Four die are thrown simultaneously. The probability that 4 and 3 appear on two of the die given that 5 and 6 have appeared on other two die is

Given that 5 and 6 have appeared on two of the dice,

 the sample space reduces to 6⁴ - 2 x 5⁴ + 4⁴ (inclusion-exclusion principle).

 Also, the number of favorable cases are 4! = 24. 

So, the required probability is $\frac{24}{302}=\frac{12}{151}$