Survival analysis with biased coins Suppose I have five biased coins $c1, c2, c3, c4$ and $c5$ each with different probabilities of getting heads. 


*

*Pr. heads for coin $c1=0.5$ 

*Pr. heads for coin $c2=0.8$

*Pr. heads for coin $c3=0.6$

*Pr. heads for coin $c4=0.4$

*Pr. heads for coin $c5=0.2$
Each coin is flipped once in numerical order, and it takes me $10$ seconds to flip each coin. When you get a head you stop and that cycle is over. 
Let $T$ be a random variable that describes the time of the event that you get a head; hence, $T$ can take on values ${10, 20, 30, 40, 50}$. I am interested in modeling the problem with survival analysis and I want to calculate the hazard function for each outcome. 
The discrete hazard function is defined as follows 
$$
\lambda_j=Pr(T=t_j|T \geq t_j) =  \frac{f_j}{S_j}  
$$
To illustrate lets calculate the hazard for $t=30$ sec - which means that I have to get a head only on the third coin $c3$
$$
\lambda_j=Pr(T=t_c3|T \geq t_c3) = \frac{Pr(C1=tails)Pr(C2=tails)Pr(C3=heads)}{Pr(C3=tails)Pr(C4=tails)Pr(C5=heads)+Pr(C3=tails)Pr(C4=heads)+Pr(C3=heads)}  
$$
At first I was confused since $p$s were different and I could not use a distribution i.e.: Geometric. So is this the correct approach to this problem I created? Also, is there another distribution or shortcut that I can use? Something vague about renewal processes comes to mind. 
Thank you  
 A: I will generalise your problem to allow an arbitrary sequence of binary variables with any probabilities.  Consider sequences of times $t_1 < t_2 < t_3 < \cdots$ and outcomes $X_1,X_2,X_3,...$ with respective probabilities $\theta_1, \theta_2, \theta_3,...$.  (In your game this represents the probability of a head on a coin.)  These outcomes are independent with $X_i \sim \text{Bern}(\theta_i)$ and the overall time for your game is the time until the first success, which is:
$$T \equiv \min \{ t_i | X_i = 1 \}.$$
The hazard function is:
$$\begin{equation} \begin{aligned}
\lambda_T(t_i) 
&= \mathbb{P}(T=t_i | T \geqslant t_i) \\[6pt]
&= \mathbb{P}(X_i=1 | X_1 = \cdots = X_{i-1} = 0 ) \\[6pt]
&= \mathbb{P}(X_i=1) = \theta_i. \\[6pt]
\end{aligned} \end{equation}$$
The marginal density is:
$$\begin{equation} \begin{aligned}
p_T(t_i) 
&= \mathbb{P}(T = t_i) \\[6pt]
&= \mathbb{P}(X_i=1 , X_1 = \cdots = X_{i-1} = 0 ) \\[6pt]
&= \mathbb{P}(X_i=1) \prod_{j=1}^{i-1} \mathbb{P}(X_j=0). \\[6pt]
&= \theta_i \prod_{j=1}^{i-1} (1-\theta_j). \\[6pt]
\end{aligned} \end{equation}$$
You can see that the hazard function is extremely simply in this case, and that is not surprising, given that conditioning on $T \geqslant t_i$ is equivalent to starting the game at the $i$th flip (which does not change the structure of the game very much).

Application to your problem: In your question you are looking for the hazard function at the third time $t_3$.  You should get:
$$\lambda_T(t_3) = \theta_3 = 0.6.$$
In your own working you have tried to use the representation of the hazard function using the survival function as the denominator.  This has been done incorrectly, since the expression you have used in the denominator is not equal to the survival probability (it is missing probabilities for coins 1 and 2).
