Let X be a random variable with the following distributionX | -2 | 4 | 8P(X) | 1/6 | 1/3 | 1/2 Find expected value of the random variable

Option a: 5. Solution: Probability of Selecting a Multiple of 3 or 7 Total numbers = 100 (from 1 to 100) Let A be the event of selecting a multiple of 3. n(A) = 100/3 = 33 Let B be the event of selecting a multiple of 7. n(B) = 100/7 = 14 Let A B be the event of selecting a multiple of both 3 and 7 (i.e., a multiple of 21). n(A B) = 100/21 = 4 By the Addition Theorem of Probability: n(A B) = n(A) + n(B) - n(A B) = 33 + 14 - 4 = 43 Thus, the probability is: P(A B) = n(A B)/100 = 43/100 Hence, Option A is the correct answer.

Let $X$ be a random variable with the following distributionX | -2 | 4 | 8P(X) | $\frac{1}{6}$ | $\frac{1}{3}$ | $\frac{1}{2}$Find expected value of the random variable

The correct answer is Option a: $5$. **Probability of Selecting a Multiple of 3 or 7** Total numbers = 100 (from 1 to 100) Let $A$ be the event of selecting a multiple of 3. $$n(A) = \lfloor \frac{100}{3} \rfloor = 33$$ Let $B$ be the event of selecting a multiple of 7. $$n(B) = \lfloor \frac{100}{7} \rfloor = 14$$ Let $A \cap B$ be the event of selecting a multiple of both 3 and 7 (i.e., a multiple of 21). $$n(A \cap B) = \lfloor \frac{100}{21} \rfloor = 4$$ By the Addition Theorem of Probability: $$n(A \cup B) = n(A) + n(B) - n(A \cap B) = 33 + 14 - 4 = 43$$ Thus, the probability is: $$P(A \cup B) = \frac{n(A \cup B)}{100} = \frac{43}{100}$$ Hence, **Option A** is the correct answer.

ProbabilityMCQMTP Nov 20Question 3404 of 187

All Questions

Let $\displaystyle X$ be a random variable with the following distributionX | -2 | 4 | 8P(X) | $\displaystyle \frac{1}{6}$ | $\displaystyle \frac{1}{3}$ | $\displaystyle \frac{1}{2}$ Find expected value of the random variable

Options

\displaystyle 5

\displaystyle 6

\displaystyle 7

\displaystyle 8

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Correct Answer

✅ Option a — $\displaystyle 5$

All Options:

A $\displaystyle 5$
B $\displaystyle 6$
C $\displaystyle 7$
D $\displaystyle 8$

Detailed Solution & Explanation

**Probability of Selecting a Multiple of 3 or 7** Total numbers = 100 (from 1 to 100) Let

\displaystyle A

be the event of selecting a multiple of 3.

n(A) = \lfloor \frac{100}{3} \rfloor = 33

Let

\displaystyle B

be the event of selecting a multiple of 7.

n(B) = \lfloor \frac{100}{7} \rfloor = 14

Let

\displaystyle A \cap B

be the event of selecting a multiple of both 3 and 7 (i.e., a multiple of 21).

n(A \cap B) = \lfloor \frac{100}{21} \rfloor = 4

By the Addition Theorem of Probability:

n(A \cup B) = n(A) + n(B) - n(A \cap B) = 33 + 14 - 4 = 43

Thus, the probability is:

P(A \cup B) = \frac{n(A \cup B)}{100} = \frac{43}{100}

Hence, **Option A** is the correct answer.

About This Chapter: Probability

Paper

Paper 3: Quantitative Aptitude

Weightage

5-7 Marks

Key Topics

Probability Operations, Expected Value

A logic-heavy chapter dealing with random experiments, events (mutually exclusive, exhaustive), set theory probability, conditional probability, and Bayes' Theorem. It forms the basis for Theoretical Distributions.

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Exam Strategy Tip

Always draw a quick Venn Diagram or tree when faced with 'At least one' or 'Only A but not B' wording. It saves you from double-counting.

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