If $ E $ and $ F $ are two mutually exclusive events such that... | Mathematics MCQ | Examoogle

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If $E$ and $F$ are two mutually exclusive events such that $P(E) = \frac{1}{6}$ and $P(F) = \frac{1}{2}$ , then find the $P(\text{not } E \text{ and not } F)$ .

$\frac{2}{3}$

$\frac{5}{6}$

$\frac{1}{2}$

$\frac{1}{3}$

Correct Answer

⚙️ TE • Technical Set Theory IdentityMathematicsProbability

Option A

$\frac{2}{3}$

Quick Summary:

Since events are mutually exclusive, P(E or F) is simply the sum. Use De Morgan's Law: P(not E and not F) is 1 minus the probability of the union.

📐MAMath SolutionSet Theory Identity

📋 Given

Two mutually exclusive events E and F with probabilities P(E) = 1/6 and P(F) = 1/2.

🔢 Formula Used

P(E∪F)=P(E)+P(F)−P(E∩F)P(E \cup F) = P(E) + P(F) - P(E \cap F)P(E∪F)=P(E)+P(F)−P(E∩F) and P(Ec∩Fc)=1−P(E∪F)P(E^c \cap F^c) = 1 - P(E \cup F)P(Ec∩Fc)=1−P(E∪F)

⚡ Exam Hall Shortcut / Speed Trick

Since events are mutually exclusive, P(E or F) is simply the sum. Use De Morgan's Law: P(not E and not F) is 1 minus the probability of the union.

⚠️ Common Student Trap / Pitfall

Students often wrongly assume P(not E and not F) = P(not E) * P(not F), which is only valid if events are independent, not mutually exclusive.

📊 Diagram / Illustration

🔢 Step-by-Step Solution

Calculate Union Probability

For mutually exclusive events, the intersection $P(E \cap F) = 0$ . Using the addition rule, calculate the union $P(E \cup F)$ .

P(E∪F)=16+12=1+36=46=23P(E \cup F) = \frac{1}{6} + \frac{1}{2} = \frac{1+3}{6} = \frac{4}{6} = \frac{2}{3}P(E∪F)=61​+21​=61+3​=64​=32​

Apply Complement Rule

We need the probability that neither event occurs, which is the complement of their union, denoted as $P((E \cup F)^c)$ .

P(not E and not F)=1−P(E∪F)P(\text{not } E \text{ and not } F) = 1 - P(E \cup F)P(not E and not F)=1−P(E∪F)

Final Calculation

Subtract the union probability from 1 to find the result.

1−23=131 - \frac{2}{3} = \frac{1}{3}1−32​=31​

✅

D is correct because the probability of the union is 2/3, and its complement is 1 - 2/3 = 1/3.

Core Concepts Used

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Mutually Exclusive Events De Morgan's Laws Complementary Probability

💡 EXAM TIP

This concept of 'not A and not B' is identical to the 'Principle of Inclusion-Exclusion' often used in sets and counting problems.

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