A) 4/7
B) 6/7
C) 3/7
D) 5/7
Correct Answer: A
Solution :
[a] Let the probability of occurrence of first event A be ?a? |
i.e., P [a] = a |
\[\therefore P(notA)=1-a\] |
And also suppose that probability of occurrence of second event B, \[P(B)=b;\] \[\therefore P(notB)=1-b\] |
Now, \[P(A\,\,and\,\,not\,\,B)+P(not\,A\,and\,B)=\frac{26}{49}\] |
\[\Rightarrow P(A)\times P(not\,\,B)+P(not\,A)\times P(B)=\frac{26}{49}\] |
\[\Rightarrow a+b-2ab=\frac{26}{49}\] ??.(i) |
And P (not A and not B) = \[\frac{15}{49}\] |
\[\Rightarrow P(not\,A)\times P(not\,B)=\frac{15}{49}\] |
\[\Rightarrow 1-b-a+ab=\frac{15}{49}\Rightarrow a+b-ab=\frac{34}{49}\] ??(ii) |
From (i) and (ii), \[a+b=\frac{42}{49}\] ???(iii) |
and \[ab=\frac{8}{49}\] |
\[{{(a-b)}^{2}}={{(a+b)}^{2}}-4ab\] |
\[=\frac{42}{49}\times \frac{42}{49}-\frac{4\times 8}{49}=\frac{196}{2401}\] |
\[\therefore a-b=\frac{14}{49}\] ??.(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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