Math 2000 - Solutions 10
Fall 2001
Unit 7A 4. Each possible hour of the day represents an outcome, so there are 24 possible outcomes. If we assume that these are equally likely outcomes, then the probability of meeting somebody born between during any one-hour interval, say, midnight and 1:00 am is $\frac{1}{24}=0.042$. 8. There are 45 possible outcomes (there being 45 M&M's). Assuming these are equally likely outcomes, 15 of these (the blue ones) are of interest, so the probability of selecting a blue M&M from the bag is ${\frac{15}{45}}=0.333$. 18. The probability of tossing two heads with two fair coins is ${\frac{1}{4}}$, and so the probability of not tossing two heads is $1-{\frac{1}{4}}= 0.75$. 22. If we assume that there are only two options - a sunny day or a rainy day - then the probability of a sunny day is $1-0.20=0.80$. 24. There are 36 equally likely outcomes (see table on page 413). Rolling a double-6 corresponds to just one of these outcomes, namely (6,6), and so the probability of rolling a double-6 is $\frac{1}{36}$, and the probability of not rolling a double-6 is $1-\frac{1}{36}=\frac{35}{36}=0.9722$. 26. In 2000, there were 34.7 million people over 65 years of age out of 274 million people total, so the chances of a randomly encountered person being over 65 were $\frac{34.7}{274}=0.13$. The corresponding population projections for 2050 lead to a probability of $\frac{78.9}{394}=0.20$; your chances would be much greater in 2050. Unit 7B 4. This is an AND problem and these events are independent. The probability that anyone has a telephone number ending in 1 is $\frac{1}{10}$ because there are ten possible last digits. The probability that all five of your best friends have telephone numbers ending in 1 is $(\frac{1}{10})^5=10^{-5}$. The probability that five friends all have a telephone number ending in 1 is 1 in 100,000. 8. This is an AND problem and these events are dependent. The probability that all three people selected are Americans is the probability that the first one is an American, times the probability that the second one is an American given that the first one is an American, times the probability that the third one is an American given that the first two are Americans, that is,

\begin{displaymath} \frac{9}{21} \times \frac{8}{20} \times \frac{7}{19} = 0.0632. \end{displaymath}

12. These events are non-overlapping, and the probability of getting a sum of 2, 3 or 4 is the sum of the individual probabilities, which were found in Example 8 in Unit 7A in the text. We get $\frac{1}{36} + \frac{2}{36} + \frac{3}{36} = \frac{1}{6} = 0.167$. $\textstyle \parbox{4in}{14. These events are overlapping. We can make a table s... ...\frac{20+25}{90} + \frac{25+20}{90} - \frac{25}{90} = \frac{13}{18} = 0.722$.}$ $\textstyle \parbox{2in}{ \begin{tabular}{\vert c\vert c\vert c\vert}\hline & W... ...hline American & 25 & 20 \\ \hline English & 20 & 25 \\ \hline \end{tabular}}$