JEE Main & Advanced Mathematics Probability Question Bank Self Evaluation Test - Probability-II

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

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 \[=\frac{4}{7}\]