What is the probability that an endorsed candidate will be selected to serve on a committee? $10$ people are being considered to serve as a representative in a committee of $3$ people. Each candidate is equally likely. The President has expressed his support for $2$ of these $10$ candidates. Assuming, ideally, that the President’s support does not influence outcomes:
i. What is the probability that exactly one of these two candidates will be selected for the committee?
ii. What is the probability that at least one of these two will be selected?
Ans:


*

*$\dfrac{2C1}{10C3}$

*$\dfrac{2C1+2C2}{10C3}$
Is it correct?
 A: Well for the first question you have to choose one of those candidates who President supports them and after that you have to choose two people from those 8 person who President doesn't support them.   So, the answer is:
$$\dfrac{{\sf^2C_1} \times {\sf^8C_2}}{{\sf^{10}C_3}}$$
For part II you have to choose one of them from those two who president supports them and choose two people from those eight who president doesn't support them, or you have to choose both two people who president support them and then choose one person from those eight people who president doesn't support them.   So the answer is: 
$$\dfrac{{\sf^2C_1}\times{\sf^8C_2}+{\sf^2C_2}\times{\sf^8C_1}}{{\sf^{10}C_3}}$$
A: $$ \newcommand{\1}[2]{{}^{#1}{\mathbb C}_{#2}}1)\frac{\121\182}{\1{10}3}2)\frac{\1{10}3-\183}{\1{10}3}=\frac{\121\182+\122\181}{\1{10}3}$$
You are correct in selecting presiden't candidates but youare forgetting to select rest of the candidates to make up for total 3.
A: Suppose A,B are the persons who have the president's support.If exactly one of them is to be selected look at the possibilities, with A included only 2 people can be selected from the other 8 since B is excluded.Therefore the number of possibilities are 8C2.Similarly the possible combinations with B selected are 8C2.Hence the total number of favorable combinations are 2*(8C2).Total number of outcomes are 10C3. So probability=(8C2+8C2)/(10C3) for your first question.
A: The probability that president's 1st preference a will get selected is $1 - 9/10 \cdot 8/9 \cdot 7/8$ or $.3$. This is because the probability of the intersection of him not being selected three times is equal to the odds of him not being selected at each stage multiplied together. If you then take $1$ minus this value you get the odds of him being selected. The president's second preference likewise has the same chance of being selected. Therefore, the odds of them both being selected (i.e. the intersection of the two events) is $.3 \cdot .3$ or $.09$. What you want, however, is the likelihood that at least one will be selected. That is the definition of the union of sets. You can now use basic probability identities to find your answer.
$P(A \cup B) = P(A) + P(B) - P(A \cap B)$
$P(\text{at least one selected}) = .3 + .3 - .09 = .51$
