Question:
If $P(E) = \frac{1}{3}$, $P(F) = \frac{2}{5}$ and $P(E \cup F) - P(E \cap F) = \frac{1}{5}$ then $P(E \cup F)$ is equal to
If $P(E) = \frac{1}{3}$, $P(F) = \frac{2}{5}$ and $P(E \cup F) - P(E \cap F) = \frac{1}{5}$ then $P(E \cup F)$ is equal to
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The expression $P(E \cup F) - P(E \cap F)$ represents the probability of "exactly one of the events occurring" (Symmetric Difference). Recognizing this helps visualize the problem on a Venn diagram immediately.
Updated On: Jun 6, 2026
- $\frac{11}{15}$
- $\frac{4}{15}$
- $\frac{8}{15}$
- $\frac{7}{15}$
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The Correct Option is D
Solution and Explanation
We are given the individual probabilities of two events and a relationship between their union and intersection.
Step 1: \color{redState the Addition Theorem of Probability
For any two events E and F, the probability of their union is:
$P(E \cup F) = P(E) + P(F) - P(E \cap F)$.
Step 2: \color{redUse the Given Condition to Express Intersection
We are given: $P(E \cup F) - P(E \cap F) = \frac{1}{5}$.
From this, we can write: $P(E \cap F) = P(E \cup F) - \frac{1}{5}$.
Step 3: \color{redSubstitute into the Addition Theorem
Now, substitute the expression for the intersection into the formula from
Step 1:
$P(E \cup F) = P(E) + P(F) - [P(E \cup F) - \frac{1}{5}]$.
$P(E \cup F) = \frac{1}{3} + \frac{2}{5} - P(E \cup F) + \frac{1}{5}$.
Step 4: \color{redSolve for P(E $\cup$ F)
Move both $P(E \cup F)$ terms to one side:
$2P(E \cup F) = \frac{1}{3} + \frac{2}{5} + \frac{1}{5}$.
$2P(E \cup F) = \frac{1}{3} + \frac{3}{5}$.
Find a common denominator (15):
$2P(E \cup F) = \frac{5 + 9}{15} = \frac{14}{15}$.
$P(E \cup F) = \frac{14}{15} \times \frac{1}{2} = \frac{7}{15}$.
The probability $P(E \cup F)$ is $\frac{7}{15}$, matching Option (4).
Step 1: \color{redState the Addition Theorem of Probability
For any two events E and F, the probability of their union is:
$P(E \cup F) = P(E) + P(F) - P(E \cap F)$.
Step 2: \color{redUse the Given Condition to Express Intersection
We are given: $P(E \cup F) - P(E \cap F) = \frac{1}{5}$.
From this, we can write: $P(E \cap F) = P(E \cup F) - \frac{1}{5}$.
Step 3: \color{redSubstitute into the Addition Theorem
Now, substitute the expression for the intersection into the formula from
Step 1:
$P(E \cup F) = P(E) + P(F) - [P(E \cup F) - \frac{1}{5}]$.
$P(E \cup F) = \frac{1}{3} + \frac{2}{5} - P(E \cup F) + \frac{1}{5}$.
Step 4: \color{redSolve for P(E $\cup$ F)
Move both $P(E \cup F)$ terms to one side:
$2P(E \cup F) = \frac{1}{3} + \frac{2}{5} + \frac{1}{5}$.
$2P(E \cup F) = \frac{1}{3} + \frac{3}{5}$.
Find a common denominator (15):
$2P(E \cup F) = \frac{5 + 9}{15} = \frac{14}{15}$.
$P(E \cup F) = \frac{14}{15} \times \frac{1}{2} = \frac{7}{15}$.
The probability $P(E \cup F)$ is $\frac{7}{15}$, matching Option (4).
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