From set of 20 natural no.s 2 are selected.

find probability that their sum is
1.odd
2.even
3.selected pair is twin prime

Please tell me if no.s are selected randomly then what is

If no.s are srlected randomly then what is probablity to above question

1. The probability that the sum of two natural numbers is odd can be calculated by considering the possible combinations. Out of the 20 natural numbers, half of them are odd and the other half are even. So, there are 10 odd numbers and 10 even numbers.

When selecting two numbers, there are three possibilities:
a) Select an odd number and an odd number. The number of combinations for this scenario is (10 choose 2) = 45.
b) Select an even number and an even number. The number of combinations for this scenario is (10 choose 2) = 45.
c) Select an odd number and an even number. The number of combinations for this scenario is 10 * 10 = 100.

The total number of possible combinations is (20 choose 2) = 190.

Since we are interested in the probability of selecting two numbers with an odd sum, we need to consider the first scenario from above (a). Therefore, the probability is 45/190 = 9/38.

2. For the probability that the sum of two numbers is even, we consider the same possibilities as above, but focus on scenarios b and c. This time, we are interested in the probability of selecting two numbers with an even sum.

The number of combinations for scenario b is 45, and for scenario c, it's 100. The total number of possible combinations is 190.

Therefore, the probability is (45 + 100)/190 = 145/190 = 29/38.

3. To calculate the probability of selecting a pair of twin prime numbers, we need to determine how many twin prime pairs exist within the given set of 20 natural numbers.

Twin primes are prime numbers that have a difference of 2 (e.g., 3 and 5, 11 and 13). From the given set, the twin prime pairs are: (3, 5), (5, 7), (11, 13), (17, 19). Therefore, there are four twin prime pairs.

The total number of possible combinations is (20 choose 2) = 190.

Thus, the probability of selecting a pair of twin prime numbers is 4/190 = 2/95.

To find the probability of the sum of two selected numbers from a set of 20 natural numbers, we need to first determine the total number of possible outcomes and the number of favorable outcomes for each scenario.

1. Probability that their sum is odd:
The set of 20 natural numbers contains 10 odd numbers (1, 3, 5, 7, 9, 11, 13, 15, 17, 19) and 10 even numbers (2, 4, 6, 8, 10, 12, 14, 16, 18, 20). To select two numbers whose sum is odd, we need to choose one odd number and one even number.

Total number of possible outcomes = Total number of ways to select two numbers from a set of 20 natural numbers =
20C2 = (20 * 19) / (2 * 1) = 190

Favorable outcomes = Number of ways to select one odd number (10 options) and one even number (10 options) = 10 * 10 = 100

Probability = Favorable outcomes / Total possible outcomes = 100 / 190 = 10/19

Therefore, the probability that the sum of two selected numbers from the set is odd is 10/19.

2. Probability that their sum is even:
To select two numbers whose sum is even, we need to choose either two even numbers or two odd numbers.

Total number of possible outcomes = Total number of ways to select two numbers from a set of 20 natural numbers =
20C2 = (20 * 19) / (2 * 1) = 190

Favorable outcomes:
- Number of ways to select two even numbers = 10C2 = (10 * 9) / (2 * 1) = 45
- Number of ways to select two odd numbers = 10C2 = (10 * 9) / (2 * 1) = 45

Favorable outcomes = Number of ways to select two even numbers + Number of ways to select two odd numbers = 45 + 45 = 90

Probability = Favorable outcomes / Total possible outcomes = 90 / 190 = 9/19

Therefore, the probability that the sum of two selected numbers from the set is even is 9/19.

3. Probability that the selected pair is twin prime:
Twin primes refer to pairs of prime numbers that are either consecutive (like 3 and 5) or differ by 2 (like 11 and 13).

To find the probability that the selected pair is twin prime from a set of 20 natural numbers, we need to count the number of favorable outcomes.

In the given set of 20 natural numbers, the twin prime pairs are: (3, 5), (5, 7), (11, 13), and (17, 19), totaling 4 pairs. Note that (2, 3) is not included since 2 is not considered a twin prime.

Total number of possible outcomes = Total number of ways to select two numbers from a set of 20 natural numbers =
20C2 = (20 * 19) / (2 * 1) = 190

Favorable outcomes = Number of twin prime pairs = 4

Probability = Favorable outcomes / Total possible outcomes = 4 / 190

Therefore, the probability that the selected pair is twin prime is 4/190, which can be simplified to 2/95.

If we add two evens, we get an even

If we add two odds, we get an even
If we add an even and an odd, we get an odd

Let's look at this simplified scenario, of 4 numbers

+1 . 2 .3 .4
1: 2 3 4 5
2: 3 4 5 6
3: 4 5 6 7
4: 5 6 7 8

clearly half the sum's are even, half are odd

for 20 numbers:
prob(odd, odd)
= (10/20)(10/20) = 1/4
prob(even,even) = 1/4

both of these events yield an even sum
prob(even sum) = 1/4 + 1/4 = 1/2

clearly prob(odd sum) = 1- 1/2 = 1/2

"twin primes" are defined as two prime numbers that are two apart.
There are just a few from out set:
3,5 5,7 11,13 17,19 and their reverses
so we only have 10 such pairs
prob(twin primes) = 10/400 = 1/40