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15 students in 6A are divided into 3 groups for a party , 3 students are organizers , 4 students are planners and 8 students are responsible for the food 

Mandy , Helen and Sue are 6A students are responsible for the party

find the probability that thy are in the same group

thanks

Guest Jan 26, 2015

Best Answer 

 #2
avatar+93665 
+10

 

(3C3 + 4C3 + 8C3)/15C3

 

$${\frac{\left({\left({\frac{{\mathtt{3}}{!}}{{\mathtt{3}}{!}{\mathtt{\,\times\,}}({\mathtt{3}}{\mathtt{\,-\,}}{\mathtt{3}}){!}}}\right)}{\mathtt{\,\small\textbf+\,}}{\left({\frac{{\mathtt{4}}{!}}{{\mathtt{3}}{!}{\mathtt{\,\times\,}}({\mathtt{4}}{\mathtt{\,-\,}}{\mathtt{3}}){!}}}\right)}{\mathtt{\,\small\textbf+\,}}{\left({\frac{{\mathtt{8}}{!}}{{\mathtt{3}}{!}{\mathtt{\,\times\,}}({\mathtt{8}}{\mathtt{\,-\,}}{\mathtt{3}}){!}}}\right)}\right)}{{\left({\frac{{\mathtt{15}}{!}}{{\mathtt{3}}{!}{\mathtt{\,\times\,}}({\mathtt{15}}{\mathtt{\,-\,}}{\mathtt{3}}){!}}}\right)}}} = {\frac{{\mathtt{61}}}{{\mathtt{455}}}} = {\mathtt{0.134\: \!065\: \!934\: \!065\: \!934\: \!1}}$$

 

There you go, Alan's answer and my answer look different but they work out the same because the 3! in mine cancels out.  

How about that!      

Melody  Jan 27, 2015
 #1
avatar+27044 
+5

This is (nbr of ways of arranging 3 people in a group of 3 + nbr of ways of arranging 3 people in a group of 4 + nbr of ways of arranging 3 people in a group of 8)/(nbr of ways of arranging 3 people in a group of 15):

 

$${\frac{\left({\left({\frac{{\mathtt{3}}{!}}{({\mathtt{3}}{\mathtt{\,-\,}}{\mathtt{3}}){!}}}\right)}{\mathtt{\,\small\textbf+\,}}{\left({\frac{{\mathtt{4}}{!}}{({\mathtt{4}}{\mathtt{\,-\,}}{\mathtt{3}}){!}}}\right)}{\mathtt{\,\small\textbf+\,}}{\left({\frac{{\mathtt{8}}{!}}{({\mathtt{8}}{\mathtt{\,-\,}}{\mathtt{3}}){!}}}\right)}\right)}{{\left({\frac{{\mathtt{15}}{!}}{({\mathtt{15}}{\mathtt{\,-\,}}{\mathtt{3}}){!}}}\right)}}} = {\frac{{\mathtt{61}}}{{\mathtt{455}}}} = {\mathtt{0.134\: \!065\: \!934\: \!065\: \!934\: \!1}}$$

.

Alan  Jan 26, 2015
 #2
avatar+93665 
+10
Best Answer

 

(3C3 + 4C3 + 8C3)/15C3

 

$${\frac{\left({\left({\frac{{\mathtt{3}}{!}}{{\mathtt{3}}{!}{\mathtt{\,\times\,}}({\mathtt{3}}{\mathtt{\,-\,}}{\mathtt{3}}){!}}}\right)}{\mathtt{\,\small\textbf+\,}}{\left({\frac{{\mathtt{4}}{!}}{{\mathtt{3}}{!}{\mathtt{\,\times\,}}({\mathtt{4}}{\mathtt{\,-\,}}{\mathtt{3}}){!}}}\right)}{\mathtt{\,\small\textbf+\,}}{\left({\frac{{\mathtt{8}}{!}}{{\mathtt{3}}{!}{\mathtt{\,\times\,}}({\mathtt{8}}{\mathtt{\,-\,}}{\mathtt{3}}){!}}}\right)}\right)}{{\left({\frac{{\mathtt{15}}{!}}{{\mathtt{3}}{!}{\mathtt{\,\times\,}}({\mathtt{15}}{\mathtt{\,-\,}}{\mathtt{3}}){!}}}\right)}}} = {\frac{{\mathtt{61}}}{{\mathtt{455}}}} = {\mathtt{0.134\: \!065\: \!934\: \!065\: \!934\: \!1}}$$

 

There you go, Alan's answer and my answer look different but they work out the same because the 3! in mine cancels out.  

How about that!      

Melody  Jan 27, 2015

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