+11 Part A: Meyer rolls two fair, ordinary dice with the numbers 1, 2, 3, 4, 5, and 6 on their sides. What is the probability that at least one of the dice shows a square number?
Answer for Part A: The only way for Meyer not to roll at least one square number is for non-square numbers to come up on both dice.
Two of the numbers on each die are squares: namely, 1 and 4. Four numbers on each die are not squares: 2, 3, 5, and 6. Thus there are 4*4 =16 ways for Meyer to roll a non-square number on each die, out of 36 equally likely outcomes for the pair of dice. The other 36 - 16 = 20 outcomes each involve a square number showing on one or both dice. So, Meyer's probability of rolling at least one square number is 20/36 = 5/9.
Part B: Mary has six cards whose front sides show the numbers 1, 2, 3, 4, 5, and 6. She turns the cards face-down, shuffles the cards until their order is random, then pulls the top two cards off the deck. What is the probability that at least one of those two cards shows a square number?
Explain your solution. Is the answer the same as in part (a), or is it different? Why?
