Probability
Probability of intersections — system of equations
MJAT_TS3_P2
Grade 12
Question:
For 3 events $A$, $B$, $C$: $P(\text{at least one})=\frac{3}{4}$, $P(\text{at least two})=\frac{1}{2}$, $P(\text{exactly two})=\frac{2}{5}$. Which of the following relations is/are CORRECT?
A) $P(A\cap B\cap C) = \dfrac{1}{10}$
B) $P(A\cap B)+P(B\cap C)+P(C\cap A) = \dfrac{7}{5}$
C) $P(A)+P(B)+P(C) = \dfrac{27}{20}$
D) $P(A\cap B'\cap C')+P(A'\cap B\cap C')+P(A'\cap B'\cap C) = \dfrac{1}{4}$
Step-by-Step Solution
Key Concept: $P(\text{at least two})=P(\text{exactly two})+P(\text{all three})\Rightarrow\frac{1}{2}=\frac{2}{5}+P(ABC)\Rightarrow P(ABC)=\frac{1}{10}$ (A ✓). $P(\text{exactly two})=P(AB)+P(BC)+P(CA)-3P(ABC)\Rightarrow\frac{2}{5}=P(AB)+P(BC)+P(CA)-\frac{3}{10}\Rightarrow$ sum $=\frac{7}{10}$ (not $\frac{7}{5}$, so B ✗).
A ✓, B ✗, C ✓, D ✓. Answer: A, C, D.
Correct Answer: ACD