On the fifth day of Christmas, Calvin gave to me

We can choose $2$ people out of $10$ in $^{10}C_{2}=45$ ways. But there are only $5$ couples. Hence the probability is $\frac{5}{45}=\boxed{\frac{1}{9}}$

There are 10 people. The Sample Space will be to select any 2 people out of 10.

Hence, $S = _{2}^{10}\textrm{C} = 45$

Now, we have to calculate the probability that the 2 people selected are a couple.

As we know there are 5 couples in the room,

So the number of ways to select any 1 couple out of 5 is $_{1}^{5}\textrm{C}$

Hence, the probability

$P = \frac{_{1}^{5}\textrm{C}}{_{2}^{10}\textrm{C}} = \frac{5}{45} = \frac{1}{9}$

That's the answer!

Let us pick the two people separately. Everyone has exactly one partner, so we can pick anyone for the first person. The second person has to be the one that corresponds with the first person out of the 9 remaining people. This means that we have 1 way of picking the right person and 9 possible people we can pick, so the answer is $\frac{1}{9}$

There are $\dbinom{10}{2} = 45$ ways to select any two people out of these 10.

There are $\dbinom{5}{1} = 5$ ways to select any couple out of these five couples.

Hence the probability is given by $\dfrac{5}{45} = \dfrac{1}{9} \Rightarrow \boxed{2}$

The first people we picked have 9 case with the second people. The second people we picked have 8 case with the second people (because it except the first people). And continue with the 3rd people... The total of group 2 people is 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = 45. The total of the couple is 5. So, the probability that they are a couple is 5/45 = 1/9.

There are 10.9 ways of picking (10 balls for 2 boxes), and probability to get they were couple is 5.2(AB is different with BA), so that could be5.2/10.9=1/9

There are 5 couples, thus the number of ways of selecting a couple is 5.(Favorable cases) The total number of ways of selecting 2 out of 10 is 10C2. (Sample space)

We have C(10,2) ways to choose 2 people, and only 5 are correct picks ( the five couples). So the answer is 5/C(10,2).

Request probability is equal to $\displaystyle \frac{5}{{10 \choose 2}}= \frac{1}{9}$

In 10 people (or in a set with 10 elements) there are C(10, 2) = 45 subsets each one having two elements. Five of these sets are our couples that they want to be married. So the probability we are looking for is 5/45 =1/9.

There are 5 couples, and 10 choose 2 ways to select the two people. 10 choose 2 equals $10 \times 9 \div 2 = 45$ . Therefore, the probability that the two that are selected are a couple is $\frac{5}{45} = \boxed{\frac{1}{9}}$ .