There are 16 balls numbered from 1 to 16 in a box. Four balls are randomly drawn from the box simultaneously. A. Find the total number of possible outcomes. B. Find the probability of the following events: * A: The numbers drawn form a geometric progression with a common ratio $q = 2$ in increasing order. * B: The numbers drawn form an arithmetic progression with a common difference $d = 3$ in increasing order.

1. Problem Description

There are 16 balls numbered from 1 to 16 in a box. Four balls are randomly drawn from the box simultaneously.

A. Find the total number of possible outcomes.

B. Find the probability of the following events:

* A: The numbers drawn form a geometric progression with a common ratio $q = 2$ in increasing order.

* B: The numbers drawn form an arithmetic progression with a common difference $d = 3$ in increasing order.

2. Solution Steps

A. Total number of possible outcomes:

Since we are choosing 4 balls out of 16 without regard to order, we can use combinations. The total number of possible outcomes is given by:

$$C(n, k) = \frac{n!}{k!(n-k)!}$$

where $n = 16$ and $k = 4$.

$$C(16, 4) = \frac{16!}{4!(16-4)!} = \frac{16!}{4!12!} = \frac{16 \times 15 \times 14 \times 13}{4 \times 3 \times 2 \times 1} = 2 \times 5 \times 14 \times 13 = 1820$$

Thus, the total number of possible outcomes is

1

8

2

0.

B. Probability of event A (Geometric Progression with $q = 2$):

We need to find the number of sets of 4 numbers in increasing order from 1 to 16 that form a geometric progression with common ratio

2. The sets must be of the form $\{a, 2a, 4a, 8a\}$. Since the maximum number is 16, we must have $8a \le 16$, so $a \le 2$.

If $a = 1$, the set is $\{1, 2, 4, 8\}$.

If $a = 2$, the set is $\{2, 4, 8, 16\}$.

There are only two possible geometric progressions with a common ratio of

2. The probability of event A is:

$$P(A) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{2}{1820} = \frac{1}{910}$$

C. Probability of event B (Arithmetic Progression with $d = 3$):

We need to find the number of sets of 4 numbers in increasing order from 1 to 16 that form an arithmetic progression with a common difference

3. The sets must be of the form $\{a, a+3, a+6, a+9\}$. Since the maximum number is 16, we must have $a+9 \le 16$, so $a \le 7$.

The possible values of $a$ are $1, 2, 3, 4, 5, 6, 7$.

The sets are:

$\{1, 4, 7, 10\}$

$\{2, 5, 8, 11\}$

$\{3, 6, 9, 12\}$

$\{4, 7, 10, 13\}$

$\{5, 8, 11, 14\}$

$\{6, 9, 12, 15\}$

$\{7, 10, 13, 16\}$

There are 7 such arithmetic progressions.

The probability of event B is:

$$P(B) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{7}{1820} = \frac{1}{260}$$

3. Final Answer

A. Total number of possible outcomes: 1820

B. Probability of event A: $\frac{1}{910}$

C. Probability of event B: $\frac{1}{260}$