If a fair die is rolled 7 times, what is the probability, rounded to the nearest
thousandth, of getting at least 4 threes?
we have that
probability that get three is
P=1/6
and the probability that get other number is P=5/6
therefore'
the probability that get at least 4 threes is
P=C(7,4)*(1/6)^4*(5/6)^3
P=C(7,4)*(1/1,296)*(125/216)
Find out C(7,4)
C(7,4)=7!/(7-4)!*3!)=