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

Answer:

a.) .0469

b.) .9961

c.) .9492

Rounded these check below for full answers

Step-by-step 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)=

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