## Category Archives: probability theory

### Solutions to Birthday Problems: IITJEE Advanced Mathematics

In the following problems, each year is assumed to be consisting of 365 days (no leap year):

Problem 1:

What is the least number of people in a room such that it is more likely than not that at least two people will share the same birthday?

Solution 1:

The probability of the second person having a different birthday from the first person is $\frac{364}{365}$. The probability of the first three persons having different birthdays is $\frac{364}{365} \times \frac{363}{365}$. In this way, the probability of all n persons in a room having different birthdays is $P(n) = \frac{364}{365} \times \frac{363}{365} \times \frac{362}{365} \times \ldots \frac{365-n+1}{365}$. For the value of n, when P(n) falls just below 1/2 is the least number of people in a room when the probability of at least two people having the same birthday becomes greater than one half (that is, more likely than not). Now, one can make the following table: $\begin{tabular}{|c|c|}\hline N & P(n) \\ \hline 2 & 364/365 \\ \hline 3 & 0.9918 \\ \hline 4 & 0.9836 \\ \hline 5 & 0.9729 \\ \hline 6 & 0.9595 \\ \hline 7 & 0.9438 \\ \hline 8 & 0.9257 \\ \hline 9 & 0.9054 \\ \hline 10 & 0.8830 \\ \hline 11 & 0.8589 \\ \hline 12 & 0.8330 \\ \hline 13 & 0.8056 \\ \hline 14 & 0.7769 \\ \hline 15 & 0.7471 \\ \hline 16 & 0.7164 \\ \hline 17 & 0.6850 \\ \hline 18 &0.6531 \\ \hline 19 & 0.6209 \\ \hline 20 & 0.5886 \\ \hline 21 & 0.5563 \\ \hline 22 & 0.5258 \\ \hline 23 & 0.4956 \\ \hline \end{tabular}$

Thus, the answer is 23. One may say that during a football match with one referee, it is more likely than not that at least two people on the field have the same birthday! 🙂 🙂 🙂

Problem 2:

You are in a conference. What is the least number of people in the conference (besides you) such that it is more likely than not that there is at least another person having the same birthday as yours?

Solution 2:

The probability of the first person having a different birthday from yours is $\frac{364}{365}$. Similarly, the probability of the first two persons not having the same birthday as yours is $\frac{(364)^{2}}{(365)^{2}}$. Thus, the probability of n persons not  having the same birthday as yours is $\frac{(364)^{n}}{(365)^{n}}$. When this value falls below 0.5, then it becomes more likely than not that at least another person has the same birthday as yours. So, the least value of n is obtained from $(\frac{364}{365})^{n}<\frac{1}{2}$. Taking log of both sides, we solve to get $n>252.65$. So, the least number of people required is 253.

Problem 3:

A theatre owner announces that the first person in the queue having the same birthday as the one who has already purchased a ticket will be given a free entry. Where (which position in the queue) should one stand to maximize the chance of earning a free entry?

Solution 3:

For the nth person to earn a free entry, first $(n-1)$ persons must have different birthdays and the nth person must have the same birthday as that of one of these previous $(n-1)$ persons. The probability of such an event can we written as $P(n) = [\frac{364}{365} \times \frac{363}{365} \times \frac{362}{365} \times \ldots \frac{365-n+2}{365}] \times \frac{n-1}{365}$

For a maximum, we need $P(n) > P(n+1)$. Alternatively, $\frac{P(n)}{P(n+1)} >1$. Using this expression for P(n), we get $\frac{365}{365-n} \times \frac{n-1}{n} >1$. Or, $n^{2}-n-365>0$. For positive n, this inequality is satisfied first for some n between 19 and 20. So, the best place in the queue to get a free entry is the 20th position.

More later,

Nalin Pithwa.

