Two fair dice are tossed. a) Find the probability of obtaining 22 times the total score of 7 if these two fair dice are tossed a total of 120 times. b) Determine the number of tosses needed if the probability of obtaining the score of $7$ at least one is $0.85$ or more.
The probability of obtaining a total score of $7$ is $1/6$.
$X\approx B(120, 1/6)$
$mean= 20$
$Standard deviation=4.08$
I have find out for a. But cannot solve for b.
The answer given for b is $11$.