# Probability about students and exams.

In exams, $80\%$ of the students go prepared and $20\%$ go unprepared.

From the prepared ones, $90\%$ passes the exams.

From the unprepared one, only $10\%$ passes the exams.

First question: If a student passes the exams, what is the probability for him to be prepared?

Second question: From $4$ randomly chosen students, what's the probability that at least 3 passed the exams?

• Hint: Law of total probability and Bayes' theorem are the tools you need to answer the first question. Sep 8 '13 at 19:31
• Why did you remove self-learning? Sep 8 '13 at 20:27
• Cause hints get me more confused than helped :( Sep 8 '13 at 20:29
• I think you are right though, in the end... it's self learning ... with a big help from this community. So this is getting retagged! :) Sep 8 '13 at 21:53

Let's write it in Terms: $$A:=\text{Student is prepared.}$$ $$B:=\text{Student passes the exam.}$$ We are given \begin{align*} P(A) & = 0.8 \\ P(\bar{A}) & = 0.2 \\ P(B|A) & = 0.9 \\ P(B|\bar{A}) & = 0.1 \end{align*} The question asks for
1. $P(A|B)$ and
2. $P(B)$ combined to a Bernoulli experiment.

1. $P(A|B)%$
By Bayes' theorem we have $$P(A|B) = \frac{P(B|A) P(A)}{P(B)}$$ And by total probability we have $$P(B) = P(B|A) P(A) + P(B|\bar{A}) P(\bar{A}) = 0.9 \cdot 0.8 + 0.1 \cdot 0.2 = 0.74$$ So $$P(A|B) = \frac{0.9 \cdot 0.8}{0.74} \approx 0.97$$

2. Bernoulli
The Question is for $$X := \sum_{i=1}^4 X_i$$ With $P(X_i = 1) = P(B), X_i \in \{0,1\}$ find $$P(X \geq 3) = P(X=3) + P(X=4)$$ For $P(X=4)$ we only have $(B,B,B,B)$, so $$P(X=4) = P(B)\cdot P(B) \cdot P(B) \cdot P(B) = P(B)^4 \approx 0.30$$ For $P(X=3)$ we have $(B,B,B,\bar{B}), (B,B,\bar{B},B), (B,\bar{B},B,B), (\bar{B},B,B,B)$, so $$P(X=3) = 4\cdot P(B)^3 \cdot P(\bar{B}) = 4\cdot 0.74^3 \cdot 0.26 \approx 0.42$$ And thus $$P(X\geq 3) \approx 0.72$$

In fact, the general formula given $P(X_i) = p$ is $$P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}$$ Where $n$ is the sample size (here $n=4$).

• I'll elaborate on 2. as well, then. Also $A\cup \bar{A} = \Omega$ if $\Omega$ denotes the sample space, yes. Usually, youuse the union sign ($\cup$) instead of the $+$. Sep 8 '13 at 20:19
• @user2692669 Is that sufficient? $$\binom{n}{k} = \frac{n!}{k!(n-k)!}$$ The binomial coefficients. The distribution of $X$ is therefor called binomial distribution Sep 8 '13 at 20:25
• $X$ is a (discrete) random variable with binomial distribution. It represents the number of Students (from the sample pick of $n=4$), who passed the exam. Sep 8 '13 at 20:34
• @user2692669 Somewhat, yes. It's an ugly way of writing $$1 = 1^n = (p + (1-p))^n = \sum_{k=0}^n \binom{n}{k} p^k (1-p)^k$$ So this proves that we, in fact, have a distribution ($\sum_{x\in\Omega} P(X=x) = 1$) Sep 8 '13 at 20:48
• @user2692669 Haha thanks a bunch. The last sentence is a note, that for a distribution $P_X$ per definition we must have, that the probability of the total event, i.e. $$P(\Omega) = P(X\in\Omega) = \sum_{x\in\Omega} P(X=x)$$ must equal to $1$. Sep 8 '13 at 20:54

The hard part is that prepared and pass both start with p. So we need to come up with letters other than $P$ as labels for events.

Let $S$ be the event the student Studied. Let $E$ be the event she passed the Exam. We are asked for the conditional probability $\Pr(S|E)$. By the definition of conditional probability, $$\Pr(S|E)=\frac{\Pr(S\cap E)}{\Pr(E)}.\tag{1}$$

Let us calculate $\Pr(E)$. The event passing the exam can happen in two ways (i) studied and passed or (ii) did not study but passed.

The probability of (i) is $(0.8)(0.9)$. The probability of (ii) is $(0.2)(0.1)$.

Add. We get $\Pr(E)=0.74$.

We found $\Pr(S\cap E)$ already while finding $\Pr(E)$. It is $(0.8)(0.9)$. Divide as per Formula (1).

For the second problem, the probability of passing is $p=0.74$. The probability at least $3$ pass is the probability that $3$ pass plus the probability that $4$ pass. That is a standard binomial distribution problem, with answer $$\binom{4}{3}p^3(1-p)+\binom{4}{4}p^4(1-p)^0.$$