32. (1989 American Mathematical Invitational) When a fair coin is tossed 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 exactly 3 heads in 5 tosses of the coin, find the value of $i+j$.
32. Let $r$ be the probability of a coin landing heads up once, then in $n$ coin tosses, the probability of getting $k$ heads is: $C_{n}^{*} r^{k}(1-r)^{n-k}$.
From the problem, we can derive the equation $C_{5}^{1} r(1-r)^{4}=C_{r} r^{2}(1-r)^{3}$.
Solving this, we get $r=0, \frac{1}{3}$ or 1.
By the problem's condition, the probability cannot be 0 or 1, so $r=\frac{1}{3}$.
And the probability of getting 3 heads in 5 tosses is
$$
\frac{i}{j}=C_{r}^{3} r^{3}(1-r)^{2}=10 \cdot\left(\frac{1}{3}\right)^{3} \cdot\left(\frac{2}{3}\right)^{2}=\frac{40}{243} .
$$