Unit 2 Probability

Review

  1. On an ACT or SAT test, a typical multiple-choice question has 5 possible answers. If you make a random guess on one such question, what is the probability that your response is wrong?

    \( P(\text {wrong})=\frac{4}{5}=0.8\)

  2. The General Motors Corporation wants to conduct a test of a new model of Corvette. A pool of 200 drivers has been recruited, 80 of whom are men. When the first person is selected from this pool, what is the probability of not getting a male driver?

    \(\begin{equation} P(\text { NOT male })=\frac{120}{200}=\frac{3}{5}=0.6 \end{equation}\)

  3. Two psychologists surveyed 478 elementary children in Michigan. They asked students whether their primary goal was to get good grades, to be popular, or to be good at sports. Below is a contingency table giving counts of the students by their goals and demographic setting of their school: rural, suburban, or urban.. Complete the table and answer the questions.
    Grades Most Important Popularity Most Important Sports Most Important Total Number
    Rural 57 50 42

    149
    Suburban 87 42 22

    151
    Urban 103 49 26

    178
    Total Number

    247

    141

    90

    478

    For one randomly selected student, find:

    1. P(Rural) = \(\begin{equation} \frac{149}{478} \approx 0.3117 \end{equation}\)
    2. P(Sports) = \(\begin{equation} \frac{90}{478} \approx 0.1883 \end{equation}\)
    3. P(Suburban and Popularity) = \(\begin{equation} \frac{42}{478} \approx 0.0879 \end{equation}\)
    4. P(Suburban or Popularity) = \(\begin{equation} \frac{250}{478} \approx 0.5230 \end{equation}\)

      P(Suburban) + P(Popularity) - P(Suburban AND Popularity) \(\begin{equation} =\frac{151}{478}+\frac{141}{478}-\frac{42}{478}=\frac{250}{478} \approx 0.5230 \end{equation}\)

    5. P(Popularity | Rural) = \(\frac{50}{149} \approx 0.3356\)
    6. P(Rural | Grades) = \(\frac{57}{247} \approx 0.2308\)

    4.  

  4. According to cancer.org, there is about a 25% success rate for those who try to stop smoking through medication alone. Find the probability that for 8 randomly selected smokers who use medication, they all successfully quit smoking.

    \((.25)^{8}=0.000015\)

  5. If Knoxville has an annual robbery rate of 0.23%, find the probability that among 3 randomly selected residents, all have been robbed during a given year. (The population of Knoxville is about 183,000.)

    Data from http://www.neighborhoodscout.com/tn/knoxville/crime/

    \((.0023)^{3}=0.000000012\)

  6. Find the probability of drawing a hand of cards that is 2 Kings 3 Aces (not replacing the cards) from a standard deck.

    \(\left(\frac{4}{52}\right)\left(\frac{3}{51}\right)\left(\frac{4}{50}\right)\left(\frac{3}{49}\right)\left(\frac{2}{48}\right)=0.00000092\)

  7. You roll two dice three times. Find the probability of rolling a seven all three times.

    \(P(7)=\frac{6}{36}=\frac{1}{6} ;\left(\frac{1}{6}\right)\left(\frac{1}{6}\right)\left(\frac{1}{6}\right)=\frac{1}{216}\)

  8. Below is a table showing the number of people with various levels of education in 5 countries.
    Post-graduate Some College Some High School Primary or Less No Answer Total Number
    China 7 315 671 506 3 1502
    France 69 388 766 309 7 1539
    India 161 514 622 227 11 1535
    U.K. 58 207 1240 32 20 1557
    USA 84 486 896 87 4 1557
    Total Number 379 1910 4195 1161 45 7690

    If we select someone randomly from the survey, what is the probability the person…

    1. is from the US? \(\frac{1557}{7690}=0.2025\)
    2. completed his or her education before college? Disjoint

      P(some HS OR Primary or Less) \(=\frac{4195}{7690}+\frac{1161}{7690}-\frac{0}{7690}=\frac{5356}{7690}=0.6965\)

    3. is from France or did some post-graduate study?

      P(France) + P(Post Grad) - P(France AND Post Grad) = \(\frac{1539}{7690}+\frac{379}{7690}-\frac{69}{7690}=\frac{1849}{7690}=.2404\)

  9. If I roll a regular fair die, what is the probability of rolling an even number or a number greater than 2?

    \(P(\operatorname{even} \text{ OR}>2)=\frac{3}{6}+\frac{4}{6}-\frac{2}{6}=\frac{5}{6}\)

    1. What is the probability of randomly selecting a 6 or a club out of a deck of cards?

      P(6 OR club) = \(\frac{4}{52}+\frac{13}{52}-\frac{1}{52}=\frac{16}{52}=\frac{4}{13}\)

    2. What is the probability of randomly selecting a queen or a 4 out of a deck of cards?

      P(Q OR 4) = \(\frac{4}{52}+\frac{4}{52}=\frac{8}{52}=\frac{2}{13}\)

    3. What is the probability of randomly selecting a face card or a diamond out of a deck of cards?

      P(Face OR Diamond)\(=\frac{12}{52}+\frac{13}{52}-\frac{3}{52}=\frac{22}{52}=\frac{11}{26}\)

  11. Complete the statement: \(P(A)+P(\overline{A})=\) 1
  12. What is the probability of randomly selecting a red pen OR a black pen from a box containing 6 red pens, 3 black pens, and 2 blue pens?

