Law of total probability occurring 
In a box we have $100$ indistinguishable coins. Each coin has a $1$ on
  one side and a $0$ on the other side. $20$ percent of the coins are
  not biased, but the rest are biased, where side $1$ has a $70$ percent
  probability of ocurring. 
a.) If we randomly grabbed a coin from the box and threw the chosen
  coin just once what is the probability to get a $1$? Also, what is the
  conditional probability that the coin we threw was biased?
In a classroom of 10 we know that 4 students cheated.
a) If we randomly choose 4 students, what is the probability that we
  get exactly the 4 students that cheated. Also, what will be the
  probability of exactly 3 students who have cheated.



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*Attempt at solution:


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*For the first part of the first question, I know that $20\%$ of the coins are not biased so they have a $50\%$ of $0$ or $1$ occurring on a toss. But then how can I calculate the probability of getting a 1? Will it just be $[.20 \dot\ \frac{1}{2} + .7]$? For the second part of the question, since they are asking the probability of choosing a bias coin then can I use binomial distribution to calculate that probability? Thus, 



$$\begin{pmatrix} 100\\1 \end{pmatrix}.7^1\left(1-.7\right)^{100-1}$$. 


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*For the second question, I know that in total their are $10$ students. So the probability of choosing the 4 that cheated is a binomial distribution. Thus, I have that 


$$\begin{pmatrix} 10\\4 \end{pmatrix}.4^4(1 -.4)^{10-4}$$
 A: You came close to the right idea for the probability of getting a $1$. With probability $0.2$, we used a fair coin, and with probability $0.8$ we used an unfair coin. So the required probability is
$$(0.2)(0.5)+(0.8)(0.7).$$
I assume that what is wanted for the second part is the probability that the coin is biased, given that we got a $1$. Let $B$ be the event the coin is biased, and let $W$ be the event we got a $1$. We want $\Pr(B|W)$. By the usual formula for conditional probabilities, we have
$$\Pr(B|W)=\frac{\Pr(B\cap W)}{\Pr(W)}.\tag{$1$}$$
We have already computed $\Pr(W)$. To find $\Pr(B\cap W)$ is not hard, we already sort of did it. We have $\Pr(B)=0.8$. Given that the coin is biased, the probability of $W$ is $0.7$, so $\Pr(B\cap W)=(0.8)(0.7)$. Now we have all the information we need to use the basic formula $(1)$.
For the cheating problem, there are $\dbinom{10}{4}$ equally likely ways to select $4$ people. There is only $1$ way (or if you prefer, $\dbinom{4}{4}$ ways) to choose cheaters only, so our probability is $\dfrac{\binom{4}{4}}{\binom{10}{4}}$.
For exactly $3$ cheaters, these can be chosen in $\dbinom{4}{3}$ ways, and for each of these ways, there are $\dbinom{6}{1}$ ways to select the non-cheater who will help make up the group of $4$. So the required probability is $\dfrac{\binom{4}{3}\binom{6}{1}}{\binom{10}{4}}$. 
