Suppose $A, B, C$ are events in a common probability space with $P(A) = 0.2, \ P(B) = 0.2, \ P(C) = 0.3$, $P(A \cap B) = 0.1, \ P(A \cap C) = 0.1, \ P(B \cap C) = 0.1$. Which of the following are possible values of $P(A \cup B \cup C)$?
To determine the possible values of \(P(A \cup B \cup C)\), we can use the principle of inclusion-exclusion in probability. This formula is given by:
\(P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A \cap B) - P(B \cap C) - P(C \cap A) + P(A \cap B \cap C)\)
We know:
- \(P(A) = 0.2\)
- \(P(B) = 0.2\)
- \(P(C) = 0.3\)
- \(P(A \cap B) = 0.1\)
- \(P(A \cap C) = 0.1\)
- \(P(B \cap C) = 0.1\)
Let's compute with the assumption that \(P(A \cap B \cap C) = 0\), as we are not given its value:
\(\begin{align*} P(A \cup B \cup C) &= 0.2 + 0.2 + 0.3 - 0.1 - 0.1 - 0.1 + 0 \\ &= 0.6 - 0.3 \\ &= 0.3 \end{align*}\)
This result shows that if \(P(A \cap B \cap C) = 0\), then \(P(A \cup B \cup C) = 0.3\).
However, the problem doesn't provide \(P(A \cap B \cap C)\), so we need to consider possible values to find other potential values of \(P(A \cup B \cup C)\). Let's check potential maximum overlap when \(P(A \cap B \cap C) = 0.1\).
\(\begin{align*} P(A \cup B \cup C) &= 0.2 + 0.2 + 0.3 - 0.1 - 0.1 - 0.1 + 0.1 \\ &= 0.6 - 0.2 \\ &= 0.4 \end{align*}\)
If we further assume \(P(A \cap B \cap C) = 0.2\), let's calculate:
\(\begin{align*} P(A \cup B \cup C) &= 0.2 + 0.2 + 0.3 - 0.1 - 0.1 - 0.1 + 0.2 \\ &= 0.6 - 0.1 \\ &= 0.5 \end{align*}\)
Therefore, the possible values of \(P(A \cup B \cup C)\) consistent with the provided information are \(0.4\) and \(0.5\).
Thus, the correct options are:
0.5
0.4
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?
