There are four boxes `B_(1), B_(2), B_(3) and B_(4)`. Box `B_(i)` has `i` cards and on each card a number is printed, the numbers are from 1 to `i`. A box is selected randomly, the probability of selecting box `B_(i)` is `(i)/(10)` and then a card is drawn.
Let `E_(i)` respresent the event that a card with number 'i' is drawn, Then :
Q. `P(E_(1))` is Equal to :

Options

A. `(1)/( 5)`
B. `(1)/(10)`
C. `(2)/(5)`
D. `(1)/(4)`

Answer

To solve the problem, we need to calculate the probability \( P(E_1) \), where \( E_1 \) is the event that a card with the number '1' is drawn. We will consider each box and the associated probabilities.

### Step-by-step Solution:

1. **Identify the Boxes and Their Contents:**
- Box \( B_1 \) has 1 card: {1}
- Box \( B_2 \) has 2 cards: {1, 2}
- Box \( B_3 \) has 3 cards: {1, 2, 3}
- Box \( B_4 \) has 4 cards: {1, 2, 3, 4}

2. **Calculate the Probability of Selecting Each Box:**
- Probability of selecting box \( B_1 \): \( P(B_1) = \frac{1}{10} \)
- Probability of selecting box \( B_2 \): \( P(B_2) = \frac{2}{10} = \frac{1}{5} \)
- Probability of selecting box \( B_3 \): \( P(B_3) = \frac{3}{10} \)
- Probability of selecting box \( B_4 \): \( P(B_4) = \frac{4}{10} = \frac{2}{5} \)

3. **Calculate the Probability of Drawing the Card with Number '1' from Each Box:**
- From box \( B_1 \): \( P(E_1 | B_1) = 1 \) (only one card)
- From box \( B_2 \): \( P(E_1 | B_2) = \frac{1}{2} \) (1 out of 2 cards)
- From box \( B_3 \): \( P(E_1 | B_3) = \frac{1}{3} \) (1 out of 3 cards)
- From box \( B_4 \): \( P(E_1 | B_4) = \frac{1}{4} \) (1 out of 4 cards)

4. **Use the Law of Total Probability to Calculate \( P(E_1) \):**
\[ P(E_1) = P(E_1 | B_1) \cdot P(B_1) + P(E_1 | B_2) \cdot P(B_2) + P(E_1 | B_3) \cdot P(B_3) + P(E_1 | B_4) \cdot P(B_4) \]
Substituting the values:
\[ P(E_1) = (1) \cdot \left(\frac{1}{10}\right) + \left(\frac{1}{2}\right) \cdot \left(\frac{2}{10}\right) + \left(\frac{1}{3}\right) \cdot \left(\frac{3}{10}\right) + \left(\frac{1}{4}\right) \cdot \left(\frac{4}{10}\right) \]
Simplifying each term:
\[ P(E_1) = \frac{1}{10} + \frac{1}{10} + \frac{1}{10} + \frac{1}{10} \]
\[ P(E_1) = \frac{4}{10} = \frac{2}{5} \]

### Final Answer:
Thus, the probability \( P(E_1) \) is \( \frac{2}{5} \).

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