The probability that a dealer will sell at least `20TV` sets during a day is `0.45` and the probability that he will sell less than `24TV` sets is `0.74`. The probability that he will sell `20,21,22` or `23` TV sets during the day is

Options

A. `0.19`
B. `0.29`
C. `0.333`
D. `0.81`

Explanation

To solve the problem, we need to find the probability that a dealer will sell 20, 21, 22, or 23 TV sets during the day.

Let's denote:
- Event A: The dealer sells at least 20 TV sets.
- Event B: The dealer sells less than 24 TV sets.

From the problem, we know:
- P(A) = 0.45 (the probability that the dealer sells at least 20 TV sets)
- P(B) = 0.74 (the probability that the dealer sells less than 24 TV sets)

We need to find the probability of the dealer selling 20, 21, 22, or 23 TV sets, which can be represented as P(A ∩ B).

### Step-by-step Solution:

1. **Understanding Events**:
- Event A includes selling 20, 21, 22, 23, or more TV sets.
- Event B includes selling 0, 1, 2, ..., 23 TV sets.

2. **Identifying the Intersection**:
- The intersection of events A and B (A ∩ B) will include the outcomes where the dealer sells 20, 21, 22, or 23 TV sets.
- Thus, A ∩ B = {20, 21, 22, 23}.

3. **Using the Addition Rule**:
- We can use the addition rule for probabilities:
\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \]
- Rearranging this gives us:
\[ P(A \cap B) = P(A) + P(B) - P(A \cup B) \]

4. **Finding P(A ∪ B)**:
- Since event B (selling less than 24) includes all sales from 0 to 23, we can conclude that:
\[ P(A \cup B) = P(B) = 0.74 \]

5. **Substituting Values**:
- Now we can substitute the known values into the equation:
\[ P(A \cap B) = P(A) + P(B) - P(A \cup B) \]
\[ P(A \cap B) = 0.45 + 0.74 - 1 \]
- Simplifying this gives:
\[ P(A \cap B) = 0.45 + 0.74 - 1 = 0.19 \]

6. **Final Calculation**:
- The probability that the dealer will sell 20, 21, 22, or 23 TV sets is:
\[ P(A \cap B) = 0.19 \]

### Conclusion:
The probability that the dealer will sell 20, 21, 22, or 23 TV sets during the day is **0.19**.