Mathematics - Probability Question with Solution | TestHub

Let the probability of occurrence of first event A, be ' $a$ ' i..e., $\mathrm{P}(\mathrm{A})=\mathrm{a}$ $$\therefore \mathrm{P}(\text{ not }\mathrm{A})=1-a$$ And also suppose that probability of occurrence of second event $\mathrm{B},\mathrm{P}(\mathrm{B})=b$, $$\therefore \mathrm{P}(\mathrm{not}\mathrm{B})=1-b$$ Now, $\mathrm{P}(\mathrm{A}$ and not $\mathrm{B})+\mathrm{P}($ not $\mathrm{A}$ and $\mathrm{B})=\frac{26}{49}$ $$\begin{array}{rl}& \Rightarrow \mathrm{P}(\mathrm{A})\times \mathrm{P}(\mathrm{not}\mathrm{B})+\mathrm{P}(\mathrm{not}\mathrm{A})\times \mathrm{P}(\mathrm{B})=\frac{26}{49}\\ & \Rightarrow a\times (1-b)+(1-a)b=\frac{26}{49}\\ & \Rightarrow a+b-2ab=\frac{26}{49}\end{array}$$ And $\mathrm{P}($ not $\mathrm{A}$ and not $\mathrm{B})=\frac{15}{49}$ $$\begin{array}{rl}& \Rightarrow \mathrm{P}(\mathrm{not}\mathrm{A})\times \mathrm{P}(\text{ not }\mathrm{B})=\frac{15}{49}\\ & \Rightarrow (1-a)\times (1-b)=\frac{15}{49}\\ & \Rightarrow 1-\mathrm{b}-\mathrm{a}+\mathrm{ab}=\frac{15}{49}\\ & \Rightarrow a+b-ab=\frac{34}{49}\end{array}$$ From (i) and (ii), $$a+b=\frac{42}{49}$$ and $ab=\frac{8}{49}$ $$\begin{array}{rl}& (a-b{)}^2=(a+b{)}^2-4ab=\frac{42}{49}\times \frac{42}{49}-\frac{4\times 8}{49}\\ & =\frac{196}{2401}\end{array}$$ $$\therefore a-b=\frac{14}{49}$$ From (iii) and (iv), $$a=\frac{4}{7},b=\frac{2}{7}$$ Hence probability of more probable of the $$\text{ two events }=\frac{4}{7}$$