Math 2000 Solutions 11

Math 2000 - Solutions 11
Fall 2001

Unit 7B 22. The probability of getting rain at least once in ten days is 1 minus the probability of not getting rain on each of the ten days, namely,

$\begin{displaymath} 1-(1-0.1)^{10}=0.651. \end{displaymath}$

26a. These events are non-overlapping, and the probability that a randomly selected person in the study was given either the drug or a placebo is the sum of the individual probabilities:

$\begin{displaymath} \frac{120}{300} + \frac{100}{300}= \frac{11}{15} = 0.733. \end{displaymath}$

26b. These events are non-overlapping, and the probability that a randomly selected person in the study either improved or did not improve is the sum of the individual probabilities:

$\begin{displaymath} \frac{138}{300} + \frac{162}{300}= 1, \end{displaymath}$

which is hardly surprising--it is a certainty that each person either improved or didn't! 26c. These events are overlapping. One way to find the probability is to count the number of people who were either given the drug or improved. We see that 55+65+42+31 = 193 people meet these conditions. Thus, the probability that a randomly selected person in the study either was given the drug or improved is $\frac{193}{300} =0.642$ . Alternatively, the probability is the sum of the individual probabilities minus the probability that the person was given the drug and improved:

$\begin{displaymath} \frac{120}{300} + \frac{138}{300} - \frac{65}{300}= \frac{193}{300} = 0.643. \end{displaymath}$

26d. As seen in (c) above, the probability that a randomly selected person was given the drug and improved is

$\begin{displaymath} \frac{65}{300}= \frac{13}{60} = 0.217. \end{displaymath}$

32a. The probability that at least one of 10 sexual partners is infected with HIV is

$\begin{displaymath} 1-(1-0.02)^{10}=0.183. \end{displaymath}$

32b. The probability that at least one of 20 sexual partners is infected with HIV is

$\begin{displaymath} 1-(1-0.02)^{20}=0.332. \end{displaymath}$

Unit 7C 4. The probability of tossing three heads in three tosses of a fair coin is $\frac 12 \times \frac 12 \times \frac 12 = \frac{1}{8}$ , and the probability of not tossing three heads is $1-\frac 18=\frac{7}{8}$ , so the expected value of the game is

$\begin{displaymath} (\$10 \times \frac{1}{8}) + (-\$1 \times \frac{7}{8}) = \$0.375, \end{displaymath}$

that is, you expect to win about 38 cents per game on average. 8. There are four events, each with a probability and value to the company: a policy purchase (which occurs with probability 1), a $5000 claim, a $10,000 claim and a $30,000 claim. Thus, the expected value to the insurance company of each policy is

$\begin{displaymath} (\$500 \times 1) + (-\$5000 \times \frac{1}{50}) + {} \end{displaymath}$

$\begin{displaymath} (-\$10,000 \times \frac{1}{100}) + (-\$30,000 \times \frac{1}{200})= \$150. \end{displaymath}$

The company can expect to gain $150 per policy. If the company sells 100,000 policies, its expected profit is $100,000 \times \$150 = \$15,000,000$ . 14. There are ten events here, only this time the jackpot is worth $25 million. Note that with probability 1, a $1 ticket must be bought. Thus, your expected win for each ticket purchased is

$\begin{displaymath} (-\$1 \times 1) + (\$25,000,000 \times \frac{1}{76,275,360}) + {} \end{displaymath}$

$\begin{displaymath} (\$150,000 \times \frac{1}{2,179,296}) + (\$5000 \times \frac{1}{339,002}) + {} \end{displaymath}$

$\begin{displaymath} (\$150 \times \frac{1}{9686}) + (\$100 \times \frac{1}{7705}) + {} \end{displaymath}$

$\begin{displaymath} (\$5 \times \frac{1}{220}) + (\$5 \times \frac{1}{538}) + (\$2 \times \frac{1}{102}) + {} \end{displaymath}$

$\begin{displaymath} (\$1 \times \frac{1}{62}) = -\$0.4924, \end{displaymath}$

that is, you expect to lose about 49 cents per ticket. Over the course of a year, your expected loss is about $365 \times \$0.49 = \$179$ . 16. If we interpret the expected value of American ages to mean the expected value of the age of a randomly selected American, then we can argue that according to the given categories, there are six possible events. The first is that the person selected is under 14, with value 6.5 years, which happens with probability 0.200, since 20% of people are in that age group; the second is that the person is between 14 and 24 years of age, with value 19 years (the midpoint of the age category), which happens with probability 0.153; and so on. The last corresponds to a value of 75 years and happens with probability 0.126. Hence, the expected value of American ages is

$\begin{displaymath} (6.5 \times 0.200) + (19 \times 0.153) + (29.5 \times 0.136) + {} \end{displaymath}$

$\begin{displaymath} (39.5 \times 0.163) + (54.5 \times 0.222) + (75 \times 0.126), \end{displaymath}$

which comes out to 36.21 years.