Question

1. blood donors a person with type o - positive blood can receive blood only from other type o donors. about 44% of the u.s. population has type o blood. at a blood drive, how many potential donors do you expect to examine in order to get three units of type o blood?

Explanation:

Step1: Identify the distribution

This is a negative - binomial distribution problem. The negative - binomial distribution formula for the expected value \(E(X)=\frac{r}{p}\), where \(r\) is the number of successes and \(p\) is the probability of success in a single trial.

Step2: Assign values to \(r\) and \(p\)

Here, the number of successes \(r = 3\) (we want 3 units of type O blood), and the probability of success \(p=0.44\) (probability that a donor has type O blood).

Step3: Calculate the expected value

Using the formula \(E(X)=\frac{r}{p}\), substitute \(r = 3\) and \(p = 0.44\). So \(E(X)=\frac{3}{0.44}\approx6.82\). Since we can't examine a fraction of a donor, and we need to get 3 successes, we use the formula for the negative - binomial distribution in terms of the number of trials.

Another way: Let \(X\) be the number of donors examined. The probability mass function of a negative - binomial distribution is \(P(X = k)=\binom{k - 1}{r-1}p^{r}(1 - p)^{k - r}\), and the expected value \(E(X)=\frac{r}{p}\)

We know that \(r = 3\) and \(p=0.44\), so \(E(X)=\frac{3}{0.44}\approx6.82\). But if we consider the number of trials \(n\) to get \(r\) successes in a sequence of independent Bernoulli trials with probability of success \(p\), we can also think in terms of the sum of geometric random variables.

A geometric random variable \(Y_i\) (the number of trials to get the \(i\) - th success after the \((i - 1)\) - th success) has \(E(Y_i)=\frac{1}{p}\). For \(r = 3\) successes, \(E(X)=E(Y_1+Y_2+Y_3)\). By the linearity of expectation \(E(X)=\frac{1}{p}+\frac{1}{p}+\frac{1}{p}=\frac{3}{p}\)

Answer:

We expect to examine about \(7\) potential donors.