If $p_1 = 0.3$ and $p_2 = 0.4$, what is the probability that it will take Jay more than 12 hours to be successful on both jobs?

Question

Jay has two jobs to do, one after the other. Each attempt at job $i$ takes one hour and is successful with probability $p_i$. If $p_1 = 0.3$ and $p_2 = 0.4$, what is the probability that it will take Jay more than 12 hours to be successful on both jobs?

I don't know how to start on this question, can someone give me some hints to help me start off? Thank you in advance.

Answer 1

To have a clear answer we need the strategy that Jay uses in choosing a job to try. Let's say he tries job $1$ until he succeeds, then tries job $2$. The chance he never gets job $1$ done is $0.7^{12}$. The chance he fails job $1$ the first $i$ times, succeeds on try $i+1$ and never gets job $2$ is $0.7^i\cdot 0.3 \cdot 0.6^{11-i}$ The total chance of failure is then $$0.7^{12}+\sum_{i=0}^{11}0.7^i\cdot 0.3 \cdot 0.6^{11-i}$$

Answer 2

Let's say he attempted 1st job $a$ times and second job $b$ times unsuccessfully. So,$a+b=11$ will serve our purpose that he'll have to attempt both jobs at least once again to successfully complete the job, and thus complete jobs in more than 12 hours.

But I have not got any way to directly inculcate all cases $(a,b)$.I think we have to add for each case one-by-one.