Five persons A,B,C,D and E are standing in a queue of a ration shop . The probability that A and E are always togethar is .

Options

A. `1/4`
B. `2/3`
C. `2/5`
D. `3/5`

Answer

To solve the problem of finding the probability that persons A and E are always together in a queue of five persons (A, B, C, D, and E), we can follow these steps:

### Step 1: Calculate the total arrangements of the five persons
The total number of arrangements of 5 persons is given by the factorial of the number of persons:
\[ \text{Total arrangements} = 5! = 120 \]

### Step 2: Treat A and E as a single unit
Since we want A and E to always be together, we can treat them as a single unit or block. Therefore, we can consider the block (AE) as one unit. Now, we have the following units to arrange:
- (AE)
- B
- C
- D

This gives us a total of 4 units to arrange.

### Step 3: Calculate the arrangements of the units
The number of arrangements of these 4 units is:
\[ \text{Arrangements of units} = 4! = 24 \]

### Step 4: Consider the internal arrangements of A and E
Within the block (AE), A and E can be arranged in 2 ways: (AE) or (EA). Therefore, we need to multiply the arrangements of the units by the arrangements of A and E within their block:
\[ \text{Favorable arrangements} = 4! \times 2! = 24 \times 2 = 48 \]

### Step 5: Calculate the probability
The probability that A and E are always together is given by the ratio of the favorable arrangements to the total arrangements:
\[ \text{Probability} = \frac{\text{Favorable arrangements}}{\text{Total arrangements}} = \frac{48}{120} \]

### Step 6: Simplify the probability
Now, we simplify the fraction:
\[ \frac{48}{120} = \frac{2}{5} \]

Thus, the probability that A and E are always together is:
\[ \text{Probability} = \frac{2}{5} \]

### Final Answer
The probability that A and E are always together is \(\frac{2}{5}\).

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