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# Probability Question, Blocks

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154
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+102

I have 120 blocks. Each block is one of two different materials, 3 different colors, 4 different sizes and 5 different shapes. No two blocks are exactly the same of all four properties. I take two blocks at random. What is the probability the two blocks have exactly two of the four properties the same?

Mar 24, 2020

#15
+109509
+2

Lets see if I cant simplify this a bit and check it in the process.

There are 120 blocks

Let the first one be  MCZS

M = material of the first one   $$n(M)= \frac{1}{2}\;\; of \;\;total, \qquad n(\bar M)=\frac{1}{2}\;\;of \;\;total$$

C = colour of the first one  $$n(C)= \frac{1}{3}\;\; of \;\;total, \qquad n(\bar C)=\frac{2}{3}\;\;of \;\;total$$

Z = size of the first one  $$n(Z)= \frac{1}{4}\;\; of \;\;total, \qquad n(\bar Z)=\frac{3}{4}\;\;of \;\;total$$

S= shape of the first one   $$n(S)= \frac{1}{5}\;\; of \;\;total, \qquad n(\bar S)=\frac{4}{5}\;\;of \;\;total$$

I will compare the second on to this one,

There are 6 ways that exactly 2 properties can be the same.

$$n(MC\bar Z \bar S)=120*\frac{1}{2}*\frac{1}{3}*\frac{3}{4}*\frac{4}{5}=12\\~\\ n(M\bar C Z \bar S)=120*\frac{1}{2}*\frac{2}{3}*\frac{1}{4}*\frac{4}{5}=8\\~\\ n(M\bar C \bar Z S)=120*\frac{1}{2}*\frac{2}{3}*\frac{3}{4}*\frac{1}{5}=6\\~\\ n(\bar M C Z \bar S)=120*\frac{1}{2}*\frac{1}{3}*\frac{1}{4}*\frac{4}{5}=4\\~\\ n(\bar MC\bar Z S)=120*\frac{1}{2}*\frac{1}{3}*\frac{3}{4}*\frac{1}{5}=3\\~\\ n(\bar M \bar C Z S)=120*\frac{1}{2}*\frac{2}{3}*\frac{1}{4}*\frac{1}{5}=2\\~\\$$

$$12+8+6+4+3+2=35$$

After the first block was taken there was 119 mot blocks to choose from so

The probability that the 2 blocks have exactly 2 properties in common is         $$\boxed{\frac{35}{119}}$$

Coding:

n(MC\bar Z \bar S)=120*\frac{1}{2}*\frac{1}{3}*\frac{3}{4}*\frac{4}{5}=12\\~\\

n(M\bar C Z \bar S)=120*\frac{1}{2}*\frac{2}{3}*\frac{1}{4}*\frac{4}{5}=8\\~\\

n(M\bar C \bar Z S)=120*\frac{1}{2}*\frac{2}{3}*\frac{3}{4}*\frac{1}{5}=6\\~\\

n(\bar M C Z \bar S)=120*\frac{1}{2}*\frac{1}{3}*\frac{1}{4}*\frac{4}{5}=4\\~\\

n(\bar MC\bar Z S)=120*\frac{1}{2}*\frac{1}{3}*\frac{3}{4}*\frac{1}{5}=3\\~\\

n(\bar M \bar C Z S)=120*\frac{1}{2}*\frac{2}{3}*\frac{1}{4}*\frac{1}{5}=2\\~\\

Mar 26, 2020

#1
+1956
+1

Equation: (2-1)((3-1)+(4-1)+(5-1)) + (2)((3)+(4)) + (12) = 35 with a total of 119 possibilities.

You can solve this now :D

Mar 24, 2020
edited by CalTheGreat  Mar 25, 2020
#2
+102
0

Do you know how you got this equation? Could you possibly break it down for me? Thank you so much for your answer, by the way. It is very elegant.

Bxtterman  Mar 24, 2020
#3
+1956
+1

It is the possible ways to get the blocks. (Two different materials, three different colors, four different sizes, five different shapes) Do you see these numbers in my equation?

Hope this makes sense!

CalTheGreat  Mar 24, 2020
#4
+109509
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I do not see offhand Cal.  (so I am not surprised that Bxtterman does not see either)

I will have to think about it.

Melody  Mar 24, 2020
#5
+483
+1

I believe Cal took their explanation from the best answer on this website, it may help you understand the context:

https://stats.stackexchange.com/questions/193554/problem-understanding-the-following-probability-problem

jfan17  Mar 24, 2020
edited by jfan17  Mar 24, 2020
#6
+109509
+1

Cal if you get an answer from somewhere else but really do not understand it properly.

It is great that you are researching things.  I applaud you for that.

