into the binomial formula:
\(P(X = 2)=C(10,2)\times(0.9)^{10 - 2}\times(0.1)^{2}\)
\(P(X = 2)=45\times(0.9)^{8}\times(0.1)^{2}\)
\((0.9)^{8}=0.43046721\), \((0.1)^{2}=0.01\)
\(P(X = 2)=45\times0.43046721\times0.01\)
\(P(X = 2)=45\times0.0043046721\approx0.19\)

Answer:

0.19