Three numbers are drawn at random successively without replacement from a set S = {1, 2,…... 10}. Let p be the probability that the minimum of the chosen is 4 or their maximum is 8, then 4p = ______

Answer

To solve the problem, we need to find the probability \( p \) that the minimum of the chosen numbers is 4 or their maximum is 8 when three numbers are drawn from the set \( S = \{1, 2, \ldots, 10\} \). We will denote the events as follows:

- Let \( A \) be the event that the minimum of the chosen numbers is 4.
- Let \( B \) be the event that the maximum of the chosen numbers is 8.

We need to find \( 4p \) where \( p = P(A \cup B) \).

### Step 1: Calculate \( P(A) \)

Event \( A \) occurs when the minimum number is 4. This means we must choose the number 4 and two other numbers from the set \( \{5, 6, 7, 8, 9, 10\} \).

- The total numbers available from 5 to 10 are \( 6 \) (i.e., \( 5, 6, 7, 8, 9, 10 \)).
- We need to choose \( 2 \) numbers from these \( 6 \).

The number of ways to choose \( 2 \) from \( 6 \) is given by \( \binom{6}{2} \).

\[ \binom{6}{2} = \frac{6!}{2!(6-2)!} = \frac{6 \times 5}{2 \times 1} = 15 \]

The total ways to choose any \( 3 \) numbers from \( 10 \) is \( \binom{10}{3} \).

\[ \binom{10}{3} = \frac{10!}{3!(10-3)!} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120 \]

Thus, the probability \( P(A) \) is:

\[ P(A) = \frac{\text{Number of favorable outcomes for A}}{\text{Total outcomes}} = \frac{15}{120} = \frac{1}{8} \]

### Step 2: Calculate \( P(B) \)

Event \( B \) occurs when the maximum number is 8. This means we can choose any two numbers from the set \( \{1, 2, 3, 4, 5, 6, 7\} \) along with the number 8.

- The total numbers available from 1 to 7 are \( 7 \).
- We need to choose \( 2 \) numbers from these \( 7 \).

The number of ways to choose \( 2 \) from \( 7 \) is given by \( \binom{7}{2} \).

\[ \binom{7}{2} = \frac{7!}{2!(7-2)!} = \frac{7 \times 6}{2 \times 1} = 21 \]

Thus, the probability \( P(B) \) is:

\[ P(B) = \frac{\text{Number of favorable outcomes for B}}{\text{Total outcomes}} = \frac{21}{120} = \frac{7}{40} \]

### Step 3: Calculate \( P(A \cap B) \)

Event \( A \cap B \) occurs when the minimum is 4 and the maximum is 8. This means we must choose the numbers 4 and 8, and one more number from the set \( \{5, 6, 7\} \).

- The total numbers available from 5 to 7 are \( 3 \).
- We need to choose \( 1 \) number from these \( 3 \).

The number of ways to choose \( 1 \) from \( 3 \) is given by \( \binom{3}{1} \).

\[ \binom{3}{1} = 3 \]

Thus, the probability \( P(A \cap B) \) is:

\[ P(A \cap B) = \frac{\text{Number of favorable outcomes for A and B}}{\text{Total outcomes}} = \frac{3}{120} = \frac{1}{40} \]

### Step 4: Calculate \( P(A \cup B) \)

Using the formula for the probability of the union of two events:

\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \]

Substituting the values we calculated:

\[ P(A \cup B) = \frac{1}{8} + \frac{7}{40} - \frac{1}{40} \]

To add these fractions, we need a common denominator. The least common multiple of \( 8 \) and \( 40 \) is \( 40 \).

Converting \( \frac{1}{8} \) to a fraction with a denominator of \( 40 \):

\[ \frac{1}{8} = \frac{5}{40} \]

Now substituting back:

\[ P(A \cup B) = \frac{5}{40} + \frac{7}{40} - \frac{1}{40} = \frac{5 + 7 - 1}{40} = \frac{11}{40} \]

### Step 5: Calculate \( 4p \)

Now we can find \( 4p \):

\[ 4p = 4 \times P(A \cup B) = 4 \times \frac{11}{40} = \frac{44}{40} = \frac{11}{10} \]

Thus, the final answer is:

\[ \boxed{\frac{11}{10}} \]