probability and probability tree 
In a cafeteria, they always serve coffee and a pastry together. Both coffee and pastry could be served hot or cold. 
  
  
*
  
*The chance to serve a hot coffee is 0.6 
  
*the chance to serve a hot pastry is 0.4 
  
*the chance to serve a hot pastry and a cold coffee is 0.2.
  
  
  What is the chance to be served a hot pastry and a hot coffee in 3 days out of 5?

I built a tree and I don't know how to put it up here but here's what I found:
Let's assign 


*

*A1- hot coffee 

*A2- Cold coffee 

*B1- hot pastry 

*B2- cold pastry

*the chance to get a hot coffee is $0.6$, then the chance to get a cold coffee is $0.4$

*the chance to get a cold pastry is $0.4$ which means P(A1 and B1) + P(A2 and B1) =0.4

*it is given that P(A2 and B1) is 0.2 which means  P(A1 and B1) = 0.2

*now P(A1 and B2) is 0.4 (because it adds up to 0.6) and P(A2 and B2) is 0.2 (because it adds up yo 0.4)


That means that what I need is the chance to 
1) choose 3 days outta 5 then $0.2^3$ (P(A1 and B1)) and $0.4^2$ ( two days with hot coffee and cold pastry)
or $0.4*0.2*{2 \choose 1}$ (one day: hot coffee cold pastry and the other day the opposite) or 0.2^2 (cold coffee and cold pastry)
which is 
$$
{5 \choose 3}  0.2^3  (0.4^2 + 0.4 \cdot 0.2 \cdot 2 + 0.2^2) = 0.0288
$$
but the answer is 0.0512. What did I miss?
 A: You did almost everything correctly, except the last point:
1) choose $3$ days outta $5$ then $0.2^3$ ($P(A1\cap B1$) and $0.4^2$ (two days with hot coffee and cold pastry) or $0.4\times0.2\times{2 \choose 1}$ (one day: hot coffee cold pastry and the other day the opposite) or $0.2^2$ (cold coffee and cold pastry) which is ${5 \choose 3}  0.2^3  (0.4^2 + 0.4 \cdot 0.2 \cdot 2 + 0.2^2) = 0.0288$.
You should say:
1) choose $3$ days outta $5$ then $0.2^3$ ($P(A1\cap B1$) and $0.8^2$ ($P(\overline{A1\cap B1})$) which is ${5 \choose 3}  (0.2)^3   (0.8)^2 = 0.0512$.
EDIT
To see where you went wrong on counting the cases for $\overline{A1\cap B1}$, you should consider:
a) two days with hot coffee and cold pastry;
b) two days with cold coffee and hot pastry;
c) two days with cold coffee and cold pastry;
d) one day with hot coffee and cold pastry and another with cold coffee and hot pastry;
e) one day with hot coffee and cold pastry and another with cold coffee and cold pastry;
f) one day with cold coffee and hot pastry and another with cold coffee and cold pastry.
See? It's a lot easier to just consider the opposite:
$$P(\overline{A1\cap B1})=1-P(A1\cap B1)=1-0.2=0.8$$
