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Elementary Statistics Supplementary Notes

David W. Lyons

Section 3.4 Probability Problems

Subsection 3.4.1 Probability Practice Problems (Ch 13–15)

Exercises Exercises

Problems 1 and 2 are about the following situation: A box contains seven tickets, labeled \(a,b,c,d,e,f,g\text{.}\) In Game 1, tickets are drawn from the box with replacement. In Game 2, tickets are drawn without replacement. On any given draw, every ticket remaining in the box is equally likely to be selected on the next draw. Consider the following events.
\begin{align*} E \amp= \text{get the ticket } a \text{ on draw 1}\\ F \amp= \text{get the ticket } b \text{ on draw 2}\\ G \amp= \text{get two vowels in the first two draws}\\ H \amp= \text{get exactly two vowels in the first seven draws} \end{align*}
1.
Find the probabilities in the table below. Give your answer as a percent, correct to the nearest \(0.01\) (that is, rounded to the two decimal places). Two entries have been filled in as examples.
\begin{align*} \amp\spacer\amp \amp\spacer\amp \text{Game 1} \amp\spacer\amp \text{Game 2} \\ \amp \amp \amp \amp \rule{.6in}{.1ex} \amp \amp \rule{.6in}{.1ex}\\ (i) \text{ (Ch 13)} \amp\spacer\amp P(E) \amp\spacer\amp \amp\spacer\amp\\ (ii) \text{ (Ch 13)} \amp\spacer\amp P(F) \amp\spacer\amp \amp\spacer\amp\\ (iii) \text{ (Ch 13)} \amp\spacer\amp P(G) \amp\spacer\amp \amp\spacer\amp\\ (iv) \text{ (Ch 13)} \amp\spacer\amp P(F|E) \amp\spacer\amp 1/7 = 14.29\% \amp\spacer\amp 1/6 = 16.67\%\\ (v) \text{ (Ch 13)} \amp\spacer\amp P(E|F) \amp\spacer\amp \amp\spacer\amp \\ (vi) \text{ (Ch 13)} \amp\spacer\amp P(E \AND F) \amp\spacer\amp \amp\spacer\amp\\ (vii) \text{ (Ch 14)} \amp\spacer\amp P(F \OR G) \amp\spacer\amp \amp\spacer\amp\\ (viii) \text{ (Ch 15)} \amp\spacer\amp P(H) \amp\spacer\amp \amp\spacer\amp \end{align*}
Answer.
\begin{align*} \amp\spacer\amp \text{Game 1} \amp\spacer\amp \text{Game 2} \\ \amp \amp \rule{.6in}{.1ex} \amp \amp \rule{.6in}{.1ex}\\ P(E) \amp\spacer\amp 1/7 = 14.29\% \amp\spacer\amp 1/7 = 14.29\%\\ P(F) \amp\spacer\amp 1/7 = 14.29\% \amp\spacer\amp 1/7 = 14.29\%\\ P(G) \amp\spacer\amp (2/7)^2 = 8.16\% \amp\spacer\amp (2/7)\cdot (1/6) = 4.76\%\\ P(F|E) \amp\spacer\amp 1/7 = 14.29\% \amp\spacer\amp 1/6 = 16.67\%\\ P(E|F) \amp\spacer\amp 1/7 = 14.29\% \amp\spacer\amp 1/6 = 16.67\%\\ P(E \AND F) \amp\spacer\amp (1/7)^2 = 2.04\% \amp\spacer\amp (1/7)\cdot (1/6) = 2.38\%\\ P(F \OR G) \amp\spacer\amp (1/7) + (2/7)^2 = 22.45\% \amp\spacer\amp (1/7) + (2/7)\cdot (1/6) = 19.05\%\\ P(H) \amp\spacer\amp {7 \choose 2} (2/7)^2 (5/7)^5 = 31.87\% \amp\spacer\amp 100.00\% \end{align*}
2.
Use the definitions of independent and dependent events given in Subsection 1.1.4 to answer the following questions.
  1. (Ch 13) Are events \(E,F\) independent or dependent in Game 1? In Game 2? Explain.
  2. (Ch 13) Are events \(E,G\) independent or dependent in Game 1? In Game 2? Explain.
  3. (Ch 15) Are events \(G,H\) independent or dependent in Game 1? In Game 2? Explain.
Answer.
  1. In Game 1, we have \(P(F)=1/7=P(F|E)\text{,}\) so \(E,F\) are independent. In Game 2, we have \(P(F)=1/7\neq 1/6=P(F|E)\text{,}\) so \(E,F\) are dependent.
  2. In Game 1, we have \(P(G)=4/49\text{,}\) while \(P(G|E)=2/7\text{.}\) Since \(P(G)\neq P(G|E)\text{,}\) we conclude that \(E,G\) are dependent in Game 1. In Game 2, we have \(P(G)=2/42\text{,}\) while \(P(G|E)=1/6\text{.}\) Since \(P(G)\neq P(G|E)\text{,}\) we conclude that \(E,G\) are dependent in Game 2.
  3. In Game 1, we have \(P(H)={7\choose 2} (2/7)^2 (5/7)^5\approx 31.87\%\text{,}\) while
    \begin{equation*} P(H|G)=P(G)\cdot P(\text{no vowels in 5 draws}) = (2/7)^2(5/7)^5\approx 1.52\% \lt P(H). \end{equation*}
    Since \(P(H)\neq P(H|G)\text{,}\) we conclude that \(G,H\) are dependent in Game 1. In Game 2, we have \(P(H)=100\% =P(H|G)\text{.}\) We conclude that \(G,H\) are independent in Game 2.
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