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 :

• A. \(4/7\)
• B. \(6/7\)
• C. \(3/7\)
• D. \(5/7\)

Hint:

Let the probabilities be \(p,q\). Use \(P(\text{none})=(1-p)(1-q)\) to get \(pq\), then form a quadratic.

Answer 1

The Correct Option is A

Let \(P(A)=p,\ P(B)=q\) (independent).
\(p(1-q)+q(1-p)=p+q-2pq=\frac{26}{49}\) and \((1-p)(1-q)=1-p-q+pq=\frac{15}{49}\).
Add the two: \(1-pq=\frac{41}{49}\Rightarrow pq=\frac{8}{49}\).
Then \(p+q=\frac{26}{49}+2\cdot\frac{8}{49}=\frac{42}{49}=\frac67\).
\(p,q\) are roots of \(t^2-\frac67t+\frac{8}{49}=0\Rightarrow49t^2-42t+8=0\Rightarrow t=\frac{42\pm14}{98}=\frac47,\frac27\).
The more probable event has probability \(\frac47\).
Answer: (1) \(4/7\)