To understand the difference, could anybody please provide solution

without

using the

complement

rule?

That's a lot of work, but I'll get the ball rolling, so someone else can finish the question.

P(at least 1 pair among the 6 selected socks) = P(1 pair and the other 4 socks are different OR 2 pair and the other 2 socks are different OR 3 pair)
= P(1 pair and the other 4 socks are different) + P(2 pair and the other 2 socks are different) + P(3 pair)

You'll see that things get VERY complicated VERY quickly.

Cheers,
Brent

In such cases it is easier to find out how not to pick a sock that does not create a pair.

The first pick can be anything, so \(P(first pick)\) = \(16\) out of \(16\)\( = \frac{16}{16} = 1\)

The second pick, to ensure that it does not pair with the first pick, must consists of picking \(14\) out of the remaining \(15\)

\(P(second pick) = \frac{14}{15}\)

The third pick, to ensure it does not pair with either of the two already picked, must consist of picking \(12\) out of the remaining \(14\)

\(P(third pick) = \frac{12}{14} \)

The fourth pick, to ensure it does not pair with any of the the three already picked, must consist of picking \(10\) out of the remaining \(13\)

\(P(fourth pick) =\frac{10}{13}\)

The fifth pick, to ensure that it does not pair with any of the four already picked, must consist of picking \(8\) out of the remaining \(12\)

\(P(fifth pick) = \frac{8}{12}\)

The sixth pick, to ensure that it does not pair with any of the five already picked, must consist of picking \(6\) out of the remaining \(11\)

\(P(sixth pick) = \frac{6}{11}\)

So the total probability of not picking a sock that pairs with the ones already picked is

\(\dfrac{16}{16} \times \dfrac{14}{15} \times \dfrac{12}{14} \times \dfrac{10}{13} \times \dfrac{8}{12} \times \dfrac{6}{11} = \dfrac{32}{143}\)

The total probability of picking at least a pair is \(1-\dfrac{32}{143} = \dfrac{111}{143}\)

The answer is Choice C.

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