Four digit numbers with different digits are formed using the digits `1,2,3,4,5,6,7,8`. One number from them is picked up at random. The chance that the selected number contains the digit '1' is

Options

A. `1//8`
B. `1//4`
C. `1//2`
D. none of these

Explanation

To solve the problem of finding the probability that a randomly selected four-digit number (formed using the digits 1, 2, 3, 4, 5, 6, 7, 8) contains the digit '1', we can follow these steps:

### Step 1: Calculate the Total Number of Four-Digit Numbers
We need to form four-digit numbers using the digits 1 to 8, ensuring that all digits are different.

- The total number of ways to choose 4 digits from 8 is given by the combination formula \( \binom{n}{r} \), where \( n \) is the total number of items to choose from, and \( r \) is the number of items to choose.
- After choosing 4 digits, we can arrange them in \( 4! \) (factorial of 4) ways.

Thus, the total number of four-digit numbers is:
\[ \text{Total numbers} = \binom{8}{4} \times 4! \]

### Step 2: Calculate the Number of Favorable Outcomes (Numbers Containing '1')
Next, we need to find the number of four-digit numbers that contain the digit '1'.

- If '1' is included, we need to choose 3 more digits from the remaining 7 digits (2, 3, 4, 5, 6, 7, 8).
- The number of ways to choose 3 digits from these 7 is \( \binom{7}{3} \).
- Again, these 4 digits (including '1') can be arranged in \( 4! \) ways.

Thus, the number of favorable outcomes is:
\[ \text{Favorable outcomes} = \binom{7}{3} \times 4! \]

### Step 3: Calculate the Probability
The probability that a randomly selected four-digit number contains the digit '1' is given by the ratio of the number of favorable outcomes to the total number of outcomes:

\[ P(\text{contains '1'}) = \frac{\text{Favorable outcomes}}{\text{Total numbers}} = \frac{\binom{7}{3} \times 4!}{\binom{8}{4} \times 4!} \]

### Step 4: Simplify the Probability
The \( 4! \) terms cancel out:

\[ P(\text{contains '1'}) = \frac{\binom{7}{3}}{\binom{8}{4}} \]

Now, we can calculate the combinations:
- \( \binom{7}{3} = \frac{7!}{3!(7-3)!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35 \)
- \( \binom{8}{4} = \frac{8!}{4!(8-4)!} = \frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1} = 70 \)

Thus, the probability becomes:

\[ P(\text{contains '1'}) = \frac{35}{70} = \frac{1}{2} \]

### Final Answer
The probability that the selected number contains the digit '1' is \( \frac{1}{2} \).

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