Question

1. from a set of 10 cards numbered 1 to 10, two cards are drawn at random without replacement. what is the probability that:

(checked) probability = 2/9
5 you got it!
(3b) one is even and one is odd?
(checked) probability = 5/9
5 nice work!
3c the sum of the two numbers is 12?
probability = enter your next step here

Explanation:

Step1: Find total number of ways to draw 2 cards

We use the combination formula \( C(n, k)=\frac{n!}{k!(n - k)!} \), where \( n = 10 \) and \( k = 2 \). So \( C(10,2)=\frac{10!}{2!(10 - 2)!}=\frac{10\times9}{2\times1}=45 \).

Step2: Find number of favorable outcomes (sum = 12)

We list the pairs of numbers from 1 - 10 that sum to 12: (2,10), (3,9), (4,8), (5,7). Note that (6,6) is not valid as we draw without replacement. So there are 4 pairs? Wait, no: (2,10), (10,2), (3,9), (9,3), (4,8), (8,4), (5,7), (7,5). Wait, no, when drawing without replacement, the order of drawing matters in terms of counting the number of ways, but in combinations, we consider unordered pairs. Wait, actually, when we draw two cards, the number of ordered pairs is \( P(10,2)=10\times9 = 90 \), but if we use combinations, the unordered pairs. Wait, let's clarify. If we use combinations (unordered), the pairs that sum to 12 are (2,10), (3,9), (4,8), (5,7). So 4 unordered pairs. But if we use permutations (ordered), it's 8 ordered pairs. Wait, but let's check the total number of ways. The problem says "drawn at random without replacement". The total number of possible outcomes is the number of ways to draw 2 cards, which is \( C(10,2)=45 \) (unordered) or \( P(10,2)=90 \) (ordered). Let's see the previous parts. In part 3a and 3b, the probabilities were \( \frac{2}{9} \) and \( \frac{5}{9} \). Let's check: \( \frac{2}{9}=\frac{10}{45} \), \( \frac{5}{9}=\frac{25}{45} \). So total outcomes are 45 (combinations). So we need to find the number of unordered pairs that sum to 12. Let's list them:
- 2 and 10: sum 12
- 3 and 9: sum 12
- 4 and 8: sum 12
- 5 and 7: sum 12
- 6 and 6: not possible (no replacement)
So there are 4 unordered pairs? Wait, no: 2+10=12, 3+9=12, 4+8=12, 5+7=12. So 4 pairs? Wait, but 12 can also be 11, but we only have up to 10. Wait, 1+11 is not possible. So the pairs are (2,10), (3,9), (4,8), (5,7). So 4 unordered pairs. Wait, but let's check the count. Wait, 2 and 10: 1 pair, 3 and 9: 1, 4 and 8:1, 5 and 7:1. So 4 pairs. Wait, but let's calculate the number of favorable outcomes. Wait, maybe I made a mistake. Let's list all possible pairs (unordered) that sum to 12:

Numbers are 1 - 10. Let \( a < b \), \( a + b = 12 \), \( a \geq1 \), \( b \leq10 \).

\( a = 2 \), \( b = 10 \) (2+10=12)

\( a = 3 \), \( b = 9 \) (3+9=12)

\( a = 4 \), \( b = 8 \) (4+8=12)

\( a = 5 \), \( b = 7 \) (5+7=12)

\( a = 6 \), \( b = 6 \) (invalid, no replacement)

So there are 4 unordered pairs. Wait, but the total number of ways is \( C(10,2)=45 \). So the probability would be \( \frac{4}{45} \)? No, that can't be. Wait, maybe I messed up the total number of outcomes. Wait, in part 3a, the probability was \( \frac{2}{9}=\frac{10}{45} \), part 3b was \( \frac{5}{9}=\frac{25}{45} \). Let's check part 3a: maybe it was "both even" or "both odd"? Wait, the first part (3a) probably had a probability of \( \frac{2}{9} \), which is \( \frac{10}{45} \). Let's see: number of even numbers from 1 - 10: 5 (2,4,6,8,10), odd numbers: 5 (1,3,5,7,9). If we draw two cards, number of ways to draw two even: \( C(5,2)=10 \), two odd: \( C(5,2)=10 \). So total for both even or both odd: 20, but maybe part 3a was something else. Anyway, back to 3c: sum is 12. Let's list all ordered pairs (since when drawing without replacement, the order matters in terms of the draw, but maybe the problem is using combinations). Wait, no, let's use permutations. Total number of ordered pairs: \( 10\times9 = 90 \). Number of ordered pairs that sum to 12:

For (2,10): (2,10) and (10,2) → 2

For (3,9): (3,9) and (9,3) → 2

For (4,8): (4,8) and (8,4) → 2

For (5,7): (5,7) and (7,5) → 2

Total ordered pairs: 2×4 = 8

So probability is \( \frac{8}{90}=\frac{4}{45} \)? But that doesn't match the previous fractions. Wait, maybe I made a mistake in the pairs. Wait, 12 can also be 1 + 11 (invalid), 2 + 10, 3 + 9, 4 + 8, 5 + 7, 6 + 6 (invalid). So that's 4 unordered pairs, 8 ordered pairs. Now, let's check the total number of ways. If we use combinations (unordered), total is 45, favorable is 4, so probability \( \frac{4}{45} \approx 0.088 \). But let's see the previous parts. In part 3b, the probability was \( \frac{5}{9}=\frac{25}{45} \). Let's check what 3b was: "one is even and one is odd". Number of ways: 5 even, 5 odd. So number of unordered pairs: 5×5 = 25. So \( \frac{25}{45}=\frac{5}{9} \), which matches. So total outcomes are 45 (combinations). So for 3c, number of unordered pairs that sum to 12: as we listed, (2,10), (3,9), (4,8), (5,7) → 4 pairs. So probability is \( \frac{4}{45} \)? Wait, no, wait: 2+10=12, 3+9=12, 4+8=12, 5+7=12. That's 4 unordered pairs. So \( \frac{4}{45} \) is the probability? Wait, but let's check again. Wait, 5 and 7: 5+7=12, yes. 4 and 8: 12, 3 and 9:12, 2 and 10:12. So 4 pairs. So the number of favorable outcomes is 4, total is 45. So probability is \( \frac{4}{45} \)? Wait, but maybe I missed a pair. Wait, 6 and 6: no, replacement not allowed. 1 and 11: no. So 4 pairs. So the probability is \( \frac{4}{45} \), which simplifies to \( \frac{4}{45} \) (can't simplify further). Wait, but let's check with ordered pairs. 8 ordered pairs, total 90, so \( \frac{8}{90}=\frac{4}{45} \), same result. So that's the probability.

Answer:

\(\frac{4}{45}\)