Calculate probability than more than three phone calls are required A representative of a market research firm contacts consumers by phone in order to conduct surveys. The specific consumer contacted by each phone call is randomly determined. The probability that a phone call produces a completed survey is 0.25. Calculate the probability that more than three phone calls are required to produce one completed survey.
Try
If I call $X$ to be the number of phones required until one completed survey (success in this case), then $X$ is geometric with $p=0.25$. Then we want $P(X>3)$ which is obviously
$$ P(X>3) = \sum_{x=3}^{\infty} 0.75^{x-1} 0.25 = \frac{0.25}{0.75} \sum_{x=3}^{\infty} 0.75^x = \frac{0.25}{0.75} \cdot (0.75)^3 \sum_{x=0}^{\infty} 0.75^x = 0.75^2\cdot0.25 \cdot \frac{1}{1-0.75} = 0.5625$$
Now, the answer key says it should $\boxed{0.42}$ but I am doing everything correct. IS it a typo in the answer key?
 A: Let $p=1/4$ be the probability of success. 
Let $q=3/4$ be the complementary probability. If we have success, we stop, else continue. Then the problem is modeled by the following tree:
  *p
 /
*    *p
 \  /
  *q    *p
    \  /
     *q
       \
        *q MORE THAN THREE...

The probability to land in MORE THAN THREE... is
$$q^3=\left(\frac 34\right)^3=\frac{27}{64}=0.421875\ .$$
A: There are several approaches possible.
1) Your approach
$$ P(X>3) = \sum_{x=\color{red}4}^{\infty} 0.75^{x-1} 0.25 =\frac{0.25}{0.75}\cdot\sum_{x=\color{red}4}^{\infty} 0.75^{x}=\frac{0.25}{0.75}\cdot\left(\sum_{x=1}^{\infty}0.75^{x}-\sum_{x=1}^{3}0.75^{x}\right)  $$
$$\frac{0.25}{0.75}\cdot \left(\frac{1}{1-0.75} -0.75\cdot\frac{1-0.75^3}{1-0.75} \right)= \frac{\require{cancel} \cancel{0.25}}{\require{cancel} \bcancel{0.75}}\cdot \left(\frac{\require{cancel} \bcancel{0.75}}{\require{cancel} \cancel{1-0.75}} -\require{cancel} \bcancel{0.75}\cdot\frac{1-0.75^3}{\require{cancel} \cancel{1-0.75}} \right)$$
$= \left(1- (1-0.75^3)\right)=0.75^3=0.421875\approx \boxed{0.42}$
2) Your approach, but applying converse probability
$$P(X>3)=1-P(X\leq 3)=1-\sum_{x=1}^{3} 0.75^{x-1} 0.25=1-0.25\cdot \sum_{x=1}^{3} 0.75^{x-1}$$
substitution: $y=x-1$
$$1-0.25\cdot \sum_{y=0}^{2} 0.75^{y}=1-\require{cancel} \cancel{0.25}\cdot \frac{1-0.75^3}{\require{cancel} \cancel{1-0.75}}=1- (1-0.75^3)=0.75^3$$
3) Just not completing the first three calls
We assume a constant probability $p$ to complete a survey.
$p=0.25\Rightarrow (1-p)=0.75$
The probability not completing the next three calls is $(1-p)^3=0.75^3$
