The numbers 1-10 are placed on cards, and the cards are put in a hat. If two cards are drawn, in order, without replacement, what is the probability that

a. both are composite?
b. one number is even and the other is odd?
c. one number is even and the other a prime?

a) composites are 6,8,9, and 10

prob(both composites) = C(4,2)/C(10,2) = 6/45 = 2/15

b) prob(one even and one odd) = C(5,1)C(5,1)/C(10,2)
= 25/45 = 5/9

4 primes: 2,3,5,7
5 even:2,4,6,8,10

prob(one even and the other odd)
Case 1: the 2 is in there
number of ways for the other number is 3
Case 2: the 2 is out
number of ways is 3x4 = 12
total is 15
prob(one even the other prime) = 15/45 = 1/3

To solve these probability questions, we need to understand the concepts of composites, primes, even numbers, and odds.

A prime number is a positive integer greater than 1 that is divisible only by 1 and itself (e.g., 2, 3, 5, 7, 11, 13, etc.).

A composite number is a positive integer greater than 1 that is divisible by more than just 1 and itself (e.g., 4, 6, 8, 9, 10, 12, etc.).

An even number is any integer that is divisible by 2 (e.g., 2, 4, 6, 8, 10, etc.).

An odd number is any integer that is not divisible by 2 (e.g., 1, 3, 5, 7, 9, etc.).

Let's solve each part of the problem separately:

a. To find the probability that both cards drawn are composite numbers, we first need to determine the total number of cards in the hat, which is 10.

Out of these 10 numbers, 4 are primes (2, 3, 5, 7) and 6 are composites (4, 6, 8, 9, 10).

We need to draw two cards without replacement, which means after drawing the first card, there will be one less card to choose from.

The probability of drawing a composite number on the first draw is 6/10 (6 composites out of 10 total cards).

After removing one card, the probability of drawing a composite number on the second draw is 5/9 (since there will be 5 composites out of 9 remaining cards).

To find the probability of both events happening, we multiply the probabilities together:

P(both composite) = (6/10) * (5/9) = 30/90 = 1/3 (or approximately 0.333)

Therefore, the probability that both cards drawn are composite numbers is 1/3 or approximately 0.333.

b. To find the probability that one number drawn is even and the other is odd, we follow a similar approach.

Out of the 10 numbers, there are 5 even numbers (2, 4, 6, 8, 10) and 5 odd numbers (1, 3, 5, 7, 9).

The probability of drawing an even number on the first draw is 5/10 (5 even numbers out of 10 total cards).

After removing one card, the probability of drawing an odd number on the second draw is 5/9 (since there are 5 odd numbers left out of 9 remaining cards).

Since we have two possible scenarios (even-odd or odd-even), we add both probabilities:

P(one even and one odd) = (5/10) * (5/9) + (5/10) * (5/9) = 25/90 + 25/90 = 50/90 = 5/9 (or approximately 0.556)

Therefore, the probability that one card drawn is even and the other is odd is 5/9 or approximately 0.556.

c. To find the probability that one number drawn is even and the other is a prime number, we again apply the same logic.

Out of the 10 numbers, there are 5 even numbers (2, 4, 6, 8, 10) and 4 prime numbers (2, 3, 5, 7).

The probability of drawing an even number on the first draw is 5/10 (5 even numbers out of 10 total cards).

After removing one card, the probability of drawing a prime number on the second draw is 4/9 (since there are 4 prime numbers left out of 9 remaining cards).

Again, since we have two possible scenarios (even-prime or prime-even), we add both probabilities:

P(one even and one prime) = (5/10) * (4/9) + (5/10) * (4/9) = 20/90 + 20/90 = 40/90 = 4/9 (or approximately 0.444)

Therefore, the probability that one card drawn is even and the other is a prime number is 4/9 or approximately 0.444.

I hope this explanation helps you understand how to solve these probability problems!