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PROBLEM SOLVING Of 162 students honored at an academic awards banquet, 48 won awards for mathematics and 78 won awards for English. There are 14 students who won awards for both mathematics and English. A newspaper chooses a student at random for an interview. What is the probability that the student interviewed won an award for English or mathematics?

In a survey, 49 people received a flu vaccine before the flu season and 63 people did not receive the vaccine. Of those who receive the flu vaccine, 16 people got the flu. Of those who did not receive the vaccine, 17 got the flu. Make a two-way table that shows the joint and marginal relative frequencies.

Determine whether the events are independent. (See Examples I and 2.) A vase contains four white roses and one red rose. You randomly select two roses to take home. Use a sample space to determine whether randomly selecting a white rose first and randomly selecting a white rose second are independent events.

A student randomly draws a number between 1 and 30. Describe and correct the error in finding the probability that the number drawn is greater than 4. The probability that the number is less than 4 is \(\frac{3}{30}\), or \(\frac{1}{10}\). So, the probability that the number is greater than 4 is \(1-\frac{1}{10}\), or \(\frac{9}{10}\)

PROBLEM SOLVING You play a game that involves drawing three numbers from a hat. There are 25 pieces of paper numbered from 1 to 25 in the hat. Each number is replaced after it is drawn. Find the probability that you will draw the 3 on your first draw and a number greater than 10 on your second draw.

Three different local hospitals in New York surveyed their patients. The survey asked whether the patient's physician communicated efficiently. The results, given as joint relative frequencies, are shown in the two-way table. $$ \begin{array}{|l|c|c|c|} \hline & \text { Glens Falls } & \text { Saratoga } & \text { Albany } \\ \hline \text { Yes } & 0.123 & 0.288 & 0.338 \\ \hline \text { No } & 0.042 & 0.077 & 0.131 \\ \hline \end{array} $$ a. What is the probability that a randomly selected patient located in Saratoga was satisfied with the communication of the physician? b. What is the probability that a randomly selected patient who was not satisfied with the physician's communication is located in Glens Falls? c. Determine whether being satisfied with the communication of the physician and living in Saratoga are independent events.

PROBLEM SOLVING A drawer contains 12 white socks and 8 black socks. You randomly choose 1 sock and do not replace it. Then you randomly choose another sock. Find the probability that both events \(A\) and \(B\) will occur. (See Example 4.) Event \(A\) : The first sock is white. Event \(\boldsymbol{B}\) : The second sock is white.

A researcher surveys a random sample of high school students in seven states. The survey asks whether students plan to stay in their home state after graduation. The results, given as joint relative frequencies, are shown in the two-way table. $$ \begin{array}{|l|c|c|c|} \hline & \text { Nebraska } & \begin{array}{c} \text { North } \\ \text { Carolina } \end{array} & \begin{array}{c} \text { Other } \\ \text { States } \end{array} \\ \hline \text { Yes } & 0.044 & 0.051 & 0.056 \\ \hline \text { No } & 0.400 & 0.193 & 0.256 \\ \hline \end{array} $$ a. What is the probability that a randomly selected student who lives in Nebraska plans to stay in his or her home state after graduation? b. What is the probability that a randomly selected student who does not plan to stay in his or her home state after graduation lives in North Carolina? c. Determine whether planning to stay in their home state and living in Nebraska are independent events.

ERROR ANALYSIS Events \(A\) and \(B\) are independent. Describe and correct the error in finding \(P(A\) and \(B)\). $$ \begin{aligned} &P(A)=0.6 \quad P(B)=0.2 \\ &P(A \text { and } B)=0.6+0.2=0.8 \end{aligned} $$

In Exercises 15 and 16, describe and correct the error in fi nding the given conditional probability. $$ \begin{array}{|l|c|c|c|c|} \hline & \text { Tokyo } & \text { London } & \begin{array}{c} \text { Washington, } \\ \text { D.C. } \end{array} & {\text { Total }} \\ \hline \text { Yes } & 0.049 & 0.136 & 0.171 & 0.356 \\ \hline \text { No } & 0.341 & 0.112 & 0.191 & 0.644 \\ \hline \text { Total } & 0.39 & 0.248 & 0.362 & 1 \\ \hline \end{array} $$ \(\begin{aligned} P(\text { London } \mid n o) &=\frac{P(\text { no and London })}{P(\text { London })} \\ &=\frac{0.112}{0.248} \approx 0.452 \end{aligned}\)