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
- 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?
- A box contains 40 numbered red balls and 60 numbered black balls. From the box, balls are drawn one by one at random without replacement till all the balls are drawn. The probability that the last ball drawn is black equals
Consider the problem of testing $H_0 : \theta = 1$ vs $H_1 : \theta = \frac{1}{2}$ where $\theta$ is the mean of a Poisson random variable. Let $X$ and $Y$ be a random sample from Poisson ($\theta$) distribution. Consider the following test procedure:
Reject $H_0$ if either $X = 0$ or $(X = 1 \text{ and } X + Y \leq 2)$; otherwise accept $H_0$.
Which of the following are true?
In a football league, the goals scored by home teams over 380 matches have the following frequency distribution.
Number of goals 0 1 2 3 4 5 Frequency 92 121 91 50 19 7 The average goals scored by home teams is 1.49. We want to test $H_0$: Goal distribution is Poisson. Based on observations the value of the $\chi^2$-statistic for goodness of fit is 1.27. Given $\chi^2_{0.05, 6} = 1.64, \chi^2_{0.05, 5} = 1.15, \chi^2_{0.95, 6} = 12.59$ and $\chi^2_{0.95, 5} = 11.07$, which of the following are true?
