Math 2000 Solutions 9

Math 2000 - Solutions 9
Spring 2002

Unit 7B 20. The probability of getting at least one tail when tossing ten coins is 1 minus the probability of getting no tails in ten tosses, namely,

$\begin{displaymath} 1-\left(\frac{1}{2}\right)^{10}=0.999. \end{displaymath}$

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, i.e.,

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

b. 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, i.e.,

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

which is hardly surprising--it is a certainty that each person either improved or didn't! c. Be sure you don't double count! There are 193 people who either were given the drug or improved. Therefore, the probability is $\frac {193}{300}=0.643$ . Alternatively, note that these events are overlapping, so the probability is

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

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

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

30a. These events are non-overlapping, so the probability of getting a $2, $5 or $10 winner is the sum of the individual probabilities, or

$\begin{displaymath} \frac{1}{10} + \frac{1}{50} + \frac{1}{500}= \frac{61}{500} = 0.122, \end{displaymath}$

which is a bit greater than the $\frac{1}{10}=0.1$ chance of only a $2 winner. b. The probability of getting at least one $5 winner in 50 tries is

$\begin{displaymath} 1-\left(1-\frac{1}{50}\right)^{50}=0.636. \end{displaymath}$

c. The probability of getting at least one $10 winner in 500 tries is

$\begin{displaymath} 1-\left(1-\frac{1}{500}\right)^{500}=0.632. \end{displaymath}$

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

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

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

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

Unit 7C 4. As noted in 3 above, the probability of tossing three heads in three tosses of a fair coin is $\frac{1}{8}$ , and the probability of not tossing three heads is $\frac{7}{8}$ , so the expected value of the game is

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

You expect to win about 38 cents per game on average. Over 100 games, you should expect to win $100 \times$ 0.38=

. 8. There are four events, each with a probability and value to the company: a policy purchase, a $5000 claim, a $10,000 claim and a $30,000 claim. Thus, the expected value to the insurance company of each policy is

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

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

If the company sells 100,000 policies, its expected profit is $100,000 \times \$150 = \$15,000,000$ . 12. There are ten events here, and the jackpot is worth $100 million. Thus, your expected win for each ticket purchased is

$\begin{displaymath} (-\$1 \times 1) + (\$100,000,000 \times \frac{1}{80,089,128}) + {} \end{displaymath}$

$\begin{displaymath} (\$100,000 \times \frac{1}{1,953,393}) + (\$5000 \times \frac{1}{364,042}) + {} \end{displaymath}$

$\begin{displaymath} + (\$100 \times \frac{1}{8879}) + (\$100 \times \frac{1}{8466}) + {} \end{displaymath}$

$\begin{displaymath} (\$7 \times \frac{1}{207}) + (\$7 \times \frac{1}{605}) + (\$4 \times \frac{1}{188}) + {} \end{displaymath}$

$\begin{displaymath} (\$3 \times \frac{1}{74}) = \$0.4438, \end{displaymath}$

You can expect to win about 44 cents per ticket. Over the course of a year, your expected win is about $365 \times \$0.44 = \$161$ . The positive expected value is due only to the fact that the jackpot is huge!