PART A -

Meyer rolls two fair, ordinary dice with the numbers 1,2,3,4,5,6 on their sides. What is the probability that AT LEAST one of the dice shows a square number?

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?

SmartMathMan
Jan 18, 2018

#1**+1 **

B) The probability that the first card drawn is a not square number is 4/6

And the probability that the second card drawn is a not square number [ given that the first was not a square number] is 3/5

So......the probability that at least one of the cards is a square number is

1 - P(that neither is a square number) =

1 - (4/6)(3/5) =

1 - 12/30 =

1 - 2/5 =

3/5 = 60%

The result is different from part A......here, we have a dependent event......the outcome of the second card drawn depends upon the frist card drawn....however.....the number appearing on one die is independent of that appearing on the other

CPhill
Jan 18, 2018