## 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).