### Birthday Probability Problems: IITJEE Advanced Mathematics

In the following problems, each year is assumed to be consisting of 365 days (no leap year):

1. What is the least number of people in a room such that it is more likely than not that at least two people will share the same birthday?
2. You are in a conference. What is the least number of people in the conference (besides you) such that it is more likely than not that there is at least another person having the same birthday as yours?
3. A theatre owner announces that the first person in the queue having the same birthday as the one who has already purchased a ticket will be given a free entry. Where (which position in the queue) should one stand to maximize the chance of earning a free entry?

I will put up the solutions on this blog tomorrow. First, you need to make a whole-hearted attempt.

Nalin Pithwa.

### Alice in Wonderland and probability stuff !!

Lewis Carroll of Alice’s Adventures in the Wonderland was a mathematician in Cambridge. He had posed the following problem in one of his books:

A box contains a handkerchief, which is rather white or black. You put a white handkerchief in this box and mix up the contents. Then you draw a handkerchief, which turns out to  be white. Now, if you draw the remaining handkerchief, what is the probability of this one being white?

How likely is that you would enjoy such questions ?! 🙂

Nalin Pithwa.

### Calculating probability — a nice question from AMC

This problem is a beautiful one. Please go through the video. Thanks to Prof. Adriana Salerno  and MAA !!

### Answers: conceptual probability questions for IITJEE Math

These answers refer to the previous two blogs.

