Question

If \(3P(A)=P(B)=\dfrac{5}{13}\) and \(P(A/B)=\dfrac{2}{5}\), then \(P(A\cup B)\) will be:

• A. \(\dfrac{20}{39}\)
• B. \(\dfrac{16}{39}\)
• C. \(\dfrac{11}{39}\)
• D. \(\dfrac{14}{39}\)

Hint:

Find \(P(A)\) from \(3P(A)=P(B)\), then \(P(A\cap B)=P(A/B)\cdot P(B)\), then use the union formula.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
Given \( 3P(A)=P(B)=\dfrac{5}{13} \), this means \( P(B)=\dfrac{5}{13} \) and \( P(A)=\dfrac{1}{3}P(B) \).
Also given is the conditional probability \( P(A/B)=\dfrac{2}{5} \), which needs the formula \( P(A/B)=\dfrac{P(A\cap B)}{P(B)} \).

Step 2: Finding P(A):
Divide \( P(B)=\dfrac{5}{13} \) by 3.
\[ P(A)=\dfrac{1}{3}\times\dfrac{5}{13}=\dfrac{5}{39} \]

Step 3: Finding P(A intersection B):
Use the conditional probability formula and substitute the known values.
\[ P(A/B)=\dfrac{P(A\cap B)}{P(B)} \quad\Rightarrow\quad \dfrac{2}{5}=\dfrac{P(A\cap B)}{5/13} \]
\[ P(A\cap B)=\dfrac{2}{5}\times\dfrac{5}{13}=\dfrac{2}{13} \]

Step 4: Applying the union formula:
Use \( P(A\cup B)=P(A)+P(B)-P(A\cap B) \), converting all fractions to a common denominator of 39.
\[ P(A\cup B)=\dfrac{5}{39}+\dfrac{15}{39}-\dfrac{6}{39}=\dfrac{14}{39} \]

Step 5: Why option A is wrong:
Option A, \( \dfrac{20}{39} \), results from adding P(A) and P(B) without subtracting \( P(A\cap B) \) at all.

Step 6: Why option B is wrong:
Option B, \( \dfrac{16}{39} \), would come from subtracting only \( \dfrac{4}{39} \) for the intersection, an incorrect value of \( P(A\cap B) \).

Step 7: Why option C is wrong:
Option C, \( \dfrac{11}{39} \), would come from wrongly subtracting P(A) instead of \( P(A\cap B) \) or a similar arithmetic slip.

Final Answer:
Substituting the correct values of P(A), P(B) and \( P(A\cap B) \) into the union formula gives 14/39. \[ \boxed{P(A\cup B)=\dfrac{14}{39}} \]