A drawer contains 8 pairs of socks. For each sock, there is exactly

Prob of choosing 6 socks so that there is no pair among them= 8C6 (2C1)^6
We are doing 2c1 coz any one one sock can b chosen among a pair
P (atleast one pair )= 1- p (no pair)

= 1- [ 8c6 * 2^6] / 16c6
=111/143

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Hi Brent,

I tried to use combinations. What am I doing wrong?

P(at least 1 pair) = 1 - P(no pairs)

P(no pairs) = (No. of ways 6 unpaired socks can be picked from 8 unpaired socks)/(total number of ways 6 socks can be picked from 16 socks)

= 8C6/16C6
= 1/286

P(at least 1 pair) = 1 - 1/286 = 285/286

Of course, answer not among choices.

Hi dramamur,

The problem with your solution is that you are treating the socks differently when calculating the numerator and denominator.
For the denominator, you are treating the 16 socks as being 100% unique. It's as though you are saying that each pair of matching socks has a LEFT sock and a RIGHT sock.
So, we can select 6 socks from 16 UNIQUE socks in 16C6 ways.

However, for the numerator, you are essentially saying that the two socks in each matching pair are identical.
You are not making a LEFT/RIGHT distinction for each pair. Instead, you are choosing 6 COLORS from a total of 8 COLORS.

Let's keep going with your solution, BUT we'll make sure we consider the fact that we have a LEFT and RIGHT sock for each pair.

So, we'll have to alter your numerator somewhat.
First, we can select 6 COLORS from a total of 8 COLORS in 6C2 ways (28 ways)
For the 1st COLOR, we can choose either the LEFT sock or the RIGHT sock. We can do this in 2 ways.
For the 2nd COLOR, we can choose either the LEFT sock or the RIGHT sock. We can do this in 2 ways.
For the 3rd COLOR, we can choose either the LEFT sock or the RIGHT sock. We can do this in 2 ways.
For the 4th COLOR, we can choose either the LEFT sock or the RIGHT sock. We can do this in 2 ways.
For the 5th COLOR, we can choose either the LEFT sock or the RIGHT sock. We can do this in 2 ways.
For the 6th COLOR, we can choose either the LEFT sock or the RIGHT sock. We can do this in 2 ways.
So, the total number of ways to select 6 UNMATCHED socks = (28)(2)(2)(2)(2)(2)(2)

We'll keep the numerator the same.
That is, we can select 6 socks from 16 UNIQUE socks in 16C6 ways.

So, P(ZERO matching socks) = (28)(2)(2)(2)(2)(2)(2) /16C6
= 32/143

P(at least 1 pair) = 1 - P(no pairs)
= 1 - 32/143
= 111/143

Cheers,
Brent

I don't get why this is wrong:

probability he won't pick a matching sock:

(8/8 * 6/7 *5/6*4/5*3/4*2/3) = 2/7 that he won't pick a matching sock.

I guess i didn't factor in the fact that we have L and R socks. I assumed they were of different colors.

in that case it would be (16/16 * 14/15 * 12/14 * 10/13 * 8/12 *6/11)= 32/143

1- 32/143= 111/43

The problem starts with the part in blue above

P(Choose any sock 1st) 16/16 = 1
P(Choose 2nd sock that does NOT match 1st sock) = 14/15
P(Choose 3rd sock that does NOT match 1st sock OR the 2nd sock) = 12/14 (there are 14 socks remaining, and 1 sock matches 1st sock, and 1 sock matches 2nd sock)
P(Choose 4th sock that does NOT match 1st sock OR the 2nd sock OR the 3rd sock) = 10/13 (there are 13 socks remaining, and 1 sock matches 1st sock, and 1 sock matches 2nd sock, and 1 sock matches 3rd sock)
etc

Cheers,
Brent

8 pairs = 16 socks
1- 16/16*14/15*12/14*10/13*8/12*6/11 = 111/143
IMO C

Hi ,

Can you tell me whats wrong with my solution
Chances of picking the 1 st sock =16
Chances of picking the 2 nd sock =14

and so on

So my final answer was 1-(16*14*12*10*8*6/16C6)
What's wrong with my approach?

Because order of the socks doesn't matter, you have to divide (16*14*12*10*8*6) by 6!.

because you are mixing two approaches probability and combination one. chance of picking the first sock is 16/16 and second is 14/15 every sock have different you can't take just 16C6

Solution:


Instead of finding the probability that Ed will have at least one pair of matching socks, let’s find the probability that he will have no pairs of matching socks. After all, the former probability is the latter probability subtracted from 1.

Since he has 8 pairs of socks, he has 16 socks. Therefore, he has 16C6 ways to choose 6 socks from 16. Let’s say each pair of socks is of a different color. In other words, there are 8 colors and 2 socks of each color. There are 8C6 ways to choose 6 colors from 8 and for each color there are 2 ways to choose a sock of that color. Therefore, there are 8C6 x 2 x 2 x 2 x 2 x 2 x 2 = 8C6 x 2^6 ways to 6 socks of different colors (i.e., no socks of matching color). Therefore, the probability that he will have no pairs of matching socks is:

8C6 x 2^6 / 14C6 = 28 x 64 / 8008 = 28 x 8 / 1001 = 4 x 8 / 143 = 32/143

Finally, the probability that he will have at least one pair of matching socks is 1 - 32/143 = 111/143.

Alternate Solution:

Instead of finding the probability that Ed will have at least one pair of matching socks, let’s find the probability that he will have no pairs of matching socks and then subtract that probability from 1.

Our approach is to use the multiplication rule for probabilities when we are drawing without replacement.

The first draw has probability 16/16 = 1. For the second draw, there are 15 socks remaining, and he can choose any sock except the match to the first sock, so there are 14 available socks, and the probability is 14/15. For the third draw, there are 14 socks remaining and he can pick any of 12 socks to avoid a match, so the probability is 12/14. Similarly, for the fourth draw, there are 13 remaining socks, and he can pick any of 10 socks to avoid a match, so the probability is 10/13. Notice that for each pick, the number of socks available to be chosen decreases by 1, but the number of socks available to pick, to avoid a match, decreases by 2. Thus, we have:

P(no matches in 6 picks) = 16/16 x 14/15 x 12/14 x 10/13 x 8/12 x 6/11

After canceling and simplifying, we have:

P(no matches in 6 picks) = 32/143

Thus, the probability of at least one match = 1 - 32/143 = 111/143.

Answer: C

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