Distribution of a convolution. 
Assume that $X_1,X_2,X_3,X_4$ are IID such that $P(X_1=0)=0.3, P(X_1=1)=0.1$ and $X_1$ has on $(0,1)$ the density $f(x)=0.6$. Calculate $P(X_1+X_2+X_3+X_4 \leq 1).$

My work so far. It seems that the first step is to calculate $P(X_1+X_2 \leq x)$ for $x\leq 1$, by means of the formula $$P(X_1+X_2 \leq x)=\int_{R} F_{X_1} (x-t) F_{X_2} \{dt\}.$$ Is it a good idea?
 A: Outline: The sum is $\le 1$ in several possible ways.
(i) One of the $X_i$ is $1$ and the rest are $0$. The probability is $\binom{4}{1}(0.1)(0.3)^3$.
(ii) There are four $0$'s. Easy.
(iii) There is no $1$, and there are three $0$'s,
(iii) There are no $1$'s, there are two $0$'s, and the sum of the remaining $2$ random variables in $\le 1$. Apart from some combinatorial stuff, we need to evaluate a double integral.
(iii) There are no $1$'s, and there is  one $0$. A triple integral calculation.
(iv) No $1$'s, and no $0$'s. For this, we need to evaluate a conventional quadruple integral.
A: Comment, not answer:
I did a simulation to try to break the problem into cases, and
have no remarkable simplifications to offer beyond the revision
of the suggestion by @AndreNicolas.
In case it helps, with $T = X_1 + X_2 + X_3 + X_4,$ it seems $P(T \le 1) \approx 0.23.$ The same
simulation shows $E(T) \approx 1.6$ and $P(T=0) \approx .0081,$ which are encouraging.
 m = 10^6
 p1 = rbinom(m, 1, .6);  x1 = (1-p1)*rbinom(m, 1, 1/4) + p1*runif(m)
 p2 = rbinom(m, 1, .6);  x2 = (1-p2)*rbinom(m, 1, 1/4) + p2*runif(m)
 p3 = rbinom(m, 1, .6);  x3 = (1-p3)*rbinom(m, 1, 1/4) + p3*runif(m)
 p4 = rbinom(m, 1, .6);  x4 = (1-p4)*rbinom(m, 1, 1/4) + p4*runif(m)
 t = x1 + x2 + x3 + x4
 mean(t);  sd(t);  mean(t <= 1)
 ## 1.599915    # Exact E(T) = 1.6 = 4[.4(1/4)+.6(1/2)]
 ## 0.7483644   # Approx SD(T)
 ## 0.22959     # Approx P(T <= 1)
 mean(t==0);  .3^4
 ## 0.008129    # Reality check
 ## 0.0081      #  for 4 0's

