Sock socks

We want a pair of socks out of 200 distinct pairs (I can't imagine how that many pairs would fit into a drawer, honestly). When we remove $3$ socks from $200$ pairs (i.e. $400$ socks), we have $\binom{400}{3}$ possible combinations.

Out of these combinations, we need $1$ pair out of $200$ pairs and another sock, which does not matter for all intents and purposes. The number of ways to select $1$ pair out of $200$ is given by $\binom{200}{1}$ while the other sock can be selected in $\binom{400 - 2}{1}$ ways (The $398$ in the numerator comes from the fact that two socks have already been withdrawn.

Hence, the probability is given by:

$\dfrac{\binom{200}{1} \times \binom{398}{1}}{\binom{400}{3}}$ = $\dfrac{1}{133}$

$a + b = 1 + 133 = \boxed{134}$

First observe there are $200 \cdot 2 = 400$ socks in total (I have answered the question wrong at first because of this...). Then, note that there are $2$ cases for the $3$ socks chosen to contain a pair of socks:

1. The first and second socks chosen are a pair, then the third sock can be chosen randomly. So the probability that this will occur is $1 \cdot \frac{1}{399} \cdot 1 = \frac{1}{399}$ . ----- first sock can be chosen randomly, so probability of $1$

2. The first and second socks chosen are not a pair. Hence, first and third socks or first and second socks can be a pair. Thus, probability of Case 2 occurring is $1 \cdot \frac{398}{399} \cdot \frac{2}{398} = \frac{2}{399}$ .

Adding up the probability of the two cases above, we obtain $\frac{1}{399} + \frac{2}{399} = \frac{1}{133}$ . Hence $a+b = 1+133 = \boxed{134}$ .

Since there are 200 pairs, there are 400 individual socks. So, let's pick a random sock for the first (the chance of not finding a pair is 1/1, because we are only picking 1 so far), and the chance we don't find its pair next is 398/399. Now, 2 socks have been picked, and the 2 pairing socks are still among the 398 yet be picked, so we have a 396/398 chance of not finding it. Multiplying these three fractions together and simplifying the result gives 132/133. Now we look at the question, it asks for the chances you do find a pair, which is the answer we obtained earlier taken from 1, which is 1/133, for a sum of 134.

Let us label the socks X, Y, Z. The chance that X,Y form a pair is $\frac{1}{399}$ .

Therefore, there is a $\frac{398}{399}$ chance that X,Y do not form a pair. If that is the case, then there is a $\frac{2}{398}$ chance that Z forms a pair with either X or Y.

Hence, the chance that any of the socks form a pair is:

$\frac{1}{399}+\frac{398}{399}\cdot\frac{2}{398} = \frac{3}{399}=\frac{1}{133}$

Which gives the answer $1 + 133 = \boxed{134}$ .

Only 3 cases.

1. AAX, where AA is a pair, and X is any sock. = 1x(1/399)x1=1/399.

2. ABA, where A and B are different socks = 1x(398/399)x(1/398) = 1/399.

3. ABB, where A and B are different socks. = 1x(398/399)x(1/398) = 1/399.

Total=3/399=1/133. So answer is 1+133=134.

There are a total of $400$ socks in the drawer. $200 \times 2$ pairs.

Now, three are picked at random. We can do this in $400 \choose 3$ ways. These are the total number of outcomes.

The favourable outcomes are those in which we have a pair of socks and then any other single sock. For example, consider the first pair denoted by 1,1

$\underline{11} \underline{2}$ or $\underline{11} \underline{3}$ or $\underline{11} \underline{4} \dots$ and so on

Thus for the pair, we have $200$ choices, and the random sock, we have $400 - 2 = 398$ choices (because for each pair, we have $398$ choices). Thus, the total number of favourable choices are $200 \times 398$

This gives us $\dfrac{200 \times 398}{400 \choose 3}$

or

$\dfrac{1}{133}$ , thus $1 + 133 = \boxed{134}$

Consider complementary counting. We know that the probability that we can select a pair of socks is the same as $1-\frac{\frac{400*398*396}{6}}{400 \choose 3}$ . Note that the numerator is the number of ways that we can't select a pair of socks. The $400*398*396$ is from the fact that we we select the first sock in $400$ ways, the second in $398$ ways (because we can't select the sock that pairs with the original sock we selected), and the last sock in $396$ ways using the same idea that we can't select the pair of our first and second sock. We divide this by $6$ because we have counted with order and wish to count without order because the numerator counts without order (remember that in counting problems we wish to count in similar ways throughout the problem). Thus, by solving this we get that the probability that we get a pair of socks is $\frac{1}{133}$ .

first we will find by how many ways we can't draw a pair of socks. now,

we have a total of 200*2=400 socks.so choose any socks out of 400. now we are left with 399 socks and since we don't

have to draw a pair so keep the sock which is pair of selected sock out of interest. so now we have to draw one sock out of

398 socks similarly avoiding its pair we are now left with 396 socks and we have to select 1.

so total no. of ways by which we can't draw a par=400 * 398 * 396.

total no. of ways of drawing 3 socks out of 400=400 * 399 * 398.

so, probability of not drawing a pair=(400 * 398 * 396) / (400 * 399 * 398)=132 / 133.

so probability of drawing a par of socks=1-(132 / 133)=1 / 133.

so , answer will 1+133=134.

200 pairs of socks = 400 pieces

total ways = C(400,3) = 200 * 133 * 398

ways in order to get a pair of sock = 1 . 1 .398 . 200 = 200. 398

so that, $\frac{a}{b}$ = $\frac{200*398}{200 * 133 * 398}$ $\frac{a}{b}$ = $\frac{1}{133}$

then, a + b = 1 + 133 = 144

There are 200 pairs, therefore 400 individual socks The total number of possible combinations of three socks would be 400C3 or 400 399 398/3 2 1 The total number among these triples that has a pair is 200 * 1 * 398 ( pick any sock for one 200, pick its pair for another, 1 and pick any other sock among the remaining 398)

We can then simplify it to be 200 * 398 / 400C3 Following that, we can make it 200 * 398 * 3 * 2 / 400 * 399 * 398 Then cancel similar terms (200*2 and 400, 398 and 398), we will have 3 / 399 as the equivalent value. The problem mentions that a/b are coprime positive integers so we reduce it to 1/133 a = 1, b = 133 therefore a+b = 134

This problem can be simplified greatly by realizing that:

P(3 socks contain a pair) = 1 - P(3 socks are distinct).

This is because there are only three possibilities for how the 3 socks are chosen.

1. All socks are distinct.

2. The socks contain one pair.

3. The socks are all the same.

The third possibility can be disregarded since a pair contains only 2 socks. Now, in order to calculate the desired probability, we first need to find out how many ways we can choose 3 socks from 400 (200 pairs).

This is equal to:

$\frac{(400!)}{(397!)(3!)} = 10,586,800$ possible combinations.

Now, we need to find the number of ways we can choose 3 distinct socks. This means that we take 3 socks out of 3 pairs of socks (one from each pair). We will also need to multiply this result by 8 since there are 8 ways $(2 * 2 * 2)$ we can choose the socks from the three pairs. This is equal to:

$\frac{(200!)}{(197!)(3!)} * 8 = 10,507,200$ possible combinations.

The desired probability is now:

$1 - \frac{10,507,200}{10,586,800} = \boxed{\frac{1}{133}}$ .

Therefore, the value of $a + b$ is equal to $(1 + 133)$ or $\boxed{134}$ .