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Ray has enrolled as a freshman at an Eastern University and the probability that he will get a scholarship is 0.35. If he gets a scholarship the probability that he will graduate is 0.82, and if he does not get a scholarship the probability that he will graduate is only 0.44. What is the probability that he will graduate?

From my understanding its the Conditional probability. So, what I did is:

enter image description here

I'm not sure it's right. Please help.

And if we suppose that a yeat later he graduated, what is the probability that he did get the scholarship? Is it .35 x .82 = .287?

Please help.

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up vote 1 down vote accepted

Let me see if I can give an alternative explanation/visualization of conditional probability and LTP.

First, a fairly straightforward example. Let's suppose that $\frac14$ of all humans live in China. Moreover, let's suppose that $\frac35$ of the people who live in China are male, and that $\frac5{12}$ of the people who don't live in China are male. To visualize this, let's start with a square with an area of $1$ to represent all the people in the world (however many there are). We'll draw a vertical line $\frac14$ of the way from the left side of the square to split it into two rectangles--the left one, representing the people in China, having area of $\frac14;$ the right one, of $\frac34.$ Now, let's draw a horizontal line in the left-hand rectangle that is $\frac35$ of the way from the top, and a horizontal line in the right-hand rectangle $\frac5{12}$ of the way from the top.

At this point, we have four rectangles. The top two rectangles represent the males; the left two, the people who live in China. Let $C$ represent the event that a randomly selected person (spot inside the square) lives in China (is in one of the left rectangles); $M,$ that the person (spot) is male (in one of the top rectangles). Now, the top-left rectangle comprises $\frac35$ of the area of the left two rectangles combined--that is, $\frac35$ of China's population is male--that is, $\Bbb P(M\mid C)=\frac35.$ In particular, it is a $\frac14$ by $\frac35$ rectangle, so has area $\frac14\cdot\frac35=\frac3{20}$--that is, $\frac3{20}$ of the people in the world are males who live in China--that is, $\Bbb P(M\cap C)=\Bbb P(C)\cdot\Bbb P(M\mid C).$

Similarly, $\frac34\cdot\frac5{12}=\frac5{16}$ of the people of the world are men who do not live in China--that is, $\Bbb P(M\cap C')=P(C')\cdot P(M\mid C').$ From this, we see that the total area of the top two rectangles is $\frac3{20}+\frac5{16}=\frac{12}{80}+\frac{25}{80}=\frac{37}{80}$--that is, $\frac{37}{80}$ of the people of the world are men--that is, $\Bbb P(M)=\Bbb P(C)\cdot\Bbb P(M\mid C)+P(C')\cdot P(M\mid C').$ This is just the LTP applied to this example.

Let's see if we can generalize this idea. Suppose that our sample space (such as the set of all people in the world) can be partitioned into events $A_1,...,A_n$ (such as particular nationalities, or the "other nationality" category) where each $\Bbb P(A_j)$ is known. We can visualize this as splitting up the unit square into rectangles having areas (from left to right) of $\Bbb P(A_1),...,\Bbb P(A_n).$

Let us suppose further that there is an event $B$ of our sample space (such as "being male"), such that we don't know $\Bbb P(B)$ values, and instead only know $\Bbb P(B\mid A_j)$ for each $j\in\{1,...,n\}.$ We can visualize this as splitting up the $j$th column rectangle into two smaller rectangles such that the upper rectangle in the $j$th column comprises $\Bbb P(B\mid A_j)$ of the area of the $j$th column. This doesn't quite tell the whole story, though, since it may be that $\Bbb P(B\mid A_j)=0$ or $1$ for some $j.$ To make the distinction clear, we will shade in the top rectangle in the case that $0<\Bbb P(B\mid A_j)<1;$ the whole rectangle in the case that $\Bbb P(B\mid A_j)=1;$ none of the rectangle in the case that $\Bbb P(B\mid A_j)=0.$ Then for any $j\in\{1,...,n\},$ we know that there is a $\Bbb P(A_j)$ by $\Bbb P(B\mid A_j)$ shaded "rectangle" in the $j$th column, which has area $\Bbb P(B\cap A_j)=\Bbb P(A_j)\cdot\Bbb P(B\mid A_j).$

Finally, to figure out the probability of $B,$ we add up the areas of the shaded "rectangles"--that is: $$\Bbb P(B)=\sum_{j=1}^n\Bbb P(B\cap A_j)=\sum_{j=1}^n\Bbb P(A_j)\cdot\Bbb P(B\mid A_j).$$

Now, let's address your questions.

Your situation is much like the first example I gave. We want the left two rectangles to represent the event that he got the scholarship; the top two, that he graduated. So, the left two rectangles are $0.35$ wide, the top-left rectangle is $0.82$ tall, and the top-right rectangle is $0.44$ tall. Then you'll find the total area of the top two rectangles precisely as shown in your calculations above, which gives us the probability that he graduates.

Your additional question then asks for the probability that he got the scholarship, given that we happen to know he graduated (a year later). In other words, it wants to know the conditional probability $\Bbb P(\text{got scholarship}\mid\text{graduated}).$ The LTP is a way to find a probability by using known conditional probabilities (among other things), but we want to find a conditional probability, so that isn't going to help us. Let's rethink our visualization, first splitting the square into a graduation rectangle on the left (which will have width $0.573$ by the work above), then make scholarship rectangles in the top portion of the graduation and non-graduation rectangles. Unfortunately, this time, we don't know how tall these top rectangles need to be--we want to know how tall the top-left rectangle is! However, we do know (in particular) that the top-left rectangle needs to have area $$\begin{align}\Bbb P(\text{got scholarship and graduated}) &= \Bbb P(\text{graduated and got scholarship}\\ &= \Bbb P(\text{got scholarship})\cdot\Bbb P(\text{graduated}\mid\text{got scholarship})\\ &= 0.35\cdot0.82\\ &= 0.287.\end{align}$$ Since the area of a rectangle is width times height, then $$0.287=0.573\cdot\Bbb P(\text{got scholarship}\mid\text{graduated}),$$ and so $$\Bbb P(\text{got scholarship}\mid\text{graduated})=\frac{0.287}{0.573}\approx0.5009.$$ In other words, $$\Bbb P(\text{got scholarship}\mid\text{graduated})=\frac{\Bbb P(\text{got scholarship and graduated})}{\Bbb P(\text{graduated})}.$$

Ultimately, your second question can be viewed as a straightforward application of Bayes' Theorem, if you want to read more about it, but really follows directly from the definition of conditional probability (depending on which one you're using).

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Excellent!! Very well explained, thank you for your help very much!! Everything makes sense now! Thanks again. – juknee Jan 28 '14 at 0:04

You need to use the law of total probability since there are two possibilities that cover the space of possible outcomes: he either gets a scholarship and graduates or he doesn't get the scholarship and still graduates. This can be represented as follows and you already have all the information needed: P(G) = P(G|S)$\cdot$P(S) + P(G|not S)$\cdot$P(not S)

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And if we suppose that a year later he graduated, what is the probability that he did get the scholarship? Is it .35 x .82 = .287? – juknee Jan 27 '14 at 19:58
this question doesn't make sense. in the original post, nothing was ever said about when he graduated and if he had the scholarship when he graduated. – rookie Jan 27 '14 at 20:01
that's why I'm confused. This question comes straight from my book...:( – juknee Jan 27 '14 at 20:03
@user122674: I think we might want P(not G | S). "Graduating a year later" is deceiving. – rookie Jan 27 '14 at 20:20
Agreed. Well, the final answer for "Graduating a year later" with the scholarship was 0.5 – juknee Jan 27 '14 at 20:54

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