If I toss two pennies, there are 4 possible outcomes:  TT, HT, TH, HH

1) If I toss 3 pennies, what is the probability that they are all heads? _1/8__

   For each penny, there is a 1/2 (or 50%) chance that it is a head. The probability that they are all heads is (1/2)³ = 1/8 . Only a couple of students got this wrong.

2) If I just rolled 6 three times in a row using a regular die, what is the chance that I roll a 6 on my next try?  __1/6______

I want the students to realize that the rolls are all independent of each other. The probability is 1/6 regardless of how many 6's or non-6's that I got previously. A surprising number of students were thrown off. (1/6)⁴ was a common wrong answer.

3) What is the probability of rolling a prime number when rolling two dice? ___5/12__

There are 5 prime numbers that can arise: 2, 3, 5, 7, 11.  I was surprised that most students got this wrong.  Its easy to think that there are 5 primes, and 12 possiblilities, and conclude that the answer is 5/12. However, that would be arriving at the correct answer with the wrong reasoning. The correct reasoning, is to consider how many different ways you can get each prime number. They are:

   2: only 1 way (snake eyes)
   3: two different ways to get 3
   5: 4 ways  (1, 4)  (4, 1), (2, 3), (3, 2)
   7: 6 ways
   11: 2 ways  (5, 6), (6, 5)

So there are 15/36 = 5/12 probability that you get a prime when rolling two dice.

4) What is the probability that the sum is 5 when rolling two dice? _____1/9_______

 There are 4 ways to roll a 5: (1, 4)  (4, 1), (2, 3), (3, 2)
   So the answer is 4/36 = 1/9

5)  If I roll 3 dice, how many different ways could they sum to 5?  ______6_________

 Six different ways:  113, 131, 311, 122, 212, 221

6)  Suppose a bowl contains 3 blue, 3 red, and 3 white marbles. What is the probability of drawing two blue marbles?  _____1/12________

  Whether you draw them one at a time or together, the result is the same. Consider one at a time. P(first blue) = 3/8, P(second blue) = 2/7 so
  3/9 * 2/8 = 1/3 * 1/4 = 1/12

7) If I plan to have 3 children, what is the probability that I will have at least one girl? __7/8___

   This involves the common "trick" of subtracting the converse (which is easier to compute) from 1. We know that the probability of having no girls is (1/2)³, so the probability of having at least one girl must be
             1 - P(no girls) = 1 - 1/8 = 7/8

8) Suppose I have 4 letters and 4 labels. Each label corresponds to one of the letters. If I apply the labels randomly, what are the chances that I get all 4 labels on their correct letters? ____1/24_____

       When I assign labels randomly, I have a 1/4 chance of getting the first one in the correct position. For the next one, I have a 1/3 change of getting it correct. For the 3rd label, I have a 1/2 chance of being correct because 2 labels are already placed correctly. For the last label, there is only one place for it to go - the correct place.
  Hence the probability must be 1/4 * 1/3 *1/2 = 1/24

9) What are the chances that I get [exactly] 3 of the 4 labels on their correct letters? ___0____

This is a bit of a trick question, and only a few got it. There is no way to affix exactly 3 labels to their correct letters. Try it!

10) What are the chances of getting exactly 2 of the 4 labels correct? ___1/4____

 There are 6 different ways to have 2 labels on the correct letters.

CCWW, WCCW, WWCC, CWCW, WCWC, CWWC

so the answer is 6/24 = 1/4    (recall that there are 4! or 24 total ways to affix the labels)