Math 2000 - Solutions 8
Spring 2002
Unit 7A 6. Each possibility {BBB, BBG, BGB ,BGG, GBB, GBG, GGB, GGG} represents an outcome; for example, BGB denotes a boy followed by a girl followed by a boy. We see that there are 8 possible outcomes. If we assume that these are equally likely outcomes, then since 3 of these--namely, BGG, GBG, GGB--represents an event of interest, the probability of a randomly selected three-child family having exactly two girls is $\frac{3}{8}=0.375$. 8. There are 45 possible outcomes (as there are 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 ${\frac{15}{45}}=0.333$. 20. There are 12 possible outcomes here, and ignoring the fact that not all months have the same length, we will assume that these are equally likely outcomes. The probability of meeting a person born in January, February or March is ${\frac{3}{12}}$, and so the probability of that person not being born in one of those three months is $1-{\frac{3}{12}}=\frac{9}{12}=0.75$. 24. There are 36 equally likely outcomes, {(1,1),(1,2),...,(1,6),(2,1),(2,2), ...,(6,6)}. 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. Senior Citizen. 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 2. These events are independent. The probability that all four rolls are 6s is the product of the probabilities that any one roll is a 6, or $\frac{1}{6} \times \frac{1}{6} \times \frac{1}{6} \times \frac{1}{6} = 0.000772$. 6. These events are independent. The probability that all five of the next births are girls is the probability that the first one is, times the probability that the second one is, times the probability that the third one is, times the probability that the fourth one is, times the probability that the fifth one is, or

\begin{displaymath} \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = 0.03125. \end{displaymath}

14. These events are overlapping. Using the formula given in the text, the probability of selecting either a woman or an American is the sum of the individual probabilities minus the probability of selecting an American woman: $\frac{20+25}{90} + \frac{25+20}{90} - \frac{25}{90} = \frac{13}{18} = 0.722$. You can also make a table to summarize the data. You find that there are 65 people who are either American or women, out of 90 people. Therefore, the probability of selecting either an American or women is $frac{65}{90}=\frac{13}{18} = 0.722$ 16. These events are non-overlapping, and the probability of drawing either an ace or a king is the sum of the individual probabilities, or $\frac{4}{52} + \frac{4}{52} = \frac{2}{13} = 0.154$.