    \(P(\text {red OR black})=\frac{6}{11}+\frac{3}{11}=\frac{9}{11}\)

  13. Eye Color: Groups of 5 babies are randomly selected. In each group, the random variable x is the number of babies with green eyes.
    x values Probability P(x)
    0 0.528
    1 0.360
    2 0.098
    3 0.013
    4 0.001
    5 0+
    1. Does it meet the requirement for a probability distribution? Explain

      \(\mathrm{Yes.} \sum P(x)=1 \text { all } P(x) \leq 1 \text{ and } \geq 0\)

    2. Expected Value: E(x) = .599          Mean \(\mu= \) .599
    3. Variance:  \(\sigma^{2}= \) .5262          Standadard Deviation \(\sigma = \) \(\sqrt{.5262}=.7254\)
    4. Minimum Usual Value: -.8518      Maximum Usual Value: 2.049
    5. Would it be unusual to have 0 babies out of 5 randomly selected babies with green eyes?

      No. 0 is between max and min of usual values (-.8518, 2.049)

    6. Would it be unusual to have 2 babies out of 5 randomly selected babies with green eyes?

      No. 2 is between (-.8518, 2.049) . Also, P(x = 2) = .098 is not less than .05.

      7. Find the following probabilities (assuming 5 randomly selected babies)

    7. Probability of exactly 2 green eyed babies: \(P(x=2)=.098\)
    8. Probability of 3 or more green eyed babies: \(P(x \geq 3)=.013+.001+0^{+}=.014\)
    9. Probability of at least 1 green eyed baby: \(P(x \geq 1)=1-P(x=0)=1-.528=.472\)
    10. Probability of at most 1 green eyed babies: \(P(x \leq 1)=P(x=0)+P(x=1)=.528+.360=.888\)
    11. Probability of no more than 1 green eyed babies: \(P(x \leq 1)=.888\)
  14. Create a probability distribution for the roll of one die, where the random variable is the value you roll. Find the mean and the standard deviation for the probability distribution.
    x value Probability P(x)

    1

    \(\frac{1}{6}\)

    2

    \(\frac{1}{6}\)

    3

    \(\frac{1}{6}\)

    4

    \(\frac{1}{6}\)

    5

    \(\frac{1}{6}\)

    6

    \(\frac{1}{6}\)

    Mean   \(\mu = \) 3.5     Standard Deviation   \(\sigma = \) 1.7

  15. Find the probability of getting exactly 3 correct responses among 5 different requests from AT&T directory assistance. Assume in general, AT&T is correct 90% of the time.

    n = 5     x = 3     p = 0.9     q = 0.1

    Use the binomial probability formula to solve:

    \(P(x)=\frac{5 !}{2 ! 3 !}(0.9)^{3}(0.1)^{2}=0.073\)

  16. A slot machine has a 1/2000 probability of winning the jackpot on any individual trial. Suppose someone plays the slot machine 5 times.
    1. Make a binomial probability distribution for the situation where x is defined as the number of wins

      \(n=5\)       \(p=\frac{1}{2000}=.0005\)

      x value Probability P(x)

      0

      .9975

      1

      .0025

      2

      .0000025

      3

      \(1.2 \times 10^{-9}\)

      4

      \(3 \times 10^{-13}\)

      5

      \(3 \times 10^{-17}\)
    2. What is the probability of hitting the jackpot exactly twice? Is it likely?

      .0000025, not likely

    3. What is the probability of hitting it 3 or fewer times? Is that likely?

      \(\approx 1,\) yes, it is likely

  17. Your brother baked a large batch of cookies. He put chocolate chips in 45% of the cookies. He randomly selects 10 cookies to give to a friend. What is the probability that 6 of the cookies contain chocolate chips?

    \(n : 10, p : 0.45 \quad P(x=6)=0.1595678\)

  18. How would you work each of the following problems using Technology?

    60% of the class likes chocolate. If 20 students are randomly selected, find the probability: n = 20      p = 0.60

    Words Probability Statement Calculated Probability Unusual (Yes or No)
    a. 4 like chocolate

    \(P(x=4)\)

    0.0003

    Yes
    b. No more than 5 like chocolate

    \(P(x \leq 5)\)

    0.0016

    Yes
    c. 7 or more like chocolate

    \(P(x \geq 7)\)

    0.9935

    No
    d. More than 7 like chocolate

    \(P(x>7)\)

    0.9790

    No
    e. Less than 13 like chocolate

    \(P(x<13)\)

    0.5841

    No
  19. What is the mean and standard deviation for the probability distribution (of 20 randomly selected students from a class in which 60% of the students like chocolate)?

    Mean: 12      Standard Deviation: 2.2