+104 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?
+118673 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\\~\\
+2095 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
+104 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.
+2095 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!
+118673 I do not see offhand Cal. (so I am not surprised that Bxtterman does not see either)
I will have to think about it.
+2095 WHAT?! I never stole anyone's solution. I got help from another PERSON, not a site! Really? I already learned my lesson from that....
+118673 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.
+118673 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\\~\\
