consider an event `E=E_(1)capE_(2)capE_(3)` find the value of P(E) if `P(E_(1))=2/5,P(E_(2)/E_(1))=1/5` and `P(E_(3)/(E_(1)E_(2)))=1/10`

Options

A. `2/125`
B. `1/125`
C. `3/125`
D. None of these

Answer

To find the value of \( P(E) \) where \( E = E_1 \cap E_2 \cap E_3 \), we will use the multiplication rule of probability. The multiplication rule states that:

\[ P(A \cap B \cap C) = P(A) \cdot P(B|A) \cdot P(C|A \cap B) \]

In our case, we can substitute \( A \) with \( E_1 \), \( B \) with \( E_2 \), and \( C \) with \( E_3 \). Therefore, we can express \( P(E) \) as:

\[ P(E) = P(E_1) \cdot P(E_2|E_1) \cdot P(E_3|E_1 \cap E_2) \]

Given the probabilities:
- \( P(E_1) = \frac{2}{5} \)
- \( P(E_2|E_1) = \frac{1}{5} \)
- \( P(E_3|E_1 \cap E_2) = \frac{1}{10} \)

Now we can substitute these values into our equation:

\[ P(E) = P(E_1) \cdot P(E_2|E_1) \cdot P(E_3|E_1 \cap E_2) \]

Substituting the values:

\[ P(E) = \left(\frac{2}{5}\right) \cdot \left(\frac{1}{5}\right) \cdot \left(\frac{1}{10}\right) \]

Now we will multiply these fractions:

1. First, multiply the first two fractions:

\[ \frac{2}{5} \cdot \frac{1}{5} = \frac{2 \cdot 1}{5 \cdot 5} = \frac{2}{25} \]

2. Now multiply this result by the third fraction:

\[ \frac{2}{25} \cdot \frac{1}{10} = \frac{2 \cdot 1}{25 \cdot 10} = \frac{2}{250} = \frac{1}{125} \]

Thus, the value of \( P(E) \) is:

\[ P(E) = \frac{1}{125} \]

### Summary of the Steps:
1. Use the multiplication rule of probability.
2. Substitute the known probabilities into the formula.
3. Perform the multiplication of the fractions step by step.
4. Simplify the final result.