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25 Apr 2026, 23:54
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
35%
(medium)
Question Stats:
63% (01:09) correct
37%
(01:45)
wrong
based on 30
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In a box, 7 different pairs of socks were kept. Two socks were chosen at random – What is the probability that a pair was chosen?
A) 1/13
B) 1/7
C) 1/14
D) 2/13
E) 2/7
A) 1/13
B) 1/7
C) 1/14
D) 2/13
E) 2/7
Doubt: Why won't the probability be 7/13? We find the probability as 14/14x 1/13 and then multiply bb 7 to account for all pairs.
26 Apr 2026, 01:16
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Good Probability question — and honestly one of those where the trap catches people who overthink it.
The concept being tested here is basic Combinatorics / Probability using combinations.
1. Count total socks: 7 pairs × 2 = 14 socks.
2. Total ways to choose any 2 socks from 14: C(14,2) = (14 × 13) / 2 = 91.
3. Favorable outcomes — how many ways to pick a matching pair? One way per pair type, so 7 ways total. (Pick either sock from pair #1, and the match is determined. Repeat for each of the 7 pairs.)
4. P(pair) = 7/91 = 1/13.
Answer is A.
The trap here is that some people calculate 7/C(14,2) and then second-guess themselves, thinking "wait, shouldn't there be more favorable outcomes?" There aren't. Once you fix which pair you're picking (7 choices), there's exactly one way to complete that pair, giving you 7 favorable outcomes.
The other classic trap is confusing this with ordered selection — but since we're just asking "did we get a pair," order doesn't matter. Stick with combinations.
Takeaway: when the problem says "chosen at random" with no mention of order, go straight to C(n,r).
The concept being tested here is basic Combinatorics / Probability using combinations.
1. Count total socks: 7 pairs × 2 = 14 socks.
2. Total ways to choose any 2 socks from 14: C(14,2) = (14 × 13) / 2 = 91.
3. Favorable outcomes — how many ways to pick a matching pair? One way per pair type, so 7 ways total. (Pick either sock from pair #1, and the match is determined. Repeat for each of the 7 pairs.)
4. P(pair) = 7/91 = 1/13.
Answer is A.
The trap here is that some people calculate 7/C(14,2) and then second-guess themselves, thinking "wait, shouldn't there be more favorable outcomes?" There aren't. Once you fix which pair you're picking (7 choices), there's exactly one way to complete that pair, giving you 7 favorable outcomes.
The other classic trap is confusing this with ordered selection — but since we're just asking "did we get a pair," order doesn't matter. Stick with combinations.
Takeaway: when the problem says "chosen at random" with no mention of order, go straight to C(n,r).
26 Apr 2026, 04:27
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Hi Edskore thanks for the reply. I had a small doubt. [color=#000000] Why won't the probability be 7/13? We find the probability as 14/14 x 1/13 and then multiply by 7 to account for all pairs.[/color]
Edskore
