Number of Outcomes for Independent Events in Probability

Let $S$ be the sample space of the experiment. The total number of equally likely outcomes is:
$n(S) = 10$

Since the outcomes are equally likely, the probability of any event $E$ is given by:
$P(E) = \frac{n(E)}{n(S)}$

Given that event $A$ consists of $4$ outcomes, we have $n(A) = 4$, so:
$P(A) = \frac{4}{10} = \frac{2}{5}$

Let $B$ be a non-empty event with $n(B) = k$ outcomes, where $1 \le k \le 10$.
Let $n(A \cap B) = m$, where $m$ represents the number of common outcomes between $A$ and $B$.

The probability of $A \cap B$ and $B$ are:
$P(B) = \frac{k}{10}$
$P(A \cap B) = \frac{m}{10}$

For events $A$ and $B$ to be independent, they must satisfy the condition:
$P(A \cap B) = P(A) \cdot P(B)$

Substituting the known probabilities into the independence equation:
$\frac{m}{10} = \left(\frac{2}{5}\right) \cdot \left(\frac{k}{10}\right)$

Multiplying both sides by $10$:
$m = \frac{2k}{5}$

Since $m$ represents the number of elements in $A \cap B$, $m$ must be a non-negative integer. For $m$ to be an integer, $2k$ must be divisible by $5$. Since $\gcd(2, 5) = 1$, $k$ must be a multiple of $5$.

Given that $B$ is a non-empty event ($k \ge 1$) and $k \le 10$, the possible integer values for $k$ are:
$k = 5 \quad \text{or} \quad k = 10$

• If $k = 5$, then $m = \frac{2 \times 5}{5} = 2$, which is valid since $0 \le 2 \le 4$.
• If $k = 10$, then $m = \frac{2 \times 10}{5} = 4$, which is also valid since $B = S$ and $A \cap B = A$.

Hence, the number of outcomes that $B$ must have so that $A$ and $B$ are independent is 5 or 10.

Correct Option: D