1. Since the pin may land up or down, we take the sample space $\Omega = (U, D)$. One may feel that the outcome D is more likely and put, for example, $P(\{D\})=0.75$ and, consequently, $P(\{U\})=0.25$. However, a more elaborate analysis of the shape of the pin is required, or we have to perform a long series of experiments and take the average frequency as the probability. The geometric argument is that if you imagine that the pin is circumscribed by a sphere, the majority of points on the sphere correspond to D.
2. $\Omega = \{ (H,H), (H,T), (T,H), (T, T)\}$, T stands for tails, H for heads, $P({\omega})=\frac{1}{4}$ for each $\omega \in \Omega$.
3. For one die: $\Omega = \{ i: i= 1, \ldots, 6\}$ with $P({\omega})=\frac{1}{6}$ for each $\omega \in \Omega$. For two dice, $\Omega = \{ (i, j) : i, j = 1, \ldots, 6\}$ with $P({\omega})=\frac{1}{36}$ for each $\omega \in \Omega$.
4. $\Omega = \{ (i, a): i=1, \ldots, 6, a = H \hspace{0.1in} or \hspace{0.1in}T\}$, $P({\omega})=\frac{1}{12}$ for each $\omega = \Omega$.
5. We take $\Omega = \{ Y, N\}$, if we are only interested in a yes or no answer to the question of whether the mug will break or not. The number of pieces after landing can be represented by $\Omega = \{ 1, 2, \ldots\}$. We can also take $\Omega = \{ U, D, R, L\}$ for the landing positions up, down, handle to the right, or to the left respectively. A more sophisticated choice is $\Omega = \{ U, D\} \bigcup \{ 0, 2\pi\}$ for a mug with no handle, but with a dot on its side. It may land up, down, or sideways, then roll, in which case we find the angle between the floor and the straight line containing the radius determined by the dot. In each case, the choice of a probability measure depends on the physical properties of the mug and the floor.
6.  There are two possibilities: (a) We draw two cards at once; $\Omega$ consists of all two elements subsets $\{ c_{1}, c_{2}\}$ of C. (b) We draw the second card after having returned the first one to the pack. Then, $\Omega$ consists of all ordered pairs $(c_{1}, c_{2})$ of cards $c_{1}, c_{2}$. Unless the cards are drawn by a magician, we can take the measure $P(\{ \omega\}) = \frac{1}{n}$, where n denotes the number of elements of $\Omega$.
7. $\Omega$ consists of all permutations of the set $\{ 1, \ldots, 6\}$. Each outcome is equally likely so we take the uniform probability $P(\{ \omega\}) = \frac{1}{6!}$ for each $\omega \in \Omega$.
8. The outcomes are all three-, four-, and five-element sequences of letters H or L, each containing exactly three letters of one kind (20 possibilities). The uniform probability does not seem appropriate. Everything depends on the probability of winning a single game. Even if we assume that it does not change from game to game and is known, it is not  obvious what the measure should be. There are ways and means to deal with such cases also.
9. $\Omega = \{ 0, 1, 2, \ldots\}$, the number of tails before the first head appears. $P(\{ 0\})=\frac{1}{2}$, $P(\{ 1\})=\frac{1}{4}$, $P(\{ 2\})=\frac{1}{8}$, and so on. We could also add the outcome corresponding to the fact that the head never appears. This is theoretically possible, so put $\Omega = \Omega \bigcup \{ \infty\}$. The above assignment, $P(\{ k\})=\frac{1}{2^{k+1}}$, remains valid. Because $P(\Omega)=1$, this leaves us with the only possible choice, $P(\{ \infty\})=0$. This is an example of an event which theoretically  cannot be excluded, but happens with probability 0. We have used the fact that $P(\Omega)=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+ \ldots + \frac{1}{2^{n}}+\ldots =1$.
10. An idealistic (but, mathematically convenient)) approach is to assume that the temperature is a real number, so $\Omega = (-\infty, +\infty)$ with probability measure given by a density supplied by the meteorological office. This density depends on the country in which you  live. If the accuracy of meteorological data is to within $1 \deg$, we may prefer $\Omega = \{ \ldots, -2, -1, 0, +1, +2, \ldots\}$, that is, the set of integers.
11. If, for example, the line is serviced by six buses and each takes 60 minutes for a complete round, then $\Omega = [0,10]$ with uniform density seems reasonable. If the bus has a timetable, a density with a peak at each appropriate time may be better.
12. A solution is $\Omega = \{ 1\}$, if you take a registered post or a speed post.
13. We may take $\Omega = \{ 0, 1, \ldots, n\}$ with n being the capacity of the page. We would not recommend the uniform measure here. $\Omega = N$ may be more convenient to avoid the dependence on n. One may expect the probabilities to decrease as k increases, where k is the number of misprints.
14. It may be convenient to take $\Omega = \{ 0, 1, 2, \ldots\}$ with a special measure such that $P(\{ n\}) = e^{\lambda} \frac{\lambda^{n}}{n!}$ for each $n = 0, 1, 2, \ldots$. This is called the Poisson measure with parameter $\lambda >0$.
15. If it is a watch with a simple digital display, we can take $\Omega_{hours}=\{1, \ldots, 12 \}$ $\Omega_{minutes}= \{ 0, \ldots, 59\}$ $\Omega_{seconds} = \{ 0, \ldots, 59\}$.

or, combining the above, $\Omega = \{ (h, m, s): h \in \Omega_{hours}, m \in \Omega_{minutes}, s \in \Omega_{seconds}\}$.

Some hours are more likely than others (daytime, evening), but all values of m and n seem to be equally        probable.

For an analog display, we can take the same set $\Omega = [0, 2\pi)$ for hours, minutes, and seconds with different densities (uniform for minutes and seconds).

Nalin Pithwa

### Hints: probability conceptual questions for IITJEE Math

This refers to the previous blog on probability conceptual questions for IITJEE Math.

Hints:

