Question

Patients are recruited onto the two arms $(0-\text{control,}1-\text{treatment})$ of a clinical trail. The probability that an adverse outcome occurs on the control arm is $p_0$ and is $p_1$ for treatment arm. Patients are allocated alternatively onto the two arms, and their outcomes are independent. The probability that the first adverse event occurs on the control arm, is

• A. $\frac{p_1}{p_0-p_1-p_0{p}_1}$
• B. $\frac{p_1}{p_0+p_1-p_0{p}_1}$
• C. $\frac{p_0}{p_0+p_1-p_0{p}_1}$
• D. $\frac{p_0}{p_0-p_1-p_0{p}_1}$

Answer 1

The correct option is C $\frac{p_0}{p_0+p_1-p_0{p}_1}$

Let,
Event $E_1:$ first patient (allocated onto the control arm) suffers an averse outcome,
Event $E_2:$ first patient (allocated onto the control arm) does not suffers an averse outcome, but the second (allocated onto the treatment arm) does suffers an averse outcome,
Event $E_0:$ niether of the first two patient suffer adverse outcomes.
Event $F:$ First adverse event occur on the control arm.
By Total probability theorem:
$P\left(F\right)=P\left(F|E_1\right)P\left(E_1\right)+P\left(F|E_2\right)P\left(E_2\right)+P\left(F|E_0\right)P\left(E_0\right)$
Now, $P\left(E_1\right)=p_0,P\left(E_2\right)=(1-p_0)p_1,P\left(E_0\right)=(1-p_0)(1-p_1)$ and $P\left(F|E_1\right)=1,P\left(F|E_2\right)=0,P\left(F|E_0\right)=P\left(F\right)$
Hence,
$P\left(F\right)=(1\times p_0)+(0\times (1-p_0\left)p_1\right)+\left(P\right(F)\times (1-p_0\left)\right(1-p_1\left)\right)$
$\Rightarrow P\left(F\right)=\frac{p_0}{p_0+p_1-p_0{p}_1}$