Melody  Mar 25, 2020
#12
+102
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This response had something to do with computers and programming, matrices and vectors (all of which are too advanced for me.) I already saw it, however I did not get ANY of it, so I came here. :)

Bxtterman  Mar 25, 2020
#7
+1956
+1

WHAT?! I never stole anyone's solution. I got help from another PERSON, not a site! Really? I already learned my lesson from that....

Mar 25, 2020
edited by CalTheGreat  Mar 25, 2020
#8
+109509
0

Ok Cat,  I do not think you got your answer from that site.

But what you have written could not be understood by you.

I know that because it cannot be understood by me.

If you do not understand an answer, you should not print it

OR you should admit straight away that you had help from someone else, go on to say that you believe it to be true, but that you cannot explain it yourself because you do not really understand.

Melody  Mar 25, 2020
#11
+109509
+1

Deleted, there was an error

Mar 25, 2020
edited by Melody  Mar 25, 2020
edited by Melody  Mar 26, 2020
#13
+102
+1

Thank you SOOOO MUCH!! This helped Enormously.  I get the problem 100% now, and I appreciate you spending the time to type all this out. I appreciate it!!

Bxtterman  Mar 25, 2020
#14
+109509
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I still want to think about it some more......

Melody  Mar 26, 2020
#18
+109509
0

Thanks  Bxtteman,

I hope you see my correct answer below.  :)

Melody  Mar 26, 2020
#15
+109509
+2

Lets see if I cant simplify this a bit and check it in the process.

There are 120 blocks

Let the first one be  MCZS

M = material of the first one   $$n(M)= \frac{1}{2}\;\; of \;\;total, \qquad n(\bar M)=\frac{1}{2}\;\;of \;\;total$$

C = colour of the first one  $$n(C)= \frac{1}{3}\;\; of \;\;total, \qquad n(\bar C)=\frac{2}{3}\;\;of \;\;total$$

Z = size of the first one  $$n(Z)= \frac{1}{4}\;\; of \;\;total, \qquad n(\bar Z)=\frac{3}{4}\;\;of \;\;total$$

S= shape of the first one   $$n(S)= \frac{1}{5}\;\; of \;\;total, \qquad n(\bar S)=\frac{4}{5}\;\;of \;\;total$$

I will compare the second on to this one,

There are 6 ways that exactly 2 properties can be the same.

$$n(MC\bar Z \bar S)=120*\frac{1}{2}*\frac{1}{3}*\frac{3}{4}*\frac{4}{5}=12\\~\\ n(M\bar C Z \bar S)=120*\frac{1}{2}*\frac{2}{3}*\frac{1}{4}*\frac{4}{5}=8\\~\\ n(M\bar C \bar Z S)=120*\frac{1}{2}*\frac{2}{3}*\frac{3}{4}*\frac{1}{5}=6\\~\\ n(\bar M C Z \bar S)=120*\frac{1}{2}*\frac{1}{3}*\frac{1}{4}*\frac{4}{5}=4\\~\\ n(\bar MC\bar Z S)=120*\frac{1}{2}*\frac{1}{3}*\frac{3}{4}*\frac{1}{5}=3\\~\\ n(\bar M \bar C Z S)=120*\frac{1}{2}*\frac{2}{3}*\frac{1}{4}*\frac{1}{5}=2\\~\\$$

$$12+8+6+4+3+2=35$$

After the first block was taken there was 119 mot blocks to choose from so

The probability that the 2 blocks have exactly 2 properties in common is         $$\boxed{\frac{35}{119}}$$

Coding:

n(MC\bar Z \bar S)=120*\frac{1}{2}*\frac{1}{3}*\frac{3}{4}*\frac{4}{5}=12\\~\\

n(M\bar C Z \bar S)=120*\frac{1}{2}*\frac{2}{3}*\frac{1}{4}*\frac{4}{5}=8\\~\\

n(M\bar C \bar Z S)=120*\frac{1}{2}*\frac{2}{3}*\frac{3}{4}*\frac{1}{5}=6\\~\\

n(\bar M C Z \bar S)=120*\frac{1}{2}*\frac{1}{3}*\frac{1}{4}*\frac{4}{5}=4\\~\\

n(\bar MC\bar Z S)=120*\frac{1}{2}*\frac{1}{3}*\frac{3}{4}*\frac{1}{5}=3\\~\\

n(\bar M \bar C Z S)=120*\frac{1}{2}*\frac{2}{3}*\frac{1}{4}*\frac{1}{5}=2\\~\\

Melody Mar 26, 2020
#16
+632
+1

Elegant!

AnExtremelyLongName  Mar 26, 2020
#17
+109509
0

Thanks LongName :)

Melody  Mar 26, 2020