Calculating entropies of attributes Can you please show the step by step calculation of Entropy(Ssun)?  I do not understand how 0.918 is arrived at.

I tried but I get the values as 0.521089678, 0.528771238, 0.521089678 for Sunny, Windy, Rainy.

I was able to calculate the target entropy (Decision) correctly as = -(6/10)*log(6/10) + -(2/10)log(2/10) + -(1/10)log(1/10) + -(1/10)log(1/10) = 1.570950594

I am totally stuck at the next step.  Request your help.

Reference: http://www.doc.ic.ac.uk/~sgc/teaching/pre2012/v231/lecture11.html
Please search for "The first thing we need to do is work out which attribute will be put into the node at the top of our tree:" to reach the line I am referring to.
Consider the formula for Entropy:

$E(S) = \sum_\limits{i=1}^{n}-p_{i}\log_2p_{i}$

Expanding that summation for the four concept decision attributes for your problem gives

$E(S) = -p_{cinema}\log_2\left(p_{cinema}\right) - p_{tennis}\log_2\left(p_{tennis}\right) - p_{stayin}\log_2\left(p_{stayin}\right) - p_{shopping}\log_2\left(p_{shopping}\right)$

There are three observations where Weather=Sunny. Of those three, one has Decision=Cinema and two have Decision=Tennis. So for Weather=Sunny you have $p_{cinema}=\frac{1}{3}$, $p_{tennis}=\frac{2}{3}$, and other decision probabilities are zero. Plugging those values into the equation above gives

$
\begin{align}
E(S_{Sun}) &amp;= -\frac{1}{3}\log_2\left(\frac{1}{3}\right)-\frac{2}{3}\log_2\left(\frac{2}{3}\right) - \left(0\right)\cdot\log_2\left(0\right) - \left(0\right)\cdot\log_2\left(0\right) \\
           &amp;= 0.38998 + 0.52832 + 0 + 0 \\
           &amp;= 0.918
\end{align}$