Question

a hospital has an urgent need for four units of type a+ blood. assume that approximately 33% of the population has this type of blood. if there are 10 people waiting to donate one unit of blood, what is the probability that the hospital will be able to meet its need? (hint: subtract the probability that fewer than four people have a+ blood from 1.)
the probability that the hospital gets at least four units of blood is
(round to four decimal places as needed.)

Explanation:

Step1: Identify the binomial probability formula

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

Here, \(n = 10\), \(p=0.33\), and we want to find \(P(X\geq4)=1 - P(X < 4)=1-(P(X = 0)+P(X = 1)+P(X = 2)+P(X = 3))\)

Step2: Calculate \(P(X = 0)\)

\(C(10,0)=\frac{10!}{0!(10 - 0)!}=1\)
\(P(X = 0)=C(10,0)\times(0.33)^{0}\times(1 - 0.33)^{10 - 0}=1\times1\times0.67^{10}\approx0.0182\)

Step3: Calculate \(P(X = 1)\)

\(C(10,1)=\frac{10!}{1!(10 - 1)!}=\frac{10!}{1!9!}=10\)
\(P(X = 1)=C(10,1)\times(0.33)^{1}\times(0.67)^{9}=10\times0.33\times0.67^{9}\approx0.0898\)

Step4: Calculate \(P(X = 2)\)

\(C(10,2)=\frac{10!}{2!(10 - 2)!}=\frac{10\times9}{2\times 1}=45\)
\(P(X = 2)=C(10,2)\times(0.33)^{2}\times(0.67)^{8}=45\times0.1089\times0.67^{8}\approx0.1916\)

Step5: Calculate \(P(X = 3)\)

\(C(10,3)=\frac{10!}{3!(10 - 3)!}=\frac{10\times9\times8}{3\times2\times1}=120\)
\(P(X = 3)=C(10,3)\times(0.33)^{3}\times(0.67)^{7}=120\times0.035937\times0.67^{7}\approx0.2559\)

Step6: Calculate \(P(X < 4)\)

\(P(X < 4)=P(X = 0)+P(X = 1)+P(X = 2)+P(X = 3)\)
\(P(X < 4)=0.0182+0.0898 + 0.1916+0.2559=0.5555\)

Step7: Calculate \(P(X\geq4)\)

\(P(X\geq4)=1 - P(X < 4)=1-0.5555 = 0.4445\)

Answer:

\(0.4445\)