Question

For the binomial distribution with n=4 and p=0.25

a)find the probability of three success
b)at the most three success
c)two or more failures​

Answer 1

a.) .0469

b.) .9961

c.) .9492

Rounded these check below for full answers

Explanation:

a.)

`{4\choose3}*.25^3*(1-.25)=.046875`

b.)

Porbability of at most 3 successes is equal to 1-p(4)

p(4)=

`{4\choose4}*.25^4=.003690625`

1-.003690625=.99609375

c.)

two or more failures is equa lto

p(0)+p(1)+p(2)=

`{4\choose0}*.25^0*(1-.25)^4+{4\choose1}*.25^1(1-.25)^3+{4\choose2}*.25^2*(1-.25)^2=.94921875`