Probability Problem Inquiry In an examination, the probability of getting a credit is 1/3. If four students are selected at random, what is the probability that at least one of them got a credit?
From the book:
$P = 1 - F$
$P$ = 1 - (2/3)(2/3)(2/3)(2/3) = 65/81
The book solved the problem by considering the complement (opposite) of the term "at least one" from the formula: 1 = P(A) + P(A'). I want to solve the problem by normal means, that is by following the favorable outcome of at least 1. Sadly, I can't seem to get the answer.
My analysis:
$P$(getting a credit) = $1$
$P$(at least one of them get a credit) = ?
$P$(Not getting a credit) = $1$ - $(\frac{1}{3}) = \frac{2}{3}$
n(objects) = 4
The problem asked us to get the probability of at least one of them got a credit; so it means that out of the 4, the possible outcomes are: 
(One of them got a credit and the other 3 didn't) OR (two of them got a credit and the other 2 didn't) OR (three of them got a credit and 1 didn't) OR (All four of them got a credit) 
So if we put it into mathematical expression:
$(\frac{1}{3})(\frac{2}{3})^3$ + $(\frac{1}{3})^2(\frac{2}{3})^2$ + $(\frac{1}{3})^3(\frac{2}{3})$ + $(\frac{1}{3})^4(\frac{2}{3})^0$ = 5/27
Question: From my analysis above, I don't know where I got wrong.. I think i got the possible outcomes right. So what is wrong with my analysis? Where did I go wrong? How can I avoid it next time? If you get the right solution, would you mind explaining it clearly?
Any help would be appreciated. Thank you in advance.
 A: Consider the case when $1$ got credit and the other $3$ didn't. You wrote the probability as $(1/3)(2/3)^3 = 8/81$, but which student got the credit? Could've been the first one, or the second, third or fourth.
You need to multiply this probability by the number of configurations that allow $1$ student to get a credit, and $3$ not to get a credit. Since there are $4$ such ways, the first term in your expression is multiplied by $4$. To elaborate a bit more, probability of:


*

*student $1$ getting credit and not the others $=8/81$,

*student $2$ getting credit and not the others $=8/81$,

*student $3$ getting credit and not the others $=8/81$,

*student $4$ getting credit and not the others $=8/81$.


So overall probability of exactly one of the students getting credit $=32/81$. Try this out with the other terms and you'll get the same answer: $65/81$.
A: Calling the students $S_1,S_2,S_3,S_4$ the probability that exactly one gets a credit equalizes:$$\sum_{i=1}^4\mathsf P(S_i\text{ gets credit and the others do not})=\sum_{i=1}^4\left(\frac13\right)^1\left(\frac23\right)^3=\mathbf4\left(\frac13\right)^1\left(\frac23\right)^3$$
So not $\left(\frac13\right)^1\left(\frac23\right)^3$ as you suggest.
You are dealing here with binomial distribution with parameters $n=4$ and $p=\frac13$. 
Then the probability that at least one of the students gets a credit is: $$\sum_{k=1}^4\binom4k\left(\frac13\right)^k\left(\frac23\right)^{4-k}=\frac{65}{81}$$
A: Since the probability for each student receiving credit is the same, the Binomial distribution applies.  The probability of exactly $k$ successes in $n$ trials, each of which has probability $p$ of success is 
$$\Pr(X = k) = \binom{n}{k}p^k(1 - p)^{n - k}$$
where $p^k$ is the probability of $k$ successes, $(1 - p)^{n - k}$ is the probability of $n - k$ failures, and $\binom{n}{k}$ is the number of ways of selecting which $k$ of the $n$ trials result in a success.  
In your attempt, you did not take into account the way the successes could be distributed among the students.  
Since you are interested in the probability that at least one student got a credit, we need to find 
$$\Pr(X \geq 1) = \Pr(X = 1) + \Pr(X = 2) + \Pr(X = 3) + \Pr(X = 4)$$
Using the formula above with $n = 4$ and $p = 1/3$ gives
\begin{align*}
\Pr(X \geq 1) & = \sum_{k = 1}^{4} \binom{4}{k}\left(\frac{1}{3}\right)^k\left(\frac{2}{3}\right)^{4 - k}\\
              & = \binom{4}{1}\left(\frac{1}{3}\right)\left(\frac{2}{3}\right)^3 + \binom{4}{2}\left(\frac{1}{3}\right)^2\left(\frac{2}{3}\right)^2 + \binom{4}{3}\left(\frac{1}{3}\right)^3\left(\frac{2}{3}\right)^1 + \binom{4}{4}\left(\frac{1}{3}\right)^4\left(\frac{2}{3}\right)^0\\
             & = \frac{32}{81} + \frac{24}{81} + \frac{8}{81} + \frac{1}{81}\\
             & = \frac{65}{81}
\end{align*}
as you found by subtracting the probability that no student received credit from $1$, which is easier.  The factor $\binom{4}{k}$ that you are missing in your calculations accounts for which $k$ of the four students receives credit.
