Independent solution

How to solve this Binomial Distribution question

Setup

Setup

A policy contributes to the counted group with probability s+t. Independence makes the five-policy all-success probability a fifth power.

c=(s+t)5c=(s+t)^5
r:=c1/5=s+tr:=c^{1/5}=s+t

Model

Model

The probabilities for all possible single-policy outcomes must sum to one.

s+t+0.75s=1s+t+0.75s=1
1.75s+t=11.75s+t=1

Compute

Compute

Subtract the equation s+t=r from the normalization equation, then recover t.

0.75s=1−r⟹s=4(1−r)30.75s=1-r\quad\Longrightarrow\quad s=\frac{4(1-r)}3
t=r−s=7r−43t=r-s=\frac{7r-4}{3}
t=7c1/5−43t=\frac{7c^{1/5}-4}{3}

Answer

Answer

The required symbolic expression is the fifth listed form.

7c1/5−43(E)\boxed{\frac{7c^{1/5}-4}{3}\quad\text{(E)}}