`Aa n dB` are two candidates seeking admission in ITT. The probability that `A` is selected is 0.5 and the probability that `Aa n dB` are selected is at most 0.3. Is it possible that the probability of `B` getting selected is 0.9?

Explanation

To solve the problem, we need to analyze the given probabilities and determine if it's possible for the probability of candidate B being selected to be 0.9.

### Step-by-Step Solution:

1. **Identify Given Probabilities:**
- Let \( P(A) = 0.5 \) (the probability that candidate A is selected).
- Let \( P(A \cap B) \leq 0.3 \) (the probability that both candidates A and B are selected).

2. **Use the Formula for Union of Two Events:**
We know that:
\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \]
where \( P(A \cup B) \) is the probability that at least one of the candidates is selected.

3. **Substituting Known Values:**
We can rewrite the equation using the known values:
\[ P(A \cup B) = 0.5 + P(B) - P(A \cap B) \]
Since \( P(A \cap B) \leq 0.3 \), we can say:
\[ P(A \cup B) \geq 0.5 + P(B) - 0.3 \]
Simplifying this gives:
\[ P(A \cup B) \geq 0.2 + P(B) \]

4. **Considering the Maximum Probability:**
The maximum value for any probability is 1. Therefore:
\[ P(A \cup B) \leq 1 \]
Combining this with the previous inequality:
\[ 0.2 + P(B) \leq 1 \]

5. **Solving for \( P(B) \):**
Rearranging the above inequality gives:
\[ P(B) \leq 1 - 0.2 \]
Thus:
\[ P(B) \leq 0.8 \]

6. **Conclusion:**
Since we have established that \( P(B) \) must be less than or equal to 0.8, it is not possible for \( P(B) \) to be 0.9. Therefore, the answer is **No**, it is not possible for the probability of B getting selected to be 0.9.