Question

a fair die is cast four times. Calculate the probability of obtaining exactly two 6's round to the nearest tenth of a percent

Answer 1

`|\Omega|=6^4=1296\\|A|=1\cdot1\cdot5\cdot5\cdot(4!)/(2!2!)=25\cdot6=150\\\\P(A)=(150)/(1296)\approx11.6\%`