GreenlightTestPrep wrote:

A drawer contains 8 pairs of socks. For each sock, there is exactly one matching sock. If Ed randomly selects 6 socks without replacement, what is the probability that he will have at least one pair of matching socks?

A) 1/4

B) 19/51

C) 111/143

D) 3/4

E) 145/189

Here's an approach that uses probability rules.

When it comes to probability questions involving "at least," it's best to try using the

complement.

That is, P(Event A happening) = 1 - P(Event A

not happening)

So, here we get: P(at least 1 matching pair) = 1 -

P(zero matching pairs)P(zero matching pairs)P(zero matching pairs) = P(pick ANY sock 1st

AND 2nd sock doesn't match 1st sock

AND 3rd sock doesn't match other selections

AND 4th sock doesn't match other selections

AND 5th sock doesn't match other selections

AND 6th sock doesn't match other selections)

= P(pick ANY sock 1st)

x P(2nd sock doesn't match 1st sock)

x P(3rd sock doesn't match other selections)

x P(4th sock doesn't match other selections)

x P(5th sock doesn't match other selections)

x P(6th sock doesn't match other selections)

= 1

x 14/15

x 12/14

x 10/13

x 8/12

x 6/11

=

32/143So, P(at least 1 pair) = 1 -

P(zero pairs)= 1 -

32/143= 111/143

Answer:

ASIDE: Once we draw the first sock, there are 15 sock s remaining, and only 1 matches the 1st selection.

This means there are 14 sock s that DON'T match the first.

So, P(no match on 2nd draw) = 14/15

Then, once we draw the second sock (without a match), there are 14 socks remaining, and 2 of them match either the 1st or 2nd selection.

This means there are 12 socks that DON'T match.

So, P(no match on 3rd draw) = 12/14

etc

Cheers,

Brent

_________________

Brent Hanneson – Creator of greenlighttestprep.com

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