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?

Bxtterman Mar 24, 2020

#15**+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 \(\boxed{\frac{35}{119}}\)

Now our answers agree.

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

#1**+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

CalTheGreat Mar 24, 2020

#2**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**+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**0 **

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

#7**+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....

CalTheGreat Mar 25, 2020

#8**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

#15**+2 **

Best Answer

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 \(\boxed{\frac{35}{119}}\)

Now our answers agree.

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