Hands-on math!

Washington State Math Olympiad
Hints and Solutions
2010 Grade 8 Probability and Statistics
Problem
Solution
1) A family has two children. At least one of them is a boy. What is the probability that both children are boys?
  1. There are 4 possibilities of 2 children
    (G = girl and B = boy):  (G G)     (G B)     (B G)     (B B)
  2. Of these 4 possibilities, only the "G G" possibility is eliminated, so the probability is 1/3
Note: I know, I know, it looks like if one of them is a boy then there is a 1/2 probability that the second one is a boy, but that assumes that the first child was a boy (in the past) and you are considering the probabilities of the second child, which is not what the problem is asking
2) At the brand new middle school lockers open with PIN numbers instead of combination locks. PIN numbers consist of three one digit numbers and digits can repeat. This year all the 8th grade PIN numbers begin with the digits 6, 7, 8, or 9, but the other two can be any number from 0 up to 9. How big does the 8th grade class have to be to guarantee at least two students have the same PIN numbers? The number of possible PIN numbers is 4x10x10 = 400.
So, with 401 students, the numbers begin to repeat.

3) Toni's teacher gives a 20 point quiz every Friday. On her first five quizzes Toni scores: 15, 4, 16, 18 , and 15. After the next two 20 point quizzes, her average on all seven quizzes is 15. What is the median?
  1. Re-arrange the scores in numeric order:
        4 15 15 16 18
  2. The sum of these scores is 68
  3. For the mean to be 15 after 2 more scores, the sum must be 15x7 = 105
  4. Therefore, the next 2 added scores must sum to 105 - 68 = 37
  5. The 2 addes scores can only be (20 and 17) or (19 and 18), both of which are higher (or as high) than the existing scores.
  6. So the scores, with the 2 added scores, are either:
        4 15 15 16 18 17 20 or
        4 15 15 16 18 18 20
  7. In either case, the median is 16

Problem
Solution
4) The letters
"M A T H O L Y M P I A D" are written one to a card to make a game. A player draws a card, records the letter and replaces the card. The player draws a second card, records the letter, and replaces the card. Points are earned as follows:
Combination of a consonant and vowel: 1 point
Combination of two different consonants: 2 points
Combination of two different vowels: 3 points
Double Letters of Any Type: 0 points and turn is over
Poison Y - Any combination with a "Y": 0 points and turn is over
What is the probability of drawing double letters or drawing the poison Y? Express your answer as a fraction.
  1. There are 12x12 = 144 possible 2-card combinations, drawing a card, replacing it and drawing again.
  2. There are 12/144 possibilities of drawing the same card twice
  3. There are 2 repeated letters, M and A. There are an extra 2 chances each for drawing a match with the other letter for a total of 2x2 = 4/144 possibilities
  4. There are 11 + 11 chances of drawing a Y = 22 extra chances = 22/144. The 12th Y was counted in the probability of drawing the same card twice, above.
  5. The total of these probabilities is:
    12⁄144 + 4⁄144 + 22⁄144 = 38⁄144 probability of drawing double letters or a Y.
This table shows the possibilities of drawing double letters or a Y:
M A T H O L Y M P   I A D
M X X X
A X X X
T X X
H X X
O X X
L X X
Y X X X X X X X X X X X X
M X X X
P X X
I X X
A X X X
D X X
5) Consider the following survey information:
Suppose the information is based on surveying 1000 people. At least how many have dogs that sometimes sit in the back seat and sometimes in the front passenger seat while they drive?
  1. For the front passenger seat dogs, there are 57% of 1000 = 570 dogs.
  2. For the back seat dogs, there are 69% of 1000 = 690 dogs.
  3. Together, these are 570 + 690 = 1280 which is 280 more than the number of dogs, so the answer is 280