In packing for a trip, Sarah puts three pairs of socks - one

Hi Brunei,

Can you help in explaining the answer in detail.
I still cant understand how you have plugged in the numbers.

\(1 - P(Exactly 1 Pair)\) \(=1 - ((\frac{2_C_1*2_C_1*2_C_2}{6_C_4})*3)\) \(=1 - (\frac{12}{15})=\) \(\frac{3}{15}\) \(=\frac{1}{5}\)

I know this is a longer approach, but is this right?


NOTE: I understood that this could happen in 3 ways. So, I multiplied by 3.

But I'm never completely sure how to always consider the NUMBER OF ARRANGEMENTS in certain questions.
I get confused when there are different elements to be considered and when there are similar elements.
Can someone help me with this?

Thanks in advance.

Hi,

I think \((\frac{2_C_1*2_C_1*2_C_2}{6_C_4})\) represents a case where in 1 type of matching socks are pulled out but there 3 pairs and any one of them can be pulled out so I think it should be

\(1 - 3*(\frac{2_C_1*2_C_1*2_C_2}{6_C_4})\)

1-12/15 or 3/15 or 1/5

Yes, but you need to be more specific about the kind of problems you are talking about. Small things change the entire question in P&C so let me know the exact questions that confuse you.

easier solution: inverse the problem. Sarah picking up 2 pair of matching socks and keeping one pair in the bag is equivalent to Sarah picking up 1 pair of socks and keeping 2 pairs in the bag. Now let's see what's the probability of that happening when done at random. Sarah picks up the 1st one, can be any color and so probability = 1. Sarah picks up the 2nd one, now for it to be of the same color as the 1st one, she has to pick 1 exact piece out of 5, probability of which is 1/5. So option A.

Favorable outcomes > selecting 2 pairs out of 3 = 3c2 = 3

Total outcomes > selecting 4 socks randomly = 6c4 = 15

Probability = 3/15 = 1/5

experts please tell whats wrong in this

6/6*1/5*4/4*1/3 *6|/2*2 =2/5

Hi Guys, can somebody explain what does this C mean? where can I learn the theory? I have never seen this terminology in the books... It is killing me

Thanks!

C stands for combinations:

Combinatorics Made Easy!

Theory on Combinations

DS questions on Combinations
PS questions on Combinations

Hope it helps.

Thank you Bunuel!!!

Can someone please explain in this format? :

[2/6*1/5*2/4*1/3]*3 = 1/30 is my answer...

How can i get a 6 in the numerator so that the answer becomes 1/5? What would the logic be?

Isn't it supposed to be \(C^4_6\) = picking 4 socks out of 6 (i.e 2 pairs)?

Hi Bunuel,

Why is the below approach incorrect?

6c1*1*4c1*1/(4!*6c4) Firstly, we are selecting one sock out of 6, then we can select second in just one way as we are assuming that it is the second sock of the pair and so on; since the combination can be any so divide by 4! and total cases are 6c4.

i did it using probability directly
(6/6 * 1/5 * 4/4 * 1/3 ) * 3C2 ways = (1/15) * 3 = 1/5

6/6 - we can choose any of the 6 socks in the first one
1/5 - we should pick the exact pair of the first sock picked so only 1 way out of the remaining 5
4/4 - again we can pick any of the remaining 4 socks
1/3 - we have to pick the exact pair of the sock picked.

3C2 - because we could have either red + blue pair or red+ green pair or blue+green pair so 3 ways.

We have three scenarios of two matching pairs: 1) a red pair and a blue pair; 2) a red pair and a green pair; 3) a blue pair and a green pair. Let’s start with the probability of selecting a red pair and a blue pair. To select a red pair and a blue pair is to select two red socks and two blue socks. So let’s assume the first two socks are red and the last two socks are blue; the probability of selecting these socks in that order is:

P(R, R, B, B) = 2/6 x 1/5 x 2/4 x 1/3 = 1/6 x 1/5 x 1/3 = 1/90.

However, the two red socks and the two blue socks, in any order, can be selected in 4!/(2! x 2!) = 24/4 = 6 ways. Thus, the probability of two red socks and two blue socks is:

P(2R and 2B) = 1/90 x 6 = 6/90 = 1/15.

Using similar logic, we see that the probability of pulling a red pair and a green pair is 1/15, and so is the probability of pulling a blue pair and a green pair. Thus, the total probability is:

1/15 + 1/15 + 1/15 = 3/15 = 1/5.

Alternate Solution:

From a total of 6 socks, two pairs, i.e., 4 socks, can be pulled in 6C4 = 6!/(4! 2!) = (6 x 5)/2 = 3 x 5 = 15 ways.

Three of these choices contain two matching pairs, namely: 1) a red pair and a blue pair, 2) a blue pair and a green pair; 3) a red pair and a green pair.

Therefore, the probability of pulling two matching pairs is 3/15 = 1/5.

Answer: A

Plz let me know if my approach is wrong-

Total number of socks = 6

Total number of ways Sarah can pick up 4 socks = 6C4 = 15 ways

No. of ways he can pick matching shoes = no.of ways 2 red socks + 2 blue socks + 2 green socks

= 2C2 + 2C2 + 2C2

= 1 + 1 + 1

= 3 ways

Hence

P(Matching socks) =3/15 = 1/15

Hi JeffTargetTestPrep,


Can you please suggest why my approach is incorrect?