1
$\begingroup$

Only one of the three years is repeat in the faculty (year $I$ or year $II$ or year $III$). The probability that the student will repeat the first year is $0.2$, the second year $0.5$, and the third year $0.25$. The probability of a student graduating a faculty while repeating the first year is $0.4$, graduating a faculty while repeating a second year is $0.5$, and graduating a faculty while repeating a third year is $0.9$.

$a)$ find the probability that any student who repeats one year (whether $I$, $II$ or $III$) will graduate from faculty

$b)$ if we know that the student has graduated from the faculty, find the probability that he / she has repeating the second year;

My attemp is: Write the set $A={\text {graduated the faculty}}$, and $B={\text {repeat the faculty}}$

$I^c,II^c,III^c,$ student repeat the first, second or third year in the faculty $$P(I^c)=0,25$$ $$P(II^c)=0,5$$ $$P(III^c)=0,25$$ and $$P(A|I^c)=0.4$$ $$P(A|II^c)=0.5$$ $$P(A|III^c)=0.9$$

I didnt know how to continue. Please help me. Thanks for your help. Thanky very much

$\endgroup$

1 Answer 1

0
$\begingroup$

Here are some hints.

Start from first principles. $$\Pr(A|B)=\frac{\Pr(A\cap B)}{\Pr(B)}$$

In part a), A is the event that the student repeats a year, and $B$ is the event that a student graduates. For $A$, we have $$A=I^c\cup II^c\cup III^c$$ and we are given that $I^c, II^c, III^c$ are mutually exclusive. Therefore,

$$\Pr(A\cap B) = \Pr(I^c\cap B)+\Pr(II^c\cap B)+\Pr(III^c\cap B)\tag1$$

Can you compute this probability?

For the denominator, we just have to add the probability that the student graduates without repeating any year to the numerator we computing in $(1)$.

What is that probability?

In part b), we want the probability that the student repeated the second year, given that he has graduated. Since we know the probability that the student graduates, given that he repeats the second year, we can use Bayes' rule.

EDIT

In reply to OP's comment.

To compute $(1)$ we use $$\Pr(X\cap Y) = \Pr(X)\Pr(Y|X).$$ Let me just comment on why this definition makes intuitive sense. In order that $X$ and $Y$ both occur, we first need $X$ to occur, and then, knowing that $X$ has occurred, we need $Y$ to occur. ("First" and "then" aren't intended to imply that $X$ precedes $Y$ in time.) So, to compute $\Pr(I^c\cap B)$ we have a choice of computing $$\Pr(I^c)\Pr(B|I^c)\tag2$$ or $$\Pr(B)\Pr(I^c|B)\tag3$$ We are given the quantities in $(2)$, but not those in $(3)$, so naturally, we choose the first alternative.

$$\Pr(I^c\cap B)=\Pr(I^c)\Pr(B|I^c)=.2\cdot.4$$

In the same way, we get $$ \Pr(A\cap B) = .2\cdot.4+.5\cdot.5+.25\cdot.95=.5675$$

To finish part a), we need to know $\Pr(B).$ We've just computed the probability that a student who repeats a year graduates, so we need to add in the probability that a student graduates without repeating any year. The probability that a student does not repeat a year is $$1-.2-.5-.25=.05$$ so that $\Pr(B)=.5675+.05=.6575+.05=.6175$ and
$$\Pr(A|B)=\frac{.5675}{.6175}\approx\boxed{.9190}$$

$\endgroup$
9
  • $\begingroup$ thank you from the bottom of your heart, but will we be able to solve it completely, i would be very grateful and thankful $\endgroup$ Commented Mar 21, 2020 at 15:29
  • 1
    $\begingroup$ @MadritZhaku You'll get a lot more value if you complete it yourself. Make an effort, and if you can't see how to finish it, ping me. $\endgroup$
    – saulspatz
    Commented Mar 21, 2020 at 15:31
  • $\begingroup$ maybe it's easy but i can't, i think $P(I^c\cap B)=P(B)P(B|I^c)$, but i didint know $P(B)$ $\endgroup$ Commented Mar 21, 2020 at 15:41
  • $\begingroup$ @MadritZhaku You've got the formula wrong. $P(I^c\cap B)=P(I^c)P(B|I^c)$ $\endgroup$
    – saulspatz
    Commented Mar 21, 2020 at 16:08
  • $\begingroup$ Please sir, continue the given example $\endgroup$ Commented Mar 21, 2020 at 16:12

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .