What is the probability that university will not have enough dormitory rooms?

A university admitted $2500$ students.However the university has room spots for only $1786$ students. If there is a $70$% chance that an admitted student will accept the offer and attend the university, what it the probability that university will not have enough dormitory rooms?

My analysis:
If there is $70$% that an admitted student accept the offer and attend the university then among the $2500$ students $1750$ would accept the offer. The probability that the university will not have enough dormitory places would be:

$1786-1785/100=0.36$

However the correct answer must be 0.0559 according to the multiple choice question.

Can you tell me what’s wrong with my analysis and what should be the correct way of thinking?

• Are you familiar with the Binomial distribution? Jul 14 '18 at 13:41
• @Chris2006 : Using Binomial law is far to be a good issue.
– Surb
Jul 14 '18 at 13:44
• use approximation with normal distribution. Jul 14 '18 at 13:45
• I edited some line breaks into the question for clarity but I have no idea how you get your $0.36$ result. Pease review, correct the formula and clarify with a little .explanation. Jul 14 '18 at 13:52

No it doesn't work. Let $X_i$ denote if the $i-$th student accept or not. Then $X_i\sim Bernoulli(0.7)$. Set $$S_n=X_1+...+X_n.$$

What you have to compute is $$\mathbb P\{S_{2500}\geq 1787\},$$ and to do this, you have to use Central Limit theorem.

• First I would like to thanks you for your explanation. However, I did not really understand, if Xi denotes if the i-th student accept or not,what is the meaning of Sn and how does a Bernoulli differ from a binomial theorem. Why will we have to compute Sn greater or equal to 1787 and what is the central limit theorem? Jul 14 '18 at 13:49
• $S_n$ denote the number of student that accepted among the $n$ admitted student. The law of $S_n$ will be indeed a Binomial, i.e. rigorously $$\mathbb P\{S_{2500}\geq 1787\}=\sum_{k=1787}^{2500}\binom{2500}{k}0.7^{k}0.3^{2500-k},$$ but good luck to compute this sum. So you need to use central limit theorem that tels you that $S_n\sim \mathcal N(2500\cdot 0.7, 50\cdot 0.7\cdot 0.3)$ when $n$ is large enough (in particular, $n=2500$ is large enough). You want to compute the probability that there is not enough room. If $S_{2500}\leq 1786$, the university will have enough room.
– Surb
Jul 14 '18 at 13:57
• So what you are interested at, is when $S_{2500}>1786$, or in an equivalent way, that $S_{2500}\geq 1787$.
– Surb
Jul 14 '18 at 13:59

Let us say $X$ to be $2500$ students who will attend this university. So, $X$ has a binomial distribution with trails $n=2500$ and $p=0.70$. Also we are given in the question that the university has dorm spots for only $1786$ freshman students. So, $X\ge1787$ and now we can use normal approximation to binomial to find the probability. $$\mu=np=2500\times0.7=1750$$ $$\sigma=\sqrt{np(1-p)}=\sqrt{2500\times0.7\times0.3}\approx23$$ Now, $$P(X>1787)=P\left(Z>\frac{1787-1750}{23}\right)=P\left(Z>\dfrac{37}{23}\right)=P(Z>1.61)=1-0.946=0.054$$

• Good clear explanation, although the "2000" in your first sentence is perhaps a remnant of an earlier draft? It strikes me that a real-world evaluation would be highly sensitive to that value of 0.7; I wonder what the error on that is... Jul 14 '18 at 15:24

$X\sim Binomial(2500,0.7)\approx Y\sim Normal(1750,525)$ Where $X$ is the number of people that atend the University.
$P(X>1786)=1-P(X\le1786)\approx1-P(Y<1786.5)$ $=1-P(Z<\frac{1786-1750}{\sqrt{525}})=1-P(Z<1.593)$

using probability tables $1-P(Z<1.593)=1-0.9441=0.0559$

• You don't explain anything here. Not even your notation. Jul 14 '18 at 14:09
• What do the variables $X$, $Y$, represent? What are the functions $B$ and $N$? What is the relation represented by $\sim$? Jul 14 '18 at 14:12