## Math Series 2

1. In a certain city, 6% of teenagers are married, 25% of married teenagers have children, and 15% of unmaried teenagers have children. If a teenager has a child, what is the probability that the teenager is not married?
1. 0.7099
2. 0.1038
3. 0.8901
4. 0.9038
Solution:
\begin{align} Probability & = \frac{\text{no. of desired outcomes}}{\text{no. of possible outcomes}} \\ & = \frac{\text{no. of desired outcomes}}{\text{desired + undesirable}} \\ \end{align} Conditional Probability

P(B/A) is the probability of B given that A has happened. One event will happen for the other event to possibly happen.

$$P(B/A) = \frac{\text{P(A and B)}}{\text{P(A)}}$$

From the problem,
• 6% are married teenagers
• 94% are unmarried teenagers
• 25% are married teenagers with a child
• 15% are unmarried teenagers with a child

The desired outcome is P(B/A), not married given that they have a child. Let A = has a child, B = not married. \begin{align} P(\text{A and B}) & = \text{(not married)} \times \text{(not married and has a child)} \\ & = (0.94)(0.15) \\ & = 0.141 \\ \\ P(A) & = { \text{(total probability of teenagers with a child)} } \\ & = 0.141 + (0.06)(0.25) \\ & = 0.156 \\ \\ P(B/A) & = \frac{0.141}{0.156} \\ & = 0.9038.\\ \end{align}

2. A box of 30 diodes is known to contain five defective ones. If two diodes are selected at random without replacement, what is the probability that at least one of the diodes is defective?
1. 27/29
2. 20/29
3. 18/29
4. 9/29
Solution:
Mutually Exclusive Events

Two or more events are said to be mutually exclusive if no two of them can possibly happen in the same trial.

$$P(\text{A or B}) = \text{P(A) + P(B) + ...}$$
It can be seen that out of 30 diodes,
• 5 are defective
• 25 are non-defective
In this trial, one of the following scenarios can happen:
• two diodes are defective
• first diode is defective and second diode is not defective
• first diode is not defective and second diode is defective
Let P(D) = defective, P(N) = not defective. \begin{align} P(\text{at least one is defective}) & = P(D)P(D) + P(D)P(N) + P(N)P(D) \\ & = \frac{5}{30} \cdot \frac{5-1}{30-1} + \frac{5}{30} \cdot \frac{25}{30-1} + \frac{25}{30} \cdot \frac{5}{30-1} \\ & = \frac{9}{29} \\ \end{align}

3. An urn contains 4 green balls and 6 blue balls. A second urn contains 16 green balls and N blue balls. A single ball is drawn at random from each urn. The probability that both balls are of the same color is 0.58. Find N.
1. 144
2. 133
3. 121
4. 123
Solution:
One of the following scenarios can happen:
• Green is picked from the first urn and another Green is picked in the second urn.
• Blue is picked from the first urn and another Blue is picked in the second urn.
\begin{align} 0.58 & = \left( \frac{4}{4+6} \right) \left( \frac{16}{16+N} \right) + \left( \frac{6}{4+6} \right) \left( \frac{N}{16+N} \right) \\ N & = 144. \\ \end{align}

4. The probability that a person will show a certain gene-transmitted trait is 0.8 if the father shows the trait and 0.06 if the father does not show the trait. Suppose that 25% of the children in a certain community come from families in which the father shows the trait. Given that a child shows the trait, what is the probability that his father shows the trait?
1. 0.7092
2. 0.8163
3. 0.8099
4. 0.2163
Solution:
From the problem,
• 0.8 = the person shows a trait because the father shows the trait
• 0.06 = the person shows a trait but the father does not show the trait
• 0.25 = father shows the trait
The desired outcome is P(B/A), the father shows the trait given that a child shows the trait.
Let A = child shows the trait, B = father shows the trait. \begin{align} P(\text{A and B}) & = \text{(child shows trait)} \times \text{(father shows trait)} \\ & = (0.8)(0.25) \\ & = 0.2 \\ \\ P(A) & = { \text{(total probability that father shows trait)} } \\ & = 0.2 + (0.06)(1-0.25) \\ & = 0.245 \\ \\ P(B/A) & = \frac{0.2}{0.245} \\ & = 0.8163.\\ \end{align}

