7.5 Toss a fair coin 5 times, the probability of getting exactly one head is not 0, and it is the same as the probability of getting exactly two heads.

Let the irreducible fraction $\frac{i}{j}$ be the probability of getting 3 heads in 5 tosses of the coin, find the value of $i+j$.
［Solution］Let $r$ be the probability of getting heads in a single coin toss, then the probability of getting $k$ heads in $n$ coin tosses is:
$$
C_{n}^{k} r^{k}(1-r)^{n-k} .
$$

From the problem, we can derive the equation
$$
C_{5}^{1} r(1-r)^{4}=C_{5}^{2} r^{2}(1-r)^{3} .
$$

Solving this, we get $r=0, \frac{1}{3}$ or 1.
By the problem's condition, the probability cannot be 0 or 1, so $r=\frac{1}{3}$.
The probability of getting 3 heads in 5 tosses is
$$
\frac{i}{j}=C_{5}^{3} r^{3}(1-r)^{2}=10 \cdot\left(\frac{1}{3}\right)^{3} \cdot\left(\frac{2}{3}\right)^{2}=\frac{40}{243} .
$$

Therefore, $i+j=283$.