this is binary ... four or not four ...
p(f) = 1/6 ... p(n) = 5/6
(n+ f)^7 = n^7 + 7 n^6 f + 21 n^5 f^2 + 35 n^4 f^3 + ... + 7 n f^6 + f^7
a) evaluate the 3rd term ... 21 * (5/6)^5 * (1/6)^5 = ?
b) add the 1st two terms
c) 1 - [result of b)]
3. You roll a fair six-sided die 7 times. What is the probability of getting a) exactly two 4’s?
b) less than (and not equal to) two 4’s? c) two or more 4’s?
1 answer
1 answer