An observed event B can occur after one of the three events `A_(1), A_(2), A_(3)`. If
`P(A_(1)) = P(A_(2)) = 0.4, P(A_(3)) = 0.2 and P(B//A_(1)) = 0.25, P(B//A_(2)) = 0.4, P(B//A_(3)) = 0.125`, what is the probability of `A_(1)` after observing B ?

Options

A. `1/3`
B. `6/19`
C. `20/57`
D. `2/5`

Explanation

To find the probability of event \( A_1 \) after observing event \( B \), we can use Bayes' theorem. The theorem states:

\[ P(A_1 | B) = \frac{P(B | A_1) \cdot P(A_1)}{P(B)} \]

### Step 1: Identify the given probabilities
We have the following probabilities given in the question:
- \( P(A_1) = 0.4 \)
- \( P(A_2) = 0.4 \)
- \( P(A_3) = 0.2 \)
- \( P(B | A_1) = 0.25 \)
- \( P(B | A_2) = 0.4 \)
- \( P(B | A_3) = 0.125 \)

### Step 2: Calculate \( P(B) \)
To find \( P(B) \), we use the law of total probability:
\[ P(B) = P(B | A_1) \cdot P(A_1) + P(B | A_2) \cdot P(A_2) + P(B | A_3) \cdot P(A_3) \]

Substituting the values:
\[ P(B) = (0.25 \cdot 0.4) + (0.4 \cdot 0.4) + (0.125 \cdot 0.2) \]

Calculating each term:
- \( 0.25 \cdot 0.4 = 0.1 \)
- \( 0.4 \cdot 0.4 = 0.16 \)
- \( 0.125 \cdot 0.2 = 0.025 \)

Now, summing these values:
\[ P(B) = 0.1 + 0.16 + 0.025 = 0.285 \]

### Step 3: Substitute into Bayes' theorem
Now we substitute \( P(B) \) back into Bayes' theorem:
\[ P(A_1 | B) = \frac{P(B | A_1) \cdot P(A_1)}{P(B)} = \frac{0.25 \cdot 0.4}{0.285} \]

Calculating the numerator:
\[ 0.25 \cdot 0.4 = 0.1 \]

Thus,
\[ P(A_1 | B) = \frac{0.1}{0.285} \]

### Step 4: Simplify the fraction
To simplify \( \frac{0.1}{0.285} \), we can multiply both the numerator and denominator by 1000 to eliminate the decimals:
\[ P(A_1 | B) = \frac{100}{285} \]

Now, simplifying \( \frac{100}{285} \):
- The greatest common divisor (GCD) of 100 and 285 is 5.
- Dividing both by 5:
\[ P(A_1 | B) = \frac{20}{57} \]

### Final Answer
Thus, the probability of \( A_1 \) after observing \( B \) is:
\[ \boxed{\frac{20}{57}} \]