Probability with two random variables Suppose two basketball players throw alternately to a basket (infinity times). Player A has probability of 0.7  to score, player B has probability of 0.4 to score. Player A is starting. What is the probability that the 2 first scores were together?
 A: In order Player A to score for first time on the $n^{th}$ attempt, he needs to miss the first $n-1$ attempts. The probabilty for miss is $\frac 3{10}$ and for a hit it's $\frac 7{10}$. So the probablity to make his firs basket after $n$ attempts is:
$$P_A(n) = \left(\frac{3}{10}\right)^{n-1}\frac7{10}$$
Simularly for Player B we have:
$$P_B(n) = \left(\frac{3}{5}\right)^{n-1}\frac2{5}$$
If you want to get the probabilty that they'll hit their first shots at specific attempt $n$ is:
$$P_A(n) \cdot P_B(n) = \left(\frac{9}{50}\right)^{n-1}\frac{14}{50}$$
If you want to know the probability of both players making the first shot at any attepmt then the probability is expresed with:
$$\left(\frac{9}{50}\right)^{1-1}\frac{14}{50} + \left(\frac{9}{50}\right)^{2-1}\frac{14}{50} + \left(\frac{9}{50}\right)^{3-1}\frac{14}{50} + \cdots$$
This is simple geometric series and finding the sum as $n \to \infty$ won't be a problem, right?
A: The chance that both score at very first attempt  = 0.7*0.4
at second attempt = (0*3*0.6)*(0.7*0.4) (Both have to fail their first attemts and succeed in second attempt)
at nth attempt is (0.3*0.6)^n *(0.7*0.4) (Both have to fail there n attempts and suceed at n+1 th attempt)
The combined probabillity hence is 0.28 [1+0.18 + 0.18^2...] ~ 0.341
