In a lottery of 10 tickets numbered 1 to 10, two tickets are drawn simultaneously. What is the probability that both the tickets drawn have prime numbers?
Probability of Drawing Prime Numbers in a Lottery
The problem involves a lottery with 10 tickets numbered from 1 to 10. We need to find the probability that when two tickets are drawn simultaneously, both tickets have prime numbers.
Understanding the Total Outcomes
First, let's determine the total number of ways to draw two tickets simultaneously from a set of 10 tickets. Since the order of drawing does not matter, this is a combination problem. The total number of ways to choose 2 tickets from 10 is given by the combination formula:
\(C(n, k) = \dfrac{n!}{k!(n-k)!}\)
Here, \(n = 10\) (total number of tickets) and \(k = 2\) (number of tickets drawn).
Total number of outcomes = \(C(10, 2) = \dfrac{10!}{2!(10-2)!} = \dfrac{10!}{2!8!} = \dfrac{10 \times 9}{2 \times 1} = 45\)
So, there are 45 possible combinations of two tickets that can be drawn from the 10 tickets.
Identifying Favorable Outcomes
Next, we need to identify the prime numbers between 1 and 10. A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself.
The prime numbers between 1 and 10 are: 2, 3, 5, 7.
There are 4 prime numbers in the set of tickets.
We want both tickets drawn to have prime numbers. This means we need to choose 2 tickets from these 4 prime numbered tickets. The number of ways to do this is also a combination problem:
Here, \(n = 4\) (total number of prime tickets) and \(k = 2\) (number of prime tickets to draw).
Number of favorable outcomes = \(C(4, 2) = \dfrac{4!}{2!(4-2)!} = \dfrac{4!}{2!2!} = \dfrac{4 \times 3}{2 \times 1} = 6\)
So, there are 6 ways to draw two tickets that are both prime numbers.
Calculating the Probability
The probability of an event is calculated as the ratio of the number of favorable outcomes to the total number of possible outcomes.
Probability (Both tickets are prime) = \(\dfrac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}\)
Probability = \(\dfrac{6}{45}\)
We can simplify this fraction by dividing both the numerator and the denominator by their greatest common divisor, which is 3.
Probability = \(\dfrac{6 \div 3}{45 \div 3} = \dfrac{2}{15}\)
Conclusion
The probability that both the tickets drawn have prime numbers is \(\dfrac{2}{15}\).
Similar Questions
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If \(\rm P(A\cup B)=\dfrac{5}{6}, P(A\cap B)=\dfrac{1}{3}\:and\:P(\bar A)=\dfrac{1}{2}\) , then which of the following is/are correct?
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Select the correct answer using the code given below.
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