# Test your mettle — a probability theory problem

Some of the most formidable and extremely interesting problems are in probability theory of exams like IITJEE, ISI entrance, CMI entrance etc. Even otherwise, probability theory quickly  becomes abstract, whether you study as an engineering student or as an applied mathematics student or as a pure mathematics or statistics student.

Let me present to you a seemingly simple question which test your conceptual understanding of probability theory basics.

Question: (Parzen). Suppose there exists a (fictitious) test for cancer with the following properties. Let A be the event that the test states that tested person has cancer and B be the event that the person has cancer.

It is known that $P(A|B)=P(A^{'}|B^{'})=0.95$ and $P(B)=0.005$Is this test a good test?

Solution:

Before reading the solution, just give it a shot! And, then, after that if you see the solution, it will erase your doubts.

Now, $A^{'}$ is the event that the test states person is free from cancer.

Also, $B^{'}$ is the event that person is free from cancer.

To answer the question we should like to know the likelihood that a person actually has cancer if the test so states, that is,

$P(B|A)$. Hence,

$P(B|A) = \frac {P(B)P(A|B)}{P(A|B)P(B)+P(A|B^{'})P(B^{'})}$

The above equals $\frac {(0.005)(0.95)}{(0.95)(0.095)+(0.05)(0.995)} = 0.087$

Hence, in only $8.7 percent$ of the cases where the tests are positive will the person actually have cancer. This test has a very high false alarm rate and in this sense cannot be regarded as a good one. The fact that , initially, the test seems like a good test is not surprising given that $P(A|B)=P(A^{'}|B^{'})=0.95$. However, when the scarcity of cancer in the general public is considered, the test is found to be vacuous.

More later…

Nalin Pithwa

# A cute problem on permutations and combinations

Here, is a problem on counting. Read it and crack it before you compare your solution with the one I present !!!

In a tennis tournament, there are $2n$ participants. In the  first round of  the tournament, each participant plays just once, so there are n games, each occupying a pair of  players. Show that the pairing for the first round can be arranged in exactly

$1 x 3 x 5 x 7 x 9 \ldots(2n-1)$

different ways.

Solution.

Call the required number of pairings of $2n$ players $P_{n}$. If you are a participant, you can be matched with any one of  the other $(2n-1)$ players. Once your antagonist is chosen, there remain

$(2n-2)=2(n-1)$

players who  can be paired in $P_{n-1}$ ways. Hence,

$P_{n}=(2n-1)P_{n-1}$.

More later

Nalin Pithwa

# everywhere differentiable function — an interesting problem

Question: Determine all the functions f, which are everywhere differentiable and satisfy

$f(x)+f(y)=f((x+y)/(1-xy))$ for all real x and y with $xy \neq 1$. —- Equation I

Solution:

Let $f(x)$ satisfy Equation I above. Differentiating Equation I partially with respect to each of x and y, we obtain

$f^{'}(x)=((1+y^{2})/(1-xy)^{2})f^{'}((x+y)/(1-xy))$ Equation II

$f^{'}(y)=((1+x^{2})/(1-xy)^{2})f^{'}((x+y)/(1-xy))$ Equation III

Eliminating common  terms in Equation II and Equation III, we deduce that

$(1+x^{2})f^{'}(x)=(1+y^{2})f^{'}(y)$

As the left side of Equation III depends only on x and the  right side only on y, each side of Equation III must be equal to a constant c. Thus, we have

$f^{'}(x)=\frac {c}{1+x^{2}}$

and so, $f(x)=c\arctan x +d$,

for some constant d. However, taking $y=0$ in Equation III, we obtain $f(x)+f(0)=f(x)$, so that $f(0)=0$ and

$d=0$. Clearly, $f(x)=c\arctan x$ satisfies Equation I and so all solutions of Equation I are given by

$f(x)=c\arctan x$ where c is a constant.

Note: In Equation II, the $f^{'}$ is w.r.t. x and in Equation III, the $f^{'}$ is w.r.t. y, yet in equation II

$f^{'}(x+y)/(1-xy)$ is the same as the corresponding $f^{'}$ in Equation III because the argument $(x+y)/(1-xy)$ is symmetric w.r.t. x and y.