Hands-on math!

Washington State Math Olympiad
Hints and Solutions
2008 Grade 8 Probability and Statistics
Problem
Solution
1) There are two coins in a bag, one that has heads on both sides and one that is weighted so that heads comes up twice as often as tails. What are the odds of randomly selecting a coin out of the bag, flipping it, and having it land on tails?
  1. The probability of getting the coin that has a tails is 1⁄2
  2. If you get the coin that has a tails, then the probability that you will get the tails is 1⁄3
  3. The combined probability is 1⁄2 x 1⁄3 = 1⁄6
2) A data set consists of seven different positive whole numbers. The mean of the data is 69. What is the largest the median of the data could be?
  1. To maximize the median, the one in the middle that has 3 numbers lower than it and 3 higher, you must minimize the numbers below it.
  2. The lowest they can be is 1,2 and 3, making the list:
    1 2 3 X X X X
  3. The 4 higher numbers must sum to 69x7 - 6 = 477
  4. The 4 higher numbers must be n, n+1, n+2 and n+3 and they must add to 477
  5. 4n + 6 = 477
  6. 4n = 471, so n = 117.75
  7. Since we can't have a decimal, n = 117

Problem
Solution
3) Joan and David invent a probability game for a homework project that has players flip a fair coin and roll a fair 6 sided number cube with the number 1 - 6 on it and move pieces on a board with 50 spaces on it. David moves ahead on the game board as many spaces as the number that comes up on the number cube when the coin lands on a heads. Joan moves ahead three spaces each time the coin lands on a tail no matter what number comes up on the cube. After two roll and flip combinations who is more likely to have moved at least 6 spaces and with what probability?


Table for the sum of 2 dice:
  1. Joan's possibilities are
    (H H)
    (H T)
    (T H)
    (T T)
    The only combination that gives her a move of 6 spaces is (T T) which has a 1⁄4 probability.

  2. David has the following possibilities:
      (H H)  2 rolls of the die (see the figure)
    which gives him a 26/36 chance of rolling a sum of 6
    (H T)1 roll of the die
    which gives him a 1/6 chance of rolling a 6
    (T H)1 roll of the die
    which gives him a 1/6 chance of rolling a 6
    (T T) No points here!
    Each of these outcomes has a 1⁄4 probability of occurring, so, David's probability is the sum of these probabilities:
    (1⁄4 x 26⁄36) + (1⁄4 x 1⁄6) + (1⁄4 x 1⁄6) =
    (1⁄4 x 13⁄18) + 1⁄24 + 1⁄24 =
    13⁄72 + 1⁄12 = 13⁄72 + 6⁄72 = 19⁄72
  3. David has the higher probability.

Problem
Solution
4) Most Washington license plates consist of a set of three numbers followed by a set of three letters. A witness sees a suspicious car and reports it to the police. He says the first two numbers were 4 and 6 but he missed the third and the first letter was C. How many possible license plates does that describe? 1 missing number and 2 missing letters =
10 x 26 x 26 = 6760 possible license plates

5) The data and the scatter plot of the data show the relationship between foot length of an adult male and his height. A footprint of a burglar is found outside the window and has a length of 30 cm. John is asked to predict a height for the burglar. He draws a line of best fit and finds it has a slope of 3.7 and passes through the data point (26, 173). What is John's prediction for the height of the burglar?
 The equation of a line is
y = m x + b, where (for our example):
y = height,
m is the slope (3.7) and
b is the y-intercept.
  1. Using the provided data point (26,173) we can find the y-intercept:
    173 = 3.7 x 26 + b
    b = 173 - 3.7 x 26 = 173 - 96.2 = 76.8
  2. Now, for the foot length of 30 cm.:
  3. y = 3.7 x 30 + 76.8 = 187.8 cm.
Note: The graph and the table were interesting but not necessary to solve the problem