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 $ab = \frac{8}{49}$
$\left(a-b\right)^{2} = \left(a+b\right)^{2} - 4ab = \frac{42 }{49} \times\frac{42}{49} - \frac{4\times8}{49} = \frac{196}{2401}$
$ \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} $