5. A bag contains three red marbles, three blue ones, three green ones and two yellow ones, all distinguishable from one another. How many sets of four are there such that each one is a different color?
1. 54
2. 100
3. 87
4. 78
Solution:
If an event can be done in m different ways, and another event can be done in n different ways, then these two things can be done one after the other in: $$N = m \times n$$ \begin{align} N & = \text{(no. of blue)} \times \text{(no. of green)} \times \text{(no. of yellow)} \times \text{(no. of red)}\\ & = 3 \times 3 \times 2 \times 3 \\ & = 54. \\ \end{align}

6. An exam has 50 multiple-choice questions, each having four choices. If a student randomly guesses on each question, how many correct answers can he or she expect to get?
1. 12.5
2. 11.9
3. 10.0
4. 15.6
Solution:
\begin{align} \text{mean} & = \text{no. of trials} \times \text{probability}\\ \mu & = np \\ \end{align}
\begin{align} \mu & = 50 \left( \frac{1}{4} \right) \\ & = 12.50. \\ \end{align}

7. If X has a Poisson distribution such that P(X=2) = 9P(X=4) + 90 P(X=6) then the variance of the distribution is
1. 1
2. 2
3. -1
4. 0
Solution:
Poisson's Distribution - the probability of avarage success per unit time. \begin{align} P & = \frac{ \lambda^x e^{- \lambda}}{x!} \\ \lambda & - \text{mean or variance} \\ x & - \text{no. of successful trials} \\ \end{align}
\begin{align} P(X=2) & = 9P(X=4) + 90 P(X=6) \\ \frac{ \lambda^2 e^{- \lambda}}{2!} & = 9 \left( \frac{ \lambda^4 e^{- \lambda}}{4!} \right) + 90 \left( \frac{ \lambda^6 e^{- \lambda}}{6!} \right) \\ \frac{ \lambda^2 }{2!} & = \frac{9 \lambda^4}{4!} + \frac{90 \lambda^6}{6!} \\ \lambda & = 1.\\ \end{align}

8. The odds that a Ph.D. thesis will be favorably reviewed by three independent examiners are 5 to 2, 4 to 3, and 3 to 4. What is the probability that a majority approve the thesis?
1. 149/343
2. 69/343
3. 209/343
4. 309/343
Solution:
\begin{align} \text{Odds} & = \frac{\text{desirable}}{\text{undesirable}} \\ \\ \text{Probability} & = \frac{\text{desirable}}{\text{possible}} \\ & = \frac{\text{desirable}}{\text{desirable} + \text{undesirable}} \\ \end{align}

Examiner Odds Probability
A $\frac{5}{2}$ $\frac{5}{5+2} = \frac{5}{7}$
B $\frac{4}{3}$ $\frac{4}{4+3} = \frac{4}{7}$
C $\frac{3}{4}$ $\frac{3}{3+4} = \frac{3}{7}$
One of the following scenarios can happen:
• All three examiners approves
• Examiner A & B approves
• Examiner A & C approves
• Examiner B & C approves
\begin{align} P & = \left( \frac{5}{7} \right)\left( \frac{4}{7} \right)\left( \frac{3}{7} \right) + \left( \frac{5}{7} \right)\left( \frac{4}{7} \right)\left( 1 - \frac{3}{7} \right) \\ & + \left( \frac{5}{7} \right)\left( 1 - \frac{4}{7} \right)\left( \frac{3}{7} \right) + \left( 1- \frac{5}{7} \right)\left( \frac{4}{7} \right)\left( \frac{3}{7} \right) \\ & = \frac{209}{343}. \\ \end{align}

