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: Set up the union formula symbolically in terms of P(B):
Since $P(A)=\dfrac{1}{3}P(B)$ and $P(A\cap B)=P(A/B)\cdot P(B)=\dfrac{2}{5}P(B)$, write everything as a multiple of $P(B)$ before plugging in numbers.

Step 2: Factor P(B) out of the union formula:
\[ P(A\cup B) = P(A)+P(B)-P(A\cap B) = \dfrac{1}{3}P(B) + P(B) - \dfrac{2}{5}P(B) \]
\[ P(A\cup B) = P(B)\left(\dfrac{1}{3}+1-\dfrac{2}{5}\right) \]

Step 3: Simplify the bracket using a common denominator of 15:
\[ \dfrac{1}{3}+1-\dfrac{2}{5} = \dfrac{5}{15}+\dfrac{15}{15}-\dfrac{6}{15} = \dfrac{14}{15} \]
So the whole union probability is $\dfrac{14}{15}$ times $P(B)$.

Step 4: Substitute the numeric value of P(B):
\[ P(A\cup B) = \dfrac{14}{15}\times\dfrac{5}{13} = \dfrac{70}{195} = \dfrac{14}{39} \]
The 5 cancels between numerator and denominator, giving the same reduced fraction as before.

Final Answer:
Factoring P(B) out first and substituting last gives the same result as the direct method. \[ \boxed{P(A\cup B)=\dfrac{14}{39}} \]