Problem Setup
We need to find the probability when choosing 3 letters from {A, B, C, D, E, F} (6 letters) randomly and with replacement. The goal is to determine the probability that the chosen sequence forms either the word $BAD$ or the word $CAD$.
Total Possible Outcomes
Since the letters are chosen with replacement, each of the 3 positions in the sequence has 6 independent choices (A, B, C, D, E, F).
The total number of possible ordered sequences is calculated as:
$ \text{Total Outcomes} = 6 \times 6 \times 6 = 6^3 = 216 $
Favorable Outcomes
The question requires the probability of forming either $BAD$ or $CAD$. Given the options, the interpretation is that the selected sequence must be one of the permutations of the letters in $BAD$ or one of the permutations of the letters in $CAD$.
1. Permutations related to $BAD$:
The specific letters required are B, A, and D. The distinct permutations of these three letters are:
- $BAD$
- $BDA$
- $ABD$
- $ADB$
- $DAB$
- $DBA$
There are $3! = 6$ such sequences.
2. Permutations related to $CAD$:
The specific letters required are C, A, and D. The distinct permutations of these three letters are:
- $CAD$
- $CDA$
- $ACD$
- $ADC$
- $DAC$
- $DCA$
There are $3! = 6$ such sequences.
These two sets of sequences (permutations of $BAD$ and permutations of $CAD$) are mutually exclusive.
Total Favorable Outcomes = (Number of permutations of $BAD$) + (Number of permutations of $CAD$) = $6 + 6 = 12$.
Calculating the Probability
The probability is the ratio of the number of favorable outcomes to the total number of possible outcomes.
$ \text{Probability} = \frac{\text{Total Favorable Outcomes}}{\text{Total Possible Outcomes}} $
$ \text{Probability} = \frac{12}{216} $
This result matches Option D.
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?
