Love Stinks? J. Fetto, in the article "Love Stinks" (American Demographics, Vol. 25. No. 1. pp. 10-11), reports that Americans split with their significant other for many reasons-including indiscretion, infidelity, and simply "growing apart." According to the article, 35% of American adults have experienced a breakup at least once during the last 10 years. Of nine randomly selected American adults, find the probability that the number, X, who have experienced a breakup at least once during the last 10 years is

(a) exactly five; at most five; at least five.

(b) at least one; at most one.

(c) between six and eight, inclusive.

(d) Determine the probability distribution of the random variable X.

(e) Strictly speaking, why is the probability distribution that you obtained in part (d) only approximately correct? What is the exact distribution called?

The given statement is:

In the last ten years, 35 percent of American adults have had at least one breakup.

Therefore, $p=0.35$.

Nine American adults were chosen at random.

$n=9$

The probability that the chosen number is exactly five is:

$$\begin{gathered} P\left(X=x\right)=\left(\begin{array}{c}n\\ x\end{array}\right)p^x{\left(1-p\right)}^{n-x} \\ P\left(X=5\right)=\left(\begin{array}{c}9\\ 5\end{array}\right){\left(0.35\right)}^5{\left(1-0.35\right)}^{9-5} \\ =0.1181 \end{gathered}$$

The probability that the chosen number is at most five is:

$$\begin{gathered} P\left(X\le 5\right)=P\left(X=0\right)+P\left(X=1\right)+........+P\left(X=5\right) \\ =\left(\begin{array}{c}9\\ 0\end{array}\right){\left(0.35\right)}^0{\left(1-0.35\right)}^{9-0}+\left(\begin{array}{c}9\\ 1\end{array}\right){\left(0.35\right)}^1{\left(1-0.35\right)}^{9-1}+......+\left(\begin{array}{c}9\\ 5\end{array}\right){\left(0.35\right)}^5{\left(1-0.35\right)}^{9-5} \\ =0.0207+0.1004+0.2162+0.2716+0.2194+0.1181 \\ =0.9464 \end{gathered}$$

The probability that the chosen number is at least five is:

$$\begin{gathered} P\left(X\ge 5\right)=1-P\left(X<5\right) \\ =1-\left[P\left(X=0\right)+P\left(X=1\right)+P\left(X=2\right)+P\left(X=3\right)+P\left(X=4\right)\right] \\ =1-\left[0.0207+0.1004+0.2162+0.2716+0.2194\right] \\ =1-0.8283 \\ =0.1717 \end{gathered}$$

The probability that the chosen number is at least one is:

$$\begin{gathered} P\left(X\le 1\right)=P\left(X=0\right)+P\left(X=1\right) \\ =\left(\begin{array}{c}9\\ 0\end{array}\right){\left(0.35\right)}^0{\left(1-0.35\right)}^{9-0}+\left(\begin{array}{c}9\\ 1\end{array}\right){\left(0.35\right)}^1{\left(1-0.35\right)}^{9-1} \\ =0.0207+0.1004 \\ =0.1211 \end{gathered}$$

The probability that the chosen number is at most one is:

$$\begin{gathered} P\left(X\ge 1\right)=1-P\left(X=0\right) \\ =1-0.0207 \\ =0.9793 \end{gathered}$$

$$\begin{gathered} P\left(6\le X\ge 8\right)=P\left(X=6\right)+P\left(X=7\right)+P\left(X=8\right) \\ =0.0424+0.0098+0.0013 \\ =0.0535 \end{gathered}$$

The random variable X has the following probability distribution:

Because the sampling from a finite population is done without replacement and the number of successes follows a hypergeometric distribution.

The hypergeometric distribution is similar to the binomial distribution since the sample is less than 5% of the population.