If i roll 2 dice what is the probability that neither of them are prime nor even?

Mathematics

1 answer:

Eduardwww [97]3 years ago

4 0

Answer:

There are 3 prime numbers shown on the die: 2, 3 and 5. The probability of showing a prime number on a single die is 1/2, hence the probability of not showing a prime number is also 1/2. Total possible outcomes when two dice are rolled = 6*6 = 36. Total possible outcomes when two dice are rolled = 6*6 = 36.

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Prove that there are infinitely many primes of the form 4k + 3, where k is a non-negative integer. [Hint: Suppose that there are

Simora [160]

Answer:

The prove is as given below

Step-by-step explanation:

Suppose there are only finitely many primes of the form 4k + 3, say {p1, . . . , pk}. Let P denote their product.

Suppose k is even. Then P ≅ 3^k (mod 4) = 9^k/2 (mod 4) = 1 (mod 4).

ThenP + 2 ≅3 (mod 4), has to have a prime factor of the form 4k + 3. But pₓ≠P + 2 for all 1 ≤ i ≤ k as pₓ| P and pₓ≠2. This is a contradiction.

Suppose k is odd. Then P ≅ 3^k (mod 4) = 9^k/2 (mod 4) = 1 (mod 4).

Then P + 4 ≅3 (mod 4), has to have a prime factor of the form 4k + 3. But pₓ≠P + 4 for all 1 ≤ i ≤ k as pₓ| P and pₓ≠4. This is a contradiction.

So this indicates that there are infinite prime numbers of the form 4k+3.