Probability Calculation for Ticket Assignment
We are given 100 tickets numbered $1, 2, \dots, 100$. Four tickets are randomly selected and given to four persons: A, B, C, and D. We need to find the probability that person A gets the ticket with the largest value among the four selected tickets, and person D gets the ticket with the smallest value among the four selected tickets.
Reasoning Using Symmetry
Consider any set of four distinct ticket values that are chosen from the 100 tickets. Let these four values be $v_1, v_2, v_3, v_4$, sorted in ascending order such that $v_1 < v_2 < v_3 < v_4$.
These four distinct ticket values are distributed among the four persons A, B, C, and D. Since the tickets are arranged at random and distributed, each person is equally likely to receive any of the four chosen tickets. The total number of ways to assign these four distinct tickets to the four persons is the number of permutations of 4 items, which is $4!$.
Total possible assignments = $4! = 4 \times 3 \times 2 \times 1 = 24$.
We are interested in the specific event where A gets the largest value ($v_4$) and D gets the smallest value ($v_1$).
If A gets $v_4$ and D gets $v_1$, the remaining two tickets ($v_2$ and $v_3$) must be distributed between persons B and C. There are two possible ways to do this:
- B gets $v_2$ and C gets $v_3$.
- B gets $v_3$ and C gets $v_2$.
So, there are 2 favorable assignments out of the 24 total possible assignments for any set of four chosen tickets.
The probability of this specific event occurring is the ratio of the number of favorable assignments to the total number of possible assignments.
Probability = $\frac{\text{Number of favorable assignments}}{\text{Total number of assignments}}$
Probability = $\frac{2}{24}$
Simplifying the fraction gives:
Probability = $\frac{1}{12}$
Important Questions from Discrete Probability
- A biased six-faced die is tossed once. Suppose that the probability of any prime number showing up is twice that of any non-prime number showing up. Then, the probability that an odd number will show up is
- Let $X$ and $Y$ be independent Poisson random variables with means $4$ and $2$, respectively. Then, which of the following statements are true?
- Consider the M/M/1 queue in which customers arrive according to a Poisson process with rate $3$ and successive service times are independent exponential random variables having mean $\frac{1}{9}$. Let $P_n$ be the long run probability that there are exactly $n$ customers in the system. Then, which of the following statements are true?
- Let $X$ be a Binomial$(n, p)$ random variable, where $n \in \{5,6\}$ and $p\in \{\frac{1}{4}, \frac{3}{4}\}$. If $X = 3$ is observed, then the maximum likelihood estimate of $(n, p)$ is
- Suppose two fair dice are thrown independently at random. Let $X$ and $Y$ be the numbers on the upper face of the first die and that of the second die, respectively. Then which of the following statements are true?
