A 5 digit number is formed by using the digits 0, 1, 2, 3, 4 and 5 without repetition. The probability that the number is divisible by 6 is

Options

A. 0.08
B. 0.17
C. 0.18
D. 0.36

Answer

To find the probability that a 5-digit number formed using the digits 0, 1, 2, 3, 4, and 5 (without repetition) is divisible by 6, we need to follow these steps:

### Step 1: Understand the conditions for divisibility by 6
A number is divisible by 6 if it is divisible by both 2 and 3.

### Step 2: Determine the conditions for divisibility by 2
For a number to be divisible by 2, its last digit must be even. The even digits available from our set are 0, 2, and 4.

### Step 3: Determine the conditions for divisibility by 3
A number is divisible by 3 if the sum of its digits is divisible by 3. The sum of the digits 0, 1, 2, 3, 4, and 5 is:
\[ 0 + 1 + 2 + 3 + 4 + 5 = 15 \]
Since 15 is divisible by 3, any combination of 5 digits from this set will also be divisible by 3.

### Step 4: Calculate the total number of 5-digit numbers
To form a 5-digit number, we cannot start with 0. Therefore, the first digit can be any of 1, 2, 3, 4, or 5 (5 options). After choosing the first digit, we have 5 remaining digits (including 0) to choose from, and we can arrange these in any order.

1. Choose the first digit (5 options: 1, 2, 3, 4, 5).
2. Choose the remaining 4 digits from the remaining 5 digits (0 and the 4 remaining digits).
3. The number of arrangements of these 4 digits is \(4!\).

Thus, the total number of 5-digit numbers is:
\[ 5 \times 5! = 5 \times 120 = 600 \]

### Step 5: Calculate the number of favorable outcomes (5-digit numbers divisible by 6)
#### Case 1: Last digit is 0
If the last digit is 0, the first digit can be any of 1, 2, 3, 4, or 5 (5 options). The remaining 3 digits can be chosen from the remaining 4 digits (1, 2, 3, 4, 5 minus the chosen first digit). The number of arrangements of these 4 digits is \(4!\).

Thus, the number of favorable outcomes when the last digit is 0 is:
\[ 5 \times 4! = 5 \times 24 = 120 \]

#### Case 2: Last digit is 2
If the last digit is 2, the first digit can be any of 1, 3, 4, or 5 (4 options, since 0 cannot be the first digit). The remaining 3 digits can be chosen from the remaining 4 digits (0 and the 3 remaining digits). The number of arrangements of these 4 digits is \(4!\).

Thus, the number of favorable outcomes when the last digit is 2 is:
\[ 4 \times 4! = 4 \times 24 = 96 \]

#### Case 3: Last digit is 4
If the last digit is 4, the first digit can be any of 1, 2, 3, or 5 (4 options, since 0 cannot be the first digit). The remaining 3 digits can be chosen from the remaining 4 digits (0 and the 3 remaining digits). The number of arrangements of these 4 digits is \(4!\).

Thus, the number of favorable outcomes when the last digit is 4 is:
\[ 4 \times 4! = 4 \times 24 = 96 \]

### Step 6: Total favorable outcomes
Now, we sum the favorable outcomes from all cases:
\[ 120 + 96 + 96 = 312 \]

### Step 7: Calculate the probability
The probability that a randomly formed 5-digit number is divisible by 6 is given by the ratio of favorable outcomes to total outcomes:
\[ P(\text{divisible by 6}) = \frac{\text{favorable outcomes}}{\text{total outcomes}} = \frac{312}{600} = \frac{26}{50} = \frac{13}{25} = 0.52 \]

### Final Answer
The probability that the number is divisible by 6 is \( \frac{13}{25} \) or 0.52.