Probability
Events
MJAT None
Grade 12
Question:
Let $\mathbf{E}$, $\mathbf{F}$, and $\mathbf{G}$ be three events having probabilities $\mathbf{P}(\mathbf{E}) = \frac{1}{8}$, $\mathbf{P}(\mathbf{F}) = \frac{1}{6}$, and $\mathbf{P}(\mathbf{G}) = \frac{1}{4}$, respectively. Let $\mathbf{P}(\mathbf{E} \cap \mathbf{F} \cap \mathbf{G}) = \frac{1}{10}$. For any event $\mathbf{H}$, if $\mathbf{H}^c$ denotes its complement, then which of the following statements is (are) TRUE ?
A) $\mathbf{P}(\mathbf{E} \cap \mathbf{F} \cap \mathbf{G}^c) \leq \frac{1}{40}$
B) $\mathbf{P}(\mathbf{E}^c \cap \mathbf{F} \cap \mathbf{G}) \leq \frac{1}{15}$
C) $\mathbf{P}(\mathbf{E} \cup \mathbf{F} \cup \mathbf{G}) \leq \frac{13}{24}$
D) $\mathbf{P}(\mathbf{E}^c \cap \mathbf{F}^c \cap \mathbf{G}^c) \leq \frac{5}{12}$
Step-by-Step Solution
Key Concept: The probability of the intersection of three events is less than or equal to the product of their individual probabilities.
$$mathbf{P}(mathbf{E} cap mathbf{F} cap mathbf{G}) = mathbf{P}(mathbf{E}) mathbf{P}(mathbf{F}) mathbf{P}(mathbf{G})$$
$$Rightarrow rac{1}{10} = rac{1}{8} cdot rac{1}{6} cdot rac{1}{4}$$
$$Rightarrow rac{1}{10} = rac{1}{96}$$
This is a false statement. The correct statement should be $$mathbf{P}(mathbf{E} cap mathbf{F} cap mathbf{G}) leq mathbf{P}(mathbf{E}) mathbf{P}(mathbf{F}) mathbf{P}(mathbf{G})$$
because the probability of the intersection of three events is less than or equal to the product of their individual probabilities.
Correct Answer: A, B, C, D