A survey of students revealed that $30\ \%$ of them have a part-time job. If $5$ students are chosen at random what is the probability of the following events: That less than $2$ have part time work? More than $3$ are working part-time? None have a part-time job?
This is a question from a mock exam paper as I prepare for exams.
I guess I could start by imagining a number of $1000$ college students surveyed with 300 having part-time jobs.
so for part (i) -> $(30/100)(70/100)(70/100)(70/100)(70/100) = 7\ \%$
That ain't right. Can someone shed some light on this?
As the 5 students are chosen at random, the probability for each of them of having a part-time job is still $30\ $%. It furthermore has to assumed that there is independence among the students.
If you let $X$ be the random variable that equals the number of students among the 5 chosen that have a part-time job, then as Jean-Claude Arbaut points out in his comment, $X$ has the binomial distribution $\text{Bin}(5,0.3)$.
I'll show you how to find the probability of less than $2$ having a part-time job. Less than $2$ means at most $1$, and we have $$P(X<2) = P(X\leq 1) = P(X=1) + P(X=0) = \binom{5}{1}(0.3)^1(0.7)^4 + \binom{5}{0}(0.3)^0(0.7)^5.$$ Since $\binom{n}{k} = \frac{n!}{k!(n-k)!}$, we have $\binom{5}{1} = 5$ and $\binom{5}{0} = 1$. Also, $x^0 = 1$ for any $x>0$, and thus $$P(X<2) = 5(0.3)^1(0.7)^4 + (0.7)^5.$$ This turns out to be approximately $52.8\ $%.
Edit: Allow me to make a comment about your approach: You suggest that the solution is $$(30/100)\cdot (70/100)\cdot (70/100)\cdot (70/100)\cdot (70/100).$$ This is the same as $$0.3\cdot 0.7\cdot 0.7\cdot 0.7\cdot 0.7,$$ or $$0.3\cdot (0.7)^4.$$ Notice how this is close to $P(X=1)$, which is $5\cdot (0.3)\cdot (0.7)^4$. The reason why your answer is off by a factor of $5$ is that you consider only the case where the first student is the one with a part-time job, and where the remaining $4$ students don't have part-time jobs. It could also have been the second student, who had a part-time job, while student $1$, $3$, $4$, and $5$ didn't. By changing which student has the part-time job, you get \begin{align*} &0.3\cdot 0.7\cdot 0.7\cdot 0.7\cdot 0.7\\ +\ &0.7\cdot 0.3\cdot 0.7\cdot 0.7\cdot 0.7\\ +\ &0.7\cdot 0.7\cdot 0.3\cdot 0.7\cdot 0.7\\ +\ &0.7\cdot 0.7\cdot 0.7\cdot 0.3\cdot 0.7\\ +\ &0.7\cdot 0.7\cdot 0.7\cdot 0.7\cdot 0.3.\\ \end{align*} Notice how all five terms are just products of the form $0.3\cdot (0.7)^4$. Since there are $5$ of them, this gives you a total of $5\cdot 0.3\cdot (0.7)^4$.
This shows you how you get from your idea to the correct expression for $P(X=1)$. But the question asks for the probability that there are less than $2$ students with a part-time job, so $0$ students with part-time jobs is also an option. This is why my approach above has an extra term compared to your suggestion.