Given two independent events, if the probability that exactly one of them occurs is 26 49 and the probability that none of them occurs is 15 49, then…

Let the probability of occurrence of first event A, be ' $a$a ' i..e., $\mathrm{P}(\mathrm{A})=\mathrm{a}$P(A)=a $$\therefore \mathrm{P}(\text { not } \mathrm{A})=1-a$$∴P( not A)=1−a And also suppose that probability of occurrence of second event $\mathrm{B}, \mathrm{P}(\mathrm{B})=b$B,P(B)=b, $$\therefore \mathrm{P}(\operatorname{not} \mathrm{B})=1-b$$∴P(notB)=1−b Now, $\mathrm{P}(\mathrm{A}$P(A and not $\mathrm{B})+\mathrm{P}($B)+P( not $\mathrm{A}$A and $\mathrm{B})=\frac{26}{49}$B)=4926​ $$\begin{aligned} & \Rightarrow \quad \mathrm{P}(\mathrm{A}) \times \mathrm{P}(\operatorname{not} \mathrm{B})+\mathrm{P}(\operatorname{not} \mathrm{A}) \times \mathrm{P}(\mathrm{B})=\frac{26}{49} \\ & \Rightarrow \quad a \times(1-b)+(1-a) b=\frac{26}{49} \\ & \Rightarrow \quad a+b-2 a b=\frac{26}{49} \end{aligned}$$​⇒P(A)×P(notB)+P(notA)×P(B)=4926​⇒a×(1−b)+(1−a)b=4926​⇒a+b−2ab=4926​​ And $\mathrm{P}($P( not $\mathrm{A}$A and not $\mathrm{B})=\frac{15}{49}$B)=4915​ $$\begin{aligned} & \Rightarrow \quad \mathrm{P}(\operatorname{not} \mathrm{A}) \times \mathrm{P}(\text { not } \mathrm{B})=\frac{15}{49} \\ & \Rightarrow \quad(1-a) \times(1-b)=\frac{15}{49} \\ & \Rightarrow 1-\mathrm{b}-\mathrm{a}+\mathrm{ab}=\frac{15}{49} \\ & \Rightarrow a+b-a b=\frac{34}{49} \end{aligned}$$​⇒P(notA)×P( not B)=4915​⇒(1−a)×(1−b)=4915​⇒1−b−a+ab=4915​⇒a+b−ab=4934​​ From (i) and (ii), $$a+b=\frac{42}{49}$$a+b=4942​ and $a b=\frac{8}{49}$ab=498​ $$\begin{aligned} & (a-b)^2=(a+b)^2-4 a b=\frac{42}{49} \times \frac{42}{49}-\frac{4 \times 8}{49} \\ & =\frac{196}{2401} \end{aligned}$$​(a−b)2=(a+b)2−4ab=4942​×4942​−494×8​=2401196​​ $$\therefore \quad a-b=\frac{14}{49}$$∴a−b=4914​ From (iii) and (iv), $$a=\frac{4}{7}, b=\frac{2}{7}$$a=74​,b=72​ Hence probability of more probable of the $$\text { two events }=\frac{4}{7}$$ two events =74​