In a Iottery of 50 tickets numbered 1 to 50 , two tickets are drawn simultaneously . Find the probability that
a tickets has prime number.

Answer

To solve the problem step by step, we will follow these instructions:

### Step 1: Identify the total number of tickets
In this lottery, there are 50 tickets numbered from 1 to 50.

**Hint:** The total number of tickets is simply the range of numbers given.

### Step 2: Determine the prime numbers between 1 and 50
The prime numbers between 1 and 50 are:
2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47.
Counting these, we find there are a total of 15 prime numbers.

**Hint:** A prime number is a number greater than 1 that has no positive divisors other than 1 and itself.

### Step 3: Calculate the number of non-prime numbers
Since there are 50 tickets in total and 15 of them are prime, the number of non-prime tickets is:
50 - 15 = 35.

**Hint:** Non-prime numbers include composite numbers and the number 1.

### Step 4: Calculate the total ways to choose 2 tickets
The total number of ways to choose 2 tickets from 50 is given by the combination formula:
\[ \text{Total ways} = \binom{50}{2} = \frac{50 \times 49}{2} = 1225. \]

**Hint:** The combination formula \(\binom{n}{r}\) is used to find the number of ways to choose \(r\) items from \(n\) items without regard to the order of selection.

### Step 5: Calculate the favorable outcomes
We want to find the probability that at least one of the drawn tickets is a prime number. We can calculate this by finding the number of ways to choose 1 prime and 1 non-prime ticket:
- The number of ways to choose 1 prime ticket from 15 is \(\binom{15}{1}\).
- The number of ways to choose 1 non-prime ticket from 35 is \(\binom{35}{1}\).

Thus, the number of favorable outcomes is:
\[ \text{Favorable outcomes} = \binom{15}{1} \times \binom{35}{1} = 15 \times 35 = 525. \]

**Hint:** The product of combinations gives the total ways to select one from each category.

### Step 6: Calculate the probability
The probability that at least one of the tickets drawn is a prime number is given by:
\[ P(\text{at least one prime}) = \frac{\text{Favorable outcomes}}{\text{Total ways}} = \frac{525}{1225}. \]

### Step 7: Simplify the probability
To simplify \(\frac{525}{1225}\), we can divide both the numerator and the denominator by their greatest common divisor (GCD), which is 175:
\[ \frac{525 \div 175}{1225 \div 175} = \frac{3}{7}. \]

### Final Answer
The probability that at least one of the drawn tickets has a prime number is \(\frac{3}{7}\).

---