Four dice are thrown simultaneously 
Four dice are thrown simultaneously. The probability that $4$ and $3$ appear on two of the dice given that $5$ and $6$ appear on the other two dice is: 
a) $1/6$
b) $1/36$
c) $12/51$
d) None of these

Since the events are independent, I feel the probability is $1/6 \times 1/6 = 1/36$
But answer is c. Why? 

The tricky part of the solution is to find the number of outcomes such that 2 dice land $5$ and $6$; it can be done using inclusions/exclusions.
Let $\Omega$ be the set of all outcomes, of size $6^4$.
Let $S_5$ be the set of outcomes that contain no $5$'s, of size $5^4$.
Let $S_6$ be the set of outcomes that contain no $6$'s, of size $5^4$.
Let $S_{5,6}$ be the set of outcomes that contain no $5$'s no $6$'s, of size $4^4$.
Using inclusion/exclusion principle the number of outcomes that contain at least one $5$ and at least one $6$ is
$$6^4-5^4-5^4+4^4=302$$
The rest is simple, and the answer is $12/151$
