Question

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 :

• A. 4/7
• B. 6/7
• C. 3/7
• D. 5/7

Answer 1

Answer: (A)

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