Probability and Random Events Short Notes for XAT: Concepts, Formulas, Worked Examples & Practice

Probability and Random Events short notes for XAT: 1 study cards covering concepts, formulas, shortcuts and exam traps, plus solved practice questions.

Summary: Random Selection Strategies

1. Simultaneous vs. Sequential

Simultaneous (at once): Use combinations $(r n)$ .

Sequential (one by one): Multiply individual probabilities step-by-step.

2. The "At Least One" Shortcut

P (at least one) = 1 - P (none)

3. Conditional Selection

When a condition is given, restrict the sample space.

P (A ∣ B) = \frac{P ( A \cap B )}{P ( B )}

4. Arrangement Constraints (Gap Method)

For "no two items adjacent", place the unconstrained items first, then place the constrained items in the gaps between them.

Probability and Random Events: Solved Questions with Step-by-Step Explanations (2 Problems)

Question 1 · Quantitative Aptitude and Data Interpretation (QA & DI) MCQ

Let $S$ be the set of all permutations of the digits ${1, 2, 3, 4, 5, 6}$ . A permutation is chosen uniformly at random from $S$ .

Let $A$ be the event that the digit 1 appears before the digit 2.

Let $B$ be the event that the digit 3 appears before the digit 4.

Let $C$ be the event that the digit 5 appears before the digit 6.

What is the probability that exactly two of these events occur?

A.
1/4
B.
3/8
C.
1/2
D.
5/8

Correct Answer:

Step-by-Step Solution

Key idea: Symmetry and Independence of relative orderings.

Step 1: Analyze individual probabilities.

In any random permutation, for any distinct pair of digits (e.g., 1 and 2), the probability that one appears before the other is $1/2$ .

So, $P (A) = 1/2$ , $P (B) = 1/2$ , $P (C) = 1/2$ .

Step 2: Analyze independence.

The relative ordering of disjoint pairs (1,2), (3,4), and (5,6) are mutually independent events.

Why? The positions of 1 and 2 are symmetric with respect to 3 and 4. There is no bias introduced by the relative order of 3 and 4 on the relative order of 1 and 2.

Thus, $A, B, C$ are independent events.

Step 3: Define the target event.

We want exactly two events to occur.

Possible scenarios:

$A \cap B \cap C^{c}$
$A \cap B^{c} \cap C$
$A^{c} \cap B \cap C$

Step 4: Calculate probability for one scenario.

$P (A \cap B \cap C^{c}) = P (A) P (B) P (C^{c})$ (due to independence)

$= (1/2) \times (1/2) \times (1/2) = 1/8$ .

Step 5: Sum the probabilities.

Since the scenarios are mutually exclusive:

$P (Exactly 2) = 1/8 + 1/8 + 1/8 = 3/8$ .

Question 2 · Quantitative Aptitude and Data Interpretation (QA & DI) NAT

A box contains 4 Red, 4 Green, and 4 Blue balls. Three balls are drawn simultaneously at random.

Let $X$ be the number of different colors among the drawn balls.

What is the expected value of $X$ ?

Correct Answer:

2.2

Step-by-Step Solution

Key idea: Linearity of Expectation with Indicator Variables.

Step 1: Define Indicator Variables.

Let $I_{R}$ be 1 if at least one Red ball is drawn, 0 otherwise.

Let $I_{G}$ be 1 if at least one Green ball is drawn, 0 otherwise.

Let $I_{B}$ be 1 if at least one Blue ball is drawn, 0 otherwise.

Then $X = I_{R} + I_{G} + I_{B}$ .

By Linearity of Expectation:

$E [X] = E [I_{R}] + E [I_{G}] + E [I_{B}]$ .

Step 2: Calculate $E [I_{R}]$ .

$E [I_{R}] = P (I_{R} = 1) = P (At least one Red)$ .

$P (At least one Red) = 1 - P (No Red)$ .

Total balls = 12. Draw 3.

Total ways = $(3 12) = \frac{12 \times 11 \times 10}{6} = 220$ .

Ways to draw No Red (only from 4G, 4B = 8 balls):

$(3 8) = \frac{8 \times 7 \times 6}{6} = 56$ .

$P (No Red) = \frac{56}{220}$ .

$P (At least one Red) = 1 - \frac{56}{220} = \frac{164}{220}$ .

Step 3: Symmetry.

By symmetry, $E [I_{G}] = E [I_{B}] = E [I_{R}] = \frac{164}{220}$ .

Step 4: Sum.

$E [X] = 3 \times \frac{164}{220}$ .

Simplify fraction:

$164/220$ . Divide by 4: $41/55$ .

$E [X] = 3 \times \frac{41}{55} = \frac{123}{55}$ .

Calculate decimal:

$123 \div 55 \approx 2.236$ .

Wait, let's re-calculate $123/55$ .

$55 \times 2 = 110$ . Remainder 13.

$130/55 = 2.36...$

$123/55 = 2.23636...$

Is the answer a clean decimal?

Let's check the calculation of $(3 12)$ .

$12 \times 11 \times 10/6 = 2 \times 11 \times 10 = 220$ . Correct.

$(3 8) = 56$ . Correct.

$1 - 56/220 = 164/220$ . Correct.

$164/4 = 41$ . $220/4 = 55$ . Correct.

$3 \times 41 = 123$ .

$123/55$ .

The question asks for NAT. Usually 2 decimal places.

Answer: 2.24.

Let's double check if "Expected Value" implies a fraction or decimal. NAT usually accepts decimals.

Let's try to verify with direct counting.

Possible values for X: 1, 2, 3.

P(X=1): All same color.

3 Reds: $(3 4) = 4$ .

3 Greens: 4.

3 Blues: 4.

Total 12.

$P (X = 1) = 12/220$ .

P(X=3): All different colors (1R, 1G, 1B).

$4 \times 4 \times 4 = 64$ .

$P (X = 3) = 64/220$ .

P(X=2): 1 - P(X=1) - P(X=3) = $1 - 12/220 - 64/220 = 144/220$ .

Check sum: $12 + 64 + 144 = 220$ . Correct.

$E [X] = 1 \times \frac{12}{220} + 2 \times \frac{144}{220} + 3 \times \frac{64}{220}$ .

Numerator: $12 + 288 + 192 = 492$ .

$E [X] = \frac{492}{220}$ .

Divide by 4: $\frac{123}{55}$ .

Matches.

$123/55 = 2.23636...$

Rounding to 2 decimal places: 2.24.

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chapter

Probability and Random Events Short Notes for XAT: Concepts, Formulas, Worked Examples & Practice

Probability and Random Events short notes for XAT: 1 study cards covering concepts, formulas, shortcuts and exam traps, plus solved practice questions.

A question from this chapter

Question 1

Let S be the set of all permutations of the digits {1, 2, 3, 4, 5, 6} . A permutation is chosen uniformly at random from S . Let A be the event that the digit 1 appears before the digit 2. Let B be the event that the digit 3 appears before the digit 4. Let C be the event that the digit 5 appears before the digit 6. What is the probability that exactly two of these events occur?

Question 2

A box contains 4 Red, 4 Green, and 4 Blue balls. Three balls are drawn simultaneously at random. Let X be the number of different colors among the drawn balls. What is the expected value of X ?

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