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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 :

Option: 1

\frac{4}{7}


Option: 2

\frac{6}{7}


Option: 3

\frac{3}{7}


Option: 4

\frac{5}{7}


Answers (1)

best_answer

Let the probability of occurrence of first event A be 'a''
i..e., P(A) = a and (P not A) = 1 - a
And also suppose that probability of occurrence of second event B be 'b' i..e., i.e., P(B) = b and (P not B) = 1 - b
 

According to the question

\begin{aligned} & P(\text { A and not } B)+P(\operatorname{not} A \text { and } B)=\frac{26}{49} \\ \Rightarrow & P(A) \times P(\operatorname{not} B)+P(\operatorname{not} A) \times P(B)=\frac{26}{49} \\ \Rightarrow & a \times(1-b)+(1-a) b=\frac{26}{49} \\ \Rightarrow & a+b-2 a b=\frac{26}{49} \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\ldots(1)\end{aligned}

\\\quad \text { And } P(\operatorname{not} A \text { and } \operatorname{not} B)=\frac{15}{49} \\ \Rightarrow \quad P(\operatorname{not} A) \times P(\operatorname{not} B)=\frac{15}{49} \\ \Rightarrow \quad(1-a) \times(1-b)=\frac{15}{49} \\ \Rightarrow \quad 1-b-a+a b=\frac{15}{49} \\ \Rightarrow \quad a+b-a b=\frac{34}{49} \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\ldots(2) \\

from (1) and (2) 

\\a+b=\frac{42}{49} \\ \text { and } a b=\frac{8}{49}

\\(a-b)^{2}=(a+b)^{2}-4 a b\\=\frac{42}{49} \times \frac{42}{49}-\frac{4 \times 8}{49} \\ =\frac{196}{2401} \\ \therefore a-b=\frac{14}{49}

we get a = 4/7 and b = 2/7

 

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mansi

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