Question

a drawer contains two pairs of red socks, three pairs of yellow socks, and one pair of black socks. one sock is taken from the drawer and put back after checking its color. a second sock is then taken out. what is the probability that the first and the second are yellow? a 1/14 b 5/6 c 1/4 d 1/6

Explanation:

Step1: Calculate total number of sock pairs

Total pairs = 2 (red) + 3 (yellow) + 1 (black) = 6 pairs. Since each pair has 2 socks, total socks = 6×2 = 12. But since we are dealing with pairs (and replacement, so probability per draw is based on pairs or socks? Wait, actually, when we take a sock, the probability of yellow: number of yellow socks is 3×2 = 6. Total socks 12. But wait, maybe easier to think in terms of pairs (since each draw is independent with replacement). Wait, no, the problem says "one sock is taken... put back... second sock". So it's sampling with replacement.

First, find number of yellow socks: 3 pairs × 2 = 6 socks. Total socks: (2+3+1) pairs × 2 = 6×2 = 12 socks.

Probability first sock is yellow: $\frac{6}{12} = \frac{1}{2}$? Wait, no, wait: 2 pairs red: 4 socks, 3 pairs yellow: 6 socks, 1 pair black: 2 socks. Total socks: 4 + 6 + 2 = 12. So yellow socks: 6. So probability first yellow: $\frac{6}{12} = \frac{1}{2}$. But wait, the answer options don't have 1/2. Wait, maybe the problem is considering pairs? Wait, the question says "two pairs of red socks" – maybe each "pair" is a single unit? Wait, maybe the problem is using "pair" as a single sock? No, that doesn't make sense. Wait, maybe the problem has a typo, or I misread. Wait, the options: A is 1/14 (wrong), B 5/6, C 1/4, D 1/6. Wait, let's re-express: maybe the problem is considering "pairs" as the unit, i.e., each "pair" is one item? So total items: 2 + 3 + 1 = 6 items (each item is a pair). Then, number of yellow items: 3. So probability first is yellow: 3/6 = 1/2. Second is yellow: 3/6 = 1/2. Then probability both yellow: 1/2 × 1/2 = 1/4. Ah! That must be it. The problem is treating each "pair" as a single unit (maybe the question is using "pair" but actually each "pair" is a sock? No, that's confusing. But given the options, let's check: if we consider pairs as the unit (so each draw is a pair, but the problem says "one sock". Wait, no, the problem says "one sock is taken". So maybe the problem has an error, but let's see the options. If we consider that the number of yellow "units" (socks) is 3 pairs, but maybe the problem is using "pairs" as the count for probability. Wait, 3 yellow pairs out of 6 total pairs. So probability first sock (from a pair) is yellow: since each pair has 2 socks, but if we consider that when you take a sock, the probability of yellow is (3 pairs / 6 pairs) × (2 socks per pair / 2 socks per pair) – no, that's not right. Wait, maybe the problem is actually about pairs (i.e., each "pair" is a single sock, which is a mistake, but let's go with the options. If total pairs: 6, yellow pairs: 3. Probability first is yellow pair: 3/6 = 1/2. Second is yellow pair: 3/6 = 1/2. Then 1/2 × 1/2 = 1/4, which is option C. So that must be the intended approach: treating each pair as a single unit (even though the problem says "sock"), maybe a translation error or wording error. So the correct answer is C. $\frac{1}{4}$.

Answer:

C. $\frac{1}{4}$