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 :

Let the two independent events be $$A$$ and $$B$$ with probabilities $$P(A)=p$$ and $$P(B)=q$$, where we take $$p \ge q$$ so that $$p$$ will be the probability of the more probable event.

Step 1 : Write equations from the given data
Because the events are independent, we have
  • Probability that none of them occurs: $$P(A^c \cap B^c)=(1-p)(1-q)=\tfrac{15}{49}$$ $$-(1)$$
  • Probability that exactly one occurs:
    $$P(A \cap B^c)+P(A^c \cap B)=p(1-q)+q(1-p)=p+q-2pq=\tfrac{26}{49}$$ $$-(2)$$

Step 2 : Convert to sums and products
Set $$S=p+q$$ and $$P=pq$$.
From $$(1)$$: $$1-S+P=\tfrac{15}{49}\;\Longrightarrow\;P-S=-\tfrac{34}{49}$$ $$-(3)$$
From $$(2)$$: $$S-2P=\tfrac{26}{49}$$ $$-(4)$$

Step 3 : Solve for $$S$$ and $$P$$
From $$(3)$$, $$P=S-\tfrac{34}{49}$$. Substitute this in $$(4)$$:
$$S-2\Bigl(S-\tfrac{34}{49}\Bigr)=\tfrac{26}{49}$$
$$S-2S+\tfrac{68}{49}=\tfrac{26}{49}$$
$$-S=-\tfrac{42}{49}\;\Longrightarrow\;S=\tfrac{42}{49}=\tfrac{6}{7}$$

Now from $$(3)$$, $$P=S-\tfrac{34}{49}=\tfrac{42}{49}-\tfrac{34}{49}=\tfrac{8}{49}$$.

Step 4 : Find individual probabilities
The numbers $$p$$ and $$q$$ are roots of $$t^2-St+P=0$$, i.e.
$$t^2-\tfrac{6}{7}t+\tfrac{8}{49}=0$$
Multiplying by $$49$$: $$49t^2-42t+8=0$$.

Using the quadratic formula:
$$t=\frac{42\pm\sqrt{(-42)^2-4\cdot49\cdot8}}{2\cdot49}=\frac{42\pm\sqrt{1764-1568}}{98}=\frac{42\pm14}{98}$$
Thus $$t_1=\tfrac{56}{98}=\tfrac{4}{7}$$ and $$t_2=\tfrac{28}{98}=\tfrac{2}{7}$$.

Step 5 : Identify the more probable event
Since $$p \ge q$$, we take $$p=\tfrac{4}{7}$$.

Answer
The probability of the more probable event is $$\tfrac{4}{7}$$, which corresponds to Option A.