An intuitive solution to this problem (Using probability tree) A group of boys has been lost several days in the dessert. This group has a phone to make phone calls. After a long way walk, they believe that the current area is suitable for phone calls; even though, the battery just allow three call attempts. Due to remote conditions of the current area, the probability of a successful call is 0.2. If the call is not successful, the network would save a register of their location with a probability of 0.375. With a probability of 0.625 the attempt was in vain. If the call attempt was successful or there are 2 registers, they would be rescued within a few hours the same day.
1.) What is the probability that they are not rescued the same day if that just depends on the result of the call attempts.
2.) Which is the probability that the group is rescued with EXACTLY two call attempts
3.) What is the probability of a successful call
4.) What is the probability that the network generates register but they are not rescued
I have try a solution by a tree, but i'm blocked with the fact that with 2 calls they are rescued. I really appreciate an explanation, since my self-learning resources are not extremely complete for wide understanding. If you think in a tree, how could it be.
I would really appreciate a tree model of the problem
 A: We do two of the four problems. Please note that multiple problems may act as a deterrent to writing an answer.  
Problem 1: The people are not rescued precisely if the $3$ calls fail, and together generate $1$ or fewer registers. The probability of unsuccessful calls $3$ times in a row is $(0.8)^3$.  Given that the calls are unsuccessful, we find the probability of $1$ or fewer registers. A register is generated with probability $0.375$, and not generated with probability $0.625$. I prefer in this case $3/8$ and $5/8$. The probability of $0$ successes is $(5/8)^3$. The probability of exactly $1$ success is $\binom{3}{1}(3/8)(5/8)^2$. Thus the probability the people are not rescued is
$$(0.8)^3\left((5/8)^3 +\binom{3}{1}(3/8)(5/8)^2\right).$$
Problem 2: It is not quite clear what this means. But presumably they did not get through the first time, since they made $2$ calls. So they are rescued if either we have failure, then success (probability $(0.8)(0.2)$ or failure, then failure, but $2$ "registers" are generated. The probability of this is $(0.8)(0.8)(3/8)^2$. So the required probability is
$$(0.8)(0.2)+ (0.8)(0.8)(3/8)^2.$$
