EX 3.37

• A pediatrician aims to recruit 5 couples for a childbirth regimen.
  • Each couple is expecting their first child.
• Let:
  • $p=P$: the probability a randomly selected couple agrees to participate.
  • Given: $p=0.2$.
• We want to find the probability that:
  • 15 couples must be asked before 5 agree to participate.
  • This means we observe 10 failures (F) before the fifth success (S).
• Denote $S=\{\text{agrees to participate}\}$.
• Relevant variables:
  • $r=5$: number of successes needed.
  • $p=0.2$: probability of success.
  • $x=10$: number of failures before the fifth success.
• Substitute these into the negative binomial formula:
  • $nb(x;r;p)$ gives the probability of this scenario. $nb\left(10;5,.2\right)=\left(\begin{array}{c}14\\ 4\end{array}\right){\left(.2\right)}^5{\left(.8\right)}^{10}=.034$

The probability that at most $10\mathrm{F}$ ‘s are observed (at most 15 couples are asked) is

$$\begin{array}{rl}P(X\le 10)& =\sum _{x=0}^{10}nb(x;5,0.2)\\ & =(0.2{)}^5\sum _{x=0}^{10}(\genfrac{}{}{0ex}{}{x+4}{4})(0.8{)}^x\\ & =0.164\end{array}$$