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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.

  • Attempt at solution:

    • 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,

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

  • 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

$\begin{pmatrix} 10\\4 \end{pmatrix}.4^4(1 -.4)^{10-4}$

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    Hints: For the first question, _read_ in your book about the _law of total probability_ and about _Bayes' formula_. For the second question, there is _no_ binomial distribution involved. The question really is asking, if we choose at random one of the sets of $4$ students from among all possible subsets of $4$ students out of $10$, what is the probability that we have chosen the specific subset of $4$ cheaters?2012-11-12

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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 $\dfrac{\binom{4}{3}\binom{6}{1}}{\binom{10}{4}}$.

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    Gotcha, thanks again!2012-11-12