Given two independent events, P(exactly one)=2649, P(none)=1549.

Question

Given two independent events, $P(	ext{exactly one})=\dfrac{26}{49}$, $P(	ext{none})=\dfrac{15}{49}$. Find the probability of the more probable event.

Accepted Answer

Let $P(A)=a,P(B)=b$. $a+b-2ab=\dfrac{26}{49}$; $(1-a)(1-b)=\dfrac{15}{49}\Rightarrow a+b-ab=\dfrac{34}{49}$.

Solving: $a+b=\dfrac{42}{49}$, $ab=\dfrac{8}{49}$, giving $(a-b)^2=\dfrac{196}{2401}\Rightarrow a-b=\dfrac{14}{49}$.

$$a=\frac{4}{7},\ b=\frac27 \implies \text{more probable} = \frac47$$

Solutions · 1