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Two independent events $\mathrm{A}$ and $\mathrm{B}$ are such that $\mathrm{P}(\mathrm{A} \cup \mathrm{B})=$
$\frac{2}{3}$ and $\mathrm{P}(\mathrm{A} \cap \mathrm{B})=\frac{1}{6}$. If $\mathrm{P}(\mathrm{B}) < \mathrm{P}(\mathrm{A})$, then what is $\mathrm{P}(\mathrm{B})$
equal to ?
Options:
$\frac{2}{3}$ and $\mathrm{P}(\mathrm{A} \cap \mathrm{B})=\frac{1}{6}$. If $\mathrm{P}(\mathrm{B}) < \mathrm{P}(\mathrm{A})$, then what is $\mathrm{P}(\mathrm{B})$
equal to ?
Solution:
1280 Upvotes
Verified Answer
The correct answer is:
$\frac{1}{3}$
$P(A \cup B)=\frac{2}{3}$
$P(A \cap B)=\frac{1}{6}$
Since, $A$ and $B$ are independent events.
$P(A \cap B)=P(A) \cdot P(B)=\frac{1}{6}$ ... (1)
$P(A \cup B)=\frac{2}{3} \Rightarrow P(A)+P(B)-P(A \cap B)=\frac{2}{3}$
$\Rightarrow P(A)+P(B)-\frac{1}{6}=\frac{2}{3}$
$\Rightarrow P(A)+P(B)=\frac{2}{3}+\frac{1}{6}=\frac{5}{6}$ .... (2)
from $(1),(2), P(B)=\frac{1}{3}$ or $\frac{1}{2}$
$\because P(B) < P(A), P(B)=\frac{1}{3}$
$P(A \cap B)=\frac{1}{6}$
Since, $A$ and $B$ are independent events.
$P(A \cap B)=P(A) \cdot P(B)=\frac{1}{6}$ ... (1)
$P(A \cup B)=\frac{2}{3} \Rightarrow P(A)+P(B)-P(A \cap B)=\frac{2}{3}$
$\Rightarrow P(A)+P(B)-\frac{1}{6}=\frac{2}{3}$
$\Rightarrow P(A)+P(B)=\frac{2}{3}+\frac{1}{6}=\frac{5}{6}$ .... (2)
from $(1),(2), P(B)=\frac{1}{3}$ or $\frac{1}{2}$
$\because P(B) < P(A), P(B)=\frac{1}{3}$
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