Question

The probability of the simultaneous occurrence of two events A and B is \(p\). If the probability that exactly one of A and B occurs is \(q\), then
A. \(P(\overline{A})+P(\overline{B})=2+2q-p\)
B. \(P(\overline{A})+P(\overline{B})=2-2p-q\)
C. \(P(\overline{A}\cap\overline{B})=1-p-q\)
D. \(P((A\cap B)|(A\cup B))=\frac{p}{p+q}\)

Choose the correct answer from the options given below:

• A. A, C and D only
• B. B, C and D only
• C. C and D only
• D. B only

Hint:

Show that \(P(A\cup B)=p+q\) and \(P(A)+P(B)=2p+q\), then test each statement.

Answer 1

The Correct Option is B

Step 1: Write down what is given:
\(P(A\cap B)=p\). The event 'exactly one of A and B occurs' is \((A\cap B')\cup(A'\cap B)\), and its probability is \(q\).
So \[ q=P(A)+P(B)-2p \] because \(P(A\cap B')=P(A)-p\) and \(P(A'\cap B)=P(B)-p\).
From this, \(P(A)+P(B)=q+2p\).

Step 2: Check statement A:
\[ P(\overline{A})+P(\overline{B})=2-\big(P(A)+P(B)\big)=2-(q+2p)=2-2p-q \]
Statement A says \(2+2q-p\), which is not equal to this in general. So A is FALSE.

Step 3: Check statement B:
From the line above, \(P(\overline{A})+P(\overline{B})=2-2p-q\). This is exactly what statement B says. So B is TRUE.

Step 4: Check statement C:
By De Morgan's law, \(\overline{A}\cap\overline{B}=\overline{A\cup B}\).
Now \(P(A\cup B)=P(A)+P(B)-p=(q+2p)-p=p+q\).
So \(P(\overline{A}\cap\overline{B})=1-(p+q)=1-p-q\). Statement C is TRUE.

Step 5: Check statement D:
\[ P\big((A\cap B)|(A\cup B)\big)=\frac{P\big((A\cap B)\cap(A\cup B)\big)}{P(A\cup B)} \]
Since \(A\cap B\) lies inside \(A\cup B\), the numerator is just \(P(A\cap B)=p\). The denominator is \(p+q\). So the value is \(\frac{p}{p+q}\). Statement D is TRUE.

Step 6: Match with the options:
B, C and D are true and A is false. Option 1 wrongly includes A. Option 3 leaves out B. Option 4 has only B. So option 2 is correct.

Final Answer:
B, C and D are correct. This is option 2. \[ \boxed{\text{B, C and D only}} \]