In the Department of Accounting, 45% of the students who attend the course of Statistics are freshmen, 35% are second-year students and the rest are third-year students and above. According to data from the Department, 15% of the first-year students who attend the course receive excellent marks, 25% of the second-year students receive excellent marks, while 30% of the third-year students and above receive excellent marks.
a) What is the probability that a student (regardless of the year of study) got an excellent grade?
b) If he got excellent, then what is the probability of being a second-year student?
c) If he didn't get excellent, then what is the probability of being a second-year student?
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a) We have that :
- $P(\text{1-Year})=45\%$
- $P(\text{2-Year})=35\%$
- $P(\text{3-Year})=(100-45-35)\%=20\%$
- $P(\text{1-Year}\cap \text{excellent})=15\%$
- $P(\text{2-Year}\cap \text{excellent})=25\%$
- $P(\text{3-Year}\cap \text{excellent})=30\%$
Is that correct so far?
The probability that a student (regardless of the year of study) got an excellent grade is equal to \begin{align*}P(\text{excellent})&=P(\text{1-Year}\cap \text{excellent})+P(\text{2-Year}\cap \text{excellent})+P(\text{3-Year}\cap \text{excellent})\\ & =15\%+25\%+30\%\\ & =70\%\end{align*}
Is that correct?
b) If he got excellent, then the probability of being a second-year student is equal to $$P(\text{2-Year}\mid \text{excellent})=\frac{P(\text{2-Year}\cap \text{excellent})}{P(\text{excellent})}=\frac{25\%}{70\%}=35.71\%$$ Is that correct?
c) If he didn't get excellent, then the probability of being a second-year student is equal to $$P(\text{2-Year}\mid \text{excellent}^c)=\frac{P(\text{2-Year}\cap \text{excellent}^c)}{P(\text{excellent}^c)}=\frac{P(\text{2-Year}\cap \text{excellent}^c)}{100\%-70\%}=\frac{P(\text{2-Year}\cap \text{excellent}^c)}{30\%}$$ Is that correct so far? How can we calculate the probability $P(\text{2-Year}\cap \text{excellent}^c)$ ?
$ \small\begin{aligned} P(Y1) &=45\% \\ P(Y2) &= 35\% \\ P(Y3) &=20\% \\ \end{aligned}$
$\small\begin{aligned} P(E~|Y1) &=15\% \\ P(E~|Y2) &= 25\% \\ P(E~|Y3) &=30\% \\ \end{aligned}$
For $(a)$,
$ \small P(E) = P(E~|Y1) ~P(Y1) + P(E~|Y2) ~P(Y2) + P(E~|Y3) ~P(Y3)$
$ \small = 0.15 \times 0.45 + 0.25 \times 0.35 + 0.30 \times 0.20$
$ = 0.215 ~ $ or $ ~ 21.5 \%$
For $(b)$, note that $ \small P(Y2|E) = \cfrac{P(E|Y2) ~P(Y2)}{P(E)}$
For $(c)$, first find $ \small P(E^c) = 1 - P(E), P(E^c|Y2) = 1 - P(E|Y2)$ and then using the same formula as $(b)$, you can find $ \small P(Y2|E^c)$.