Given two independent events, if the probability that exactly one of them occurs is $\frac{26}{49}$ and the…
Given two independent events, if the probability that exactly one of them occurs is $\frac{26}{49}$ and the probability that none of them occurs is $\frac{15}{49}$, then the probability of more probable of the two events is :
$4 / 7$
$6 / 7$
3/7
$5 / 7$
Solution
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}(\operatorname{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{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}
$
And $\mathrm{P}($ not $\mathrm{A}$ and not $\mathrm{B})=\frac{15}{49}$
$
\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}
$
From (i) and (ii),
$
a+b=\frac{42}{49}
$
and $a b=\frac{8}{49}$
$
\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}
$
$
\therefore \quad 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}
$