Probability of Independent Events Occurring Exactly Once or None

Let $P(E) = x$ and $P(F) = y$.

Since $E$ and $F$ are given to be independent events, the events $E$ and $F'$, $E'$ and $F$, as well as $E'$ and $F'$ are also independent.

1. Probability that exactly one event occurs:
   $P(\text{exactly one}) = P(E \cap F') + P(E' \cap F)$
   Since $E$ and $F$ are independent:
   $P(\text{exactly one}) = P(E)P(F') + P(E')P(F)$
   $x(1-y) + y(1-x) = \frac{11}{25}$
   $x + y - 2xy = \frac{11}{25} \quad \text{--- (Equation 1)}$

2. Probability that none of the events occur:
   $P(\text{none}) = P(E' \cap F') = P(E')P(F')$
   $(1-x)(1-y) = \frac{2}{25}$
   $1 - (x + y) + xy = \frac{2}{25}$
   $x + y - xy = 1 - \frac{2}{25} = \frac{23}{25} \quad \text{--- (Equation 2)}$

3. Solving for $x + y$ and $xy$:
   Subtracting Equation (1) from Equation (2):
   $(x + y - xy) - (x + y - 2xy) = \frac{23}{25} - \frac{11}{25}$
   $xy = \frac{12}{25}$

   Substitute $xy = \frac{12}{25}$ into Equation (2):
   $x + y - \frac{12}{25} = \frac{23}{25}$
   $x + y = \frac{35}{25} = \frac{7}{5}$

4. Finding $x$ and $y$:
   $x$ and $y$ are the roots of the quadratic equation:
   $t^2 - (x+y)t + xy = 0$
   $t^2 - \frac{7}{5}t + \frac{12}{25} = 0$
   Multiply the entire equation by $25$:
   $25t^2 - 35t + 12 = 0$
   $(5t - 4)(5t - 3) = 0$
   $t = \frac{4}{5} \quad \text{or} \quad t = \frac{3}{5}$

Thus, the possible values for $P(E)$ and $P(F)$ are:

• Case 1: $P(E) = \frac{4}{5}$ and $P(F) = \frac{3}{5}$
• Case 2: $P(E) = \frac{3}{5}$ and $P(F) = \frac{4}{5}$

Therefore, the correct options are (A) and (D).