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# URGENT!! plz i just need help with the part b but i put part a there so it would help you solve part b

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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?

Jan 18, 2018

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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

Jan 18, 2018