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(a)

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

Solution:

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 5/9.

 

 

 

 

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?

 May 16, 2018
edited by sageatron2000  May 17, 2018
 #1
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+2

probability of a square number is 2/6 = 1/3 
probability of not a square is 4/6 = 2/3 
at least one die shows a square number is 
let 
S = square number 
N = not a square number 

outcomes: 
SS = (1/3)^2 
SN = (1/3)(2/3) 
NS = (2/3)(1/3) 
NN = (2/3)^2 

so at least one would be (SS + SN + NS) = 1/9 + 2/9 + 2/9 = 5/9 
or everything but NN (1 - NN) = 1 - 4/9 = 5/9 welcomes!

 Jul 10, 2018

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