1. The pin may land point up or down. Do you think these cases are equally likely?
2. This is the same as if we tossed two coins once. Then, the outcome is a pair $(a,b)$ where a and b may be tails or heads.
3. Assuming that the die is regular (fair), we have six equally likely outcomes/ For two dice, an outcome is a pair $(a,b)$, where a is the number shown on the first die and b on the second.
4. Again, we have pairs $(a,b)$ here, but we have to be careful, since the meaning and the range of a and b are different.
5. In this example, there are many possibilities. We may want to know whether mug breaks or not, into how many pieces, which side up it lands (if in one piece).
6. The description of the experiment is not precise. We can draw two cards at once or draw one, return it to the pack, and draw again.
7. The result can be written as a sequence of six numbers.
8. What is the maximum length of the match? How many games may be played?
9. Is there a number N such that in N throws we shall certainly have heads up?
10. We need to choose the scale and then decide what accuracy is possible.
11. Is this a bus stop in the first place (in some cities they have trams)? Does the bus have a time table?
12. Do we include the possibility that it will never arrive?
13. In the extreme case, each letter on the page you consider may be a misprint. (We certainly hope it is not so). Thus, there are at most finitely many misprints.
14. The number of accidents is not necessarily limited by the number of vehicles in the city. A vehicle may be involved in more than one accident on the same day.
15. Depends on the type of display and the individual working habits.

Now, you can try the solutions.

Auf wiedersehen,

Nalin Pithwa

### Probability: conceptual questions for IITJEE Math

Try to propose the set of outcomes and the measures in the following situations. (There is no such thing as one correct solution in these set of problems):

1. A drawing pin is tossed.
2. A coin is tossed twice.
3. A die is rolled. What about two dice?
4. A die and a coin are thrown simultaneously.
5. A mug falls down from the table. (luckily, an empty one).
6. From a pack of 52 cards, we draw 2.
7. A pack of six numbered cards is shuffled and the numbers are revealed one by one.
8. We play a series of chess games. The winner is the one who first scores three points, where one point is obtained for a single win and draws do not come.
9. We keep throwing a coin until it lands heads up.
10. The temperature outdoors is measured.
11. How long shall we wait for a bus at a stop?
12. How many days does a letter posted in Mumbai take to reach Bangalore?
13. What is the number of misprints on a page of your math text book?
15. What time does your watch show now?

You are most welcome to share your solutions with explanations.

I will supply hints in a later blog,

Nalin Pithwa

### Rabbits in the Hat

The Great Whodunni, a stage magician, placed his top hat on the table.

‘In this hat are two rabbits,’ he announced. ‘Each of them is either black of white, with equal probability. I am now going to convince you, with the aid of my lovely assistant Grumpelina, that I can deduce their colours without looking inside the hat!’

He turned to his assistant, and extracted a black rabbit from her costume. ‘Please place this rabbit in the hat.’ She did.

Whodunnii now turned to the audience. ‘Before Grumpelina added the third rabbit, there were four equally likely combinations of rabbits. ‘ He wrote a list on a small blackboard:BB, BW, WB and WW. ‘Each combination is equally likely — the probability is 1/4.

But, then I added a black rabbit. So, the possibilities are BBB, BWB, BBW and BWW — again, each with probability 1/4.

‘Suppose —- I won’t do it, this is hypothetical — suppose I were to pull a rabbit from the hat. What is the probability that it is black? If the rabbits are BBB, that probability is 1. If BWB or BBW, it is 2/3. If BWW, it is 1/3. So the overall probability of pulling out a black rabbit is $\frac{1}{4} \times 1 + \frac{1}{4} \times \frac{2}{3}+\frac{1}{4} \times \frac{2}{3}+\frac{1}{4} \times \frac{1}{3}$

which is exactly 2/3.

‘But. If there are three rabbits in a hat, of which exactly r are black and the rest white, the probability of extracting a black rabbit is r/3. Therefore, $r=2$, so there are two black rabbits in the hat.’ He reached into the hat and pulled out a black rabbit. ‘Since I added this black rabbit, the original pair must have been one black and one white!’

The Great Whodunni bowed to tumultous applause. Then, he pulled out two rabbits fwrom the hat — one pale lilac and the other shocking pink.

It seems evident that you can’t deduce the contents of a hat without finding out what’s inside. Adding the extra rabbit and then removing it again (was it the same black rabbit? Do we care?) is a clever piece of misdirection. But, why is the calculation wrong?

More later,

Nalin Pithwa