9. If X is a binomial variate with p = 1/5, for the experiment of 50 trials, then the standard deviation is equal to
1. 6
2. 8
3. -8
4. 2 sqrt(2)
Solution:
Variance
\begin{align} \sigma^2 & = npq \\ \sigma & - \text{standard deviation} \\ n & - \text{no. of trials} \\ p & - \text{probability of success} \\ q & - \text{probability of failure} \\ \end{align}
\begin{align} \sigma^2 & = 50 \left( \frac{1}{5} \right) \left( 1 - \frac{1}{5} \right) \\ \sigma & = \sqrt{8} \\ & = 2 \sqrt{2}. \\ \end{align}

10-13. Refer to the data set below, the number of patient visits per week at a chiropractor's office over a ten-week period.
Number of patients per ten-week
75 86 87 90 94 102 105 109 110 120

10. Calculate the first quartile of the data using index method.
1. 88
2. 90
3. 75
4. 87
Solution:
Index Method
$$z^{\text{th}} \text{ percentile} = \left( \frac{\text{ percentile}}{100} \times \text{total no. of terms} \right)^{\text{th}} \text{ term}$$
• If answer is with decimal, round up. $$\Rightarrow (n+1)^{th} \text{ term}$$
• If answer is a whole number: $$\Rightarrow \frac{ n^{th} \text{ term} + (n+1)^{th} \text{ term} }{2}$$
\begin{align} 1^{\text{st}} \text{quartile} & = 25^{\text{th}} \text{ percentile} \\ & \Rightarrow \frac{25}{100} \times 10 \\ & \Rightarrow 2.5 \\ & \approx 3 \\ 1^{\text{st}} \text{ quartile} & = 3^{\text{rd}} \text{ term} \\ & = 87. \end{align}

11. Calculate the second quartile of the data using index method.
1. 94
2. 109
3. 105
4. 98
Solution:
\begin{align} 2^{\text{nd}} \text{ quartile} & = 50^{\text{th}} \text{ percentile} \\ & \Rightarrow \frac{50}{100} \times 10 \\ & \Rightarrow 5 \\ 2^{\text{nd}} \text{ quartile} & = \frac{ 5^{\text{th}} \text{ term} + 6^{\text{th}} \text{ term} }{2} \\ & = \frac{94+102}{2} \\ & = 98. \end{align}

12. Calculate the third quartile of the data using index method.
1. 104
2. 87
3. 98
4. 109
Solution:
\begin{align} 3^{\text{rd}} \text{ quartile} & = 75^{\text{th}} \text{ percentile} \\ & \Rightarrow \frac{75}{100} \times 10 \\ & \Rightarrow 7.5 \\ & \approx 8 \\ 3^{\text{rd}} \text{ quartile} & = 8^{\text{th}} \text{ term} \\ & = 109. \end{align}

13. Calculate the interquartile range (IQR) of the data
1. 13
2. 21
3. 24
4. 22
Solution:
Interquartile Range (IQR)
$$\text{IQR} = 3^{\text{rd}} \text{ quartile} - 1^{\text{st}} \text{ quartile}$$
From the results above, \begin{align} IQR & = 109 - 87\\ & = 22. \\ \end{align}

14. A statistics professor administered an exam to three different classes. The 8:00 class had 20 students and an average exam score of 82.4. The 9:00 class had 24 students and an average of 78.9. The 29 students in the 10:00 class averaged 87.3. Calculate the overall average for all three classes.
1. 87.6
2. 88.2
3. 83.2
4. 84.5
$$\text{mean} = \frac{ \sum \text{ of all terms} }{ \text{total no. of terms} }$$
From the 8:00 class, \begin{align} 82.4 & = \frac{a}{20} \\ a & = 20(82.4) \\ \end{align} From 9:00 class, \begin{align} 78.9 & = \frac{b}{24} \\ b & = 24(78.9) \\ \end{align} From 10:00 class, \begin{align} 87.3 & = \frac{c}{29} \\ c & = 29(87.3) \\ \end{align} Solving for the overall average, \begin{align} \text{mean} & = \frac{a+b+c}{20+24+29} \\ & = 83.2. \\ \end{align}