A) \[4/7\]
B) \[6/7\]
C) \[3/7\]
D) \[5/7\]
Correct Answer: A
Solution :
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\[=\frac{4}{7}\]
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