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

asked Aug 10, 2020 153k views

4 votes

6.7k points

Please log in or register to add a comment.

4 votes

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

Peter Tsung answered Aug 14, 2020

7.0k points

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

7.6m questions

10.2m answers