• # question_answer Given two independent events, if the probability that exactly one of them occurs is $\frac{26}{49}$ and the probability that nine of them occurs is $\frac{15}{49},$ then the probability of more probable of the two events is :     JEE Main  Online Paper (Held On 22 April 2013) A)  $4/7$                   B) $6/7$ C)  $3/7$                   D)  $5/7$

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