Linear combinations of random variables — practice question

A spinner has $5$ equal sectors, labelled $1$, $2$, $3$, $4$ and $5$. The number on the sector where it stops is taken as the score. This score is represented by the random variable $X$, and its probability distribution is listed in the table: $P(X=1)=0.3$, $P(X=2)=0.15$, $P(X=3)=3p$, $P(X=4)=2p$, $P(X=5)=0.05$. There is also a second spinner with $3$ sectors, labelled $1$, $2$ and $3$. The score from this spinner is represented by the random variable $Y$. It is known that $P(Y=1)=0.3$, $P(Y=2)=0.5$ and $P(Y=3)=0.2$.