The question is as follows:
Two friends, Mike and Robert, each take a $9$ question exam for a chemistry competition. Mike answers each question correctly with probability $\frac{2}{3}$ and Robert answers each question correctly with probability $\frac{3}{5}.$ All answers are independent of other answers. Each friend receives a point for each question he answers correctly and which the other friend does not answer correctly. Let $X$ be the total number of points at the end of the competition. What is $\mathbb{E}\left [ X \right ]?$
My initial thought was to find the p.m.f. of the random variable $X$ and then use the fact that $$\mathbb{E}\left [ X \right ]=\sum_{i=0}^{9}x_{i}p(x_{i}).$$ However, the probabilities seem way too complicated. For example, $$\mathbb{P}(X=7)=\binom{9}{7}\left ( \frac{1}{3} \right )^{2}\left ( \frac{2}{3} \right )^{7}\left ( \frac{2}{5} \right )^{9}+ \ ...$$ Adding all of these would take a long time - and this is just for $\mathbb{P}(X=7)!$ Plus, since the question says nothing about ties, I am not sure what to do with those as well.
Is there any easier way to appraoch this? Am I overthinking it?
Thanks in advance!
I think what the question means is that on any given question Mike only gets a point if he gets it correct and Robert gets it wrong.
The probability if this happening is always $$P_{mike} = \frac 23 \times \frac 25=\frac 4{15} $$ Mike's score is distributed binomially with $n=9;p=\frac 4{15}$
Robert's score score is distributed binomially with $n=9;p=\frac 35\times \frac1{3}=\frac 15 $
for binomial distributions ...$$\mathbb{E}\left [ X \right ] = np $$