+0  
 
0
741
7
avatar+73 

http://i.imgur.com/gzVLEnW.png

Please show workings so i can understand thanks.

 

 

 

 

Last AIW Q.

 Mar 11, 2015

Best Answer 

 #7
avatar+118687 
+5

Yes it would be very difficult.  Maybe you select the book off the shelf but then read it off a kindle.  That way the selected book would never leave the shelf and you could select 4 of the same book.  You could almost recite it after reading it 4 times. 

Then your probability would be correct. :)

 Mar 12, 2015
 #1
avatar+118687 
+5

Since you have told me that you are 12 you will not understand this but I will show you.

 

The number of ways that 4 books can be selected from 15 is  

 

    $$15C4 = \frac{15!}{4!11!}$$           on the calc this is          $${\left({\frac{{\mathtt{15}}{!}}{{\mathtt{4}}{!}{\mathtt{\,\times\,}}({\mathtt{15}}{\mathtt{\,-\,}}{\mathtt{4}}){!}}}\right)} = {\mathtt{1\,365}}$$

 

The number of ways that  2 Lewis Carroll books can be chosen from the 5 is     5C2 =    $${\left({\frac{{\mathtt{5}}{!}}{{\mathtt{2}}{!}{\mathtt{\,\times\,}}({\mathtt{5}}{\mathtt{\,-\,}}{\mathtt{2}}){!}}}\right)} = {\mathtt{10}}$$

 

The number of ways the other 2 can be chosen from the other 10 books is   10C2 =    $${\left({\frac{{\mathtt{10}}{!}}{{\mathtt{2}}{!}{\mathtt{\,\times\,}}({\mathtt{10}}{\mathtt{\,-\,}}{\mathtt{2}}){!}}}\right)} = {\mathtt{45}}$$

 

so the number of ways that the 2 Lewis Carroll books can be chosen is  45*10 = 450

 

The probability of choosing exactly 2 Lewis Carroll books is     $${\frac{{\mathtt{450}}}{{\mathtt{1\,365}}}} = {\frac{{\mathtt{30}}}{{\mathtt{91}}}} = {\mathtt{0.329\: \!670\: \!329\: \!670\: \!329\: \!7}}$$

 

so the answer is     $${\frac{{\mathtt{30}}}{{\mathtt{91}}}}$$

 Mar 11, 2015
 #2
avatar+1038 
0

Probability of selecting one Lewis Carroll book (1/3)

Probability of NOT selecting one Lewis Carroll book (1-(1/3))

Probability of choosing exactly 2 Lewis Carroll books from 4 books

 

$$\Text {Solution }\\\\
\left( {\begin{array}{*{20}c} n \\ k \\ \end{array}} \right) = \dfrac{{n!}}{{k!\left( {n - k} \right)!}}\; \hspace{15pt}| \hspace{15pt} \Text {n=4 \; k=2\\$$

 

$$\left( {\begin{array}{*{20}c} 4 \\ 2 \\ \end{array}} \right) * \; \left ( \dfrac {1}{3} \right )^2 \; *\; \left ( \dfrac {2}{3} \right )^2 = \; \dfrac {8}{27} \; \approx \; 0.29629 \\$$

 Mar 11, 2015
 #3
avatar+1038 
+5

Nauseated's Selected Editions . . . 

 Mar 11, 2015
 #4
avatar+118687 
+5

Hi Nauseated,

I do not believe your answer is correct.

Binomial expansion are only used when the trials are independent. :)

 Mar 11, 2015
 #5
avatar+118687 
0

I like your book shelves. :)

 Mar 11, 2015
 #6
avatar+1038 
+5

Oh, that’s right. The book would have to be replaced after it was selected.

 

It softly and suddenly vanishes away after each selection like the baker, after the boojum variety of snark selects him.

 

Makes it kind of difficult to select 4 books for reading, doesn’t it?

 Mar 11, 2015
 #7
avatar+118687 
+5
Best Answer

Yes it would be very difficult.  Maybe you select the book off the shelf but then read it off a kindle.  That way the selected book would never leave the shelf and you could select 4 of the same book.  You could almost recite it after reading it 4 times. 

Then your probability would be correct. :)

Melody Mar 12, 2015

2 Online Users