Question

Solve please and photo send brother

Answer 1

Answer:

given

n(s)=729

let E₁ be the probability of getting no tail

n(E₁)=189

p(E₁)=n(E₁)/n(s)=189/729=7/9

let E₂ be the event of getting one tail

n(E₂)=297

p(E₂)=n(E₂)/n(s)=297/729=11/9

let E₃ be the event of getting two tails

n(E₃)=243

p(E₃)=n(E₃)/n(s)=243/729=1/3

I hope this answer helps you