Probability Question 330
Question: 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:
Options:
A) 4/7
B) 6/7
C) 3/7
D) 5/7
Show Answer
Answer:
Correct Answer: A
Solution:
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(AandnotB)+P(notAandB)=\frac{26}{49} $
$ \Rightarrow P(A)\times P(notB)+P(notA)\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(notA)\times P(notB)=\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} $
BETA
