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?
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?
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