Question 1194853
<br>
(1) The probability of rolling an 8 is 5/36; the probability of rolling a 7 is 1/6; the probability of rolling any other sum is 25/36.<br>
(2) The probability of rolling an 8 3 times, a 7 2 times, and any other sum 5 times in 10 throws, IN ANY PARTICULAR ORDER, is {{{((5/36)^3)((1/6)^2)((25/36)^5)}}}.<br>
(3) The number of different ORDERS in which you can get those outcomes on 10 rolls is {{{10!/((3!)(2!)(5!))}}}.<br>
So the probability asked for in the problem is<br>
{{{(10!/((3!)(2!)(5!)))(((5/36)^3)((1/6)^2)((25/36)^5))}}}<br>
Use a calculator...!<br>