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Jan 31, 2018
 #1
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Jan 31, 2018
 #6
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+5

Solution:

\(\text {The easy parts first: }\\ \text {There are }\dbinom{52}{5} = 2598960 \text{ ways to select 5 from 52 cards. } \\ \)

 

\(\text{Single suit probability – five cards of the same suit. }\\ \dbinom{4}{1}\dbinom{5}{13} = 5148\\ \rho(1) = \dfrac{5148}{2598960} = 0.19808 \% \\ ------------------- \)

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\(\text{Four suit probability – five cards four suits. }\\ \text {A five-card hand will have at least two cards of the same suit. } \\ \text {Select two cards with matching suits. }\\ \text{There are }\underbrace {\dbinom{4}{1}}_ {\small \text { ways to choose suit }} * \underbrace {\dbinom{13}{2}}_ {\small \text { ways to choose 2 of 13 }} \\ \text{Then for the 3 remaining cards }\\ \text{ there are } \dbinom{13}{1}*\dbinom{13}{1}*\dbinom{13}{1} \text { ways to choose unique suit for each of remaining cards. } \\ \text {The product of these counts gives the total number of five-card hands with four unique suites. }\\ \dbinom{4}{1} * \dbinom{13}{2} * \dbinom{13}{1} * \dbinom{13}{1}* \dbinom{13}{1} = 685464 \\ \rho(4) = \dfrac {685464}{2598960} \text { 26.3746}\% \\ \)

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\(\text {Two suits. Five cards }\\ \text {There are } \dbinom{4}{2} \text { ways to chooses two suits. Twenty-six (26) cards represent these two suits. }\\ \text{There are } \dbinom{26}{5} \text { ways to choose five cards from the 26. }\\ \text{These selections include single suits, }\\ \text{so subtract the single-suit counts of three for the five-card selections. } -(3)\dbinom{4}{1} * \dbinom{13}{5}\\ \dbinom{4}{2} * \dbinom{26}{5} – (3)\dbinom{4}{1} * \dbinom{13}{5}= 379236\\ \rho(2) = \dfrac{379236}{2598960} = 14.5918\% \)

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\(\text{For three suits, subtract the counts for 1, 2, and 4 suits from } \dbinom{52}{5}\\ \dbinom{52}{5} - 5148 - 379236 - 685464 = 1529112\\ \rho(3) = \dfrac{1529112}{2598960} = 0.14.5918\% \\ \)

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\(\rho(1) = \dfrac{5148}{2598960}\\ \rho(2) = \dfrac{379236}{2598960}\\ \rho(3) = \dfrac{1529112}{2598960}\\ \rho(4) = \dfrac {685464}{2598960}\\ -----------------\\ \text{The probability of a five-card hand having at least three (3) unique suits is }\\ \dfrac{1529112}{2598960} + \dfrac {685464}{2598960} = \dfrac {2214576}{2598960} = \dfrac {507}{595}\\ \text{ }\\ \text{ }\\ \small \text { Source: Lancelot Link & Co. Solutions for AoPS probability questions. } \)

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The result matches Melody’s answer.  This proves we monkey around until we get it right. laugh

 

 

 

GA

Jan 31, 2018
 #2
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Jan 31, 2018
Jan 30, 2018
 #2
avatar+299 
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Jan 30, 2018
 #2
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Jan 30, 2018
 #1
avatar+2446 
+1

Question 1:

 

This question definitely suggests a percent increase since the number of books increased. In other words, 15 plus some percent of 15 equals 21 books. 

 

15 + some percent of 15 = 21

 

Let's let x = some percent because we do not know what that is. 

 

15 + x of 15 = 21

 

"Of" in mathematics means multiplication. 

 

15+15x=21

 

Now, solve for x:
 

\(15+15x=21\) Now, isolate x in this equation.
\(15x=6\) Divide by 15 from both sides.
\(x=\frac{6}{15}=\frac{2}{5}=0.4\) Of course, we want to the percent increase, so we must convert 0.4, currently in decimal format, to a percent. To do that, just multiply by 100 and slap a percent sign behind the number. 
\(x=0.4\Rightarrow 40\%\) This, of course, is the increase. 
   

 

Question 2:

 

The same concept can be used to calculate the percent decrease. I know that this problem involves that since the original amount is greater than the ending amount. 

 

60 cars - some percent of 60 cars =  24 cars

 

Let's let x equal some percent.

 

60 cars - x of 60 cars = 24 cars

 

As aforementioned, "of" is a direct indicator of multiplication in mathematics. 

 

60-60x=24

 

\(60-60x=24\)  
\(-60x=-36\) Divide by -60 on both sides. 
\(x=\frac{-36}{-60}=\frac{3}{5}=0.6\) Of course, this needs to be a percentage. Let's convert the answer into one.
\(x=0.6\Rightarrow 60\%\) Remember that this is a decrease.
   

 

Note: I realize that I could have used the formula \(\frac{y_2-y_1}{y_1}*100\) to get the percent changed, but I think that the following methods allow you to understand what is occurring; the formula, on the other hand, does not.

Jan 30, 2018

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