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I have two questions

 

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?

 

b) Alejandra's Tapas Bar offers a menu consisting of $9$ savory and $5$ sweet dishes. You can also get a mix-and-match plate consisting of two different dishes on the menu. How many different mix-and-match plates can you get consisting of one savory and one sweet dish?

 May 9, 2021
 #1
avatar+2401 
+1

a)

1 and 4 are the square numbers. 

So the chance that neither dice show them is 2/3*2/3 = 4/9, meaning the chance that at least one was shown is 5/9. 

 

b)

9 * 5

9 savory options, and 5 sweet ones. 

 

=^._.^=

 May 9, 2021
 #2
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+2

 

Just illustrating catmg's answer about the dice.  

The following shows all possible rolls of the dice.  

Anything with a 1 or a 4 satisfies the condition.  

You can see by counting, there are 20 of them.  

And 20/36 reduces to 5/9.   Good work, catmg.  

 

     1,1   1,2   1,3   1,4   1,5   1,6  

     2,1   2,2   2,3   2,4   2,5   2,6  

     3,1   3,2   3,3   3,4   3,5   3,6  

     4,1   4,2   4,3   4,4   4,5   4,6  

     5,1   5,2   5,3   5,4   5,5   5,6  

     6,1   6,2   6,3   6,4   6,5   6,6  

.

 May 9, 2021
 #3
avatar+2401 
+1

Nice :))

 

=^._.^=

catmg  May 9, 2021

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