CT1 Sample Questions — Batch 2: Fundamental Probability

This document provides rigorous, textbook-style solutions for the fundamental probability questions (Q4, Q5, Q6, Q15, Q16, Q22, and Q23) from the IFS UNIT - 1 & 2 sample questions sheet. All formulas are rendered in LaTeX, with step-by-step mathematical derivations and formal academic inferences.

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Question 4: Successive Draws (Replacement vs. Non-Replacement)

Given Data

A bag contains:

• Red balls ($R$): $3$
• Black balls ($B$): $6$
• Total balls ($N$): $9$

Let $R_1$ denote the event that the first ball drawn is red, and $R_2$ denote the event that the second ball drawn is red.

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Step-by-Step Derivation

(i) Case 1: With Replacement

If the first ball is replaced before the second draw, the trials are independent. The composition of the bag remains unchanged for both draws.

• Probability of first red ball: $P(R_1) = \frac{3}{9} = \frac{1}{3}$
• Since the ball is replaced, the conditional probability of the second draw is equal to its marginal probability: $P(R_2 \mid R_1) = P(R_2) = \frac{3}{9} = \frac{1}{3}$
• Joint probability of drawing two red balls in succession: $P(R_1 \cap R_2) = P(R_1) \cdot P(R_2 \mid R_1) = \frac{1}{3} \times \frac{1}{3} = \frac{1}{9} \approx 0.1111$

Inference: Under the independent trial condition (with replacement), the probability of drawing two red balls in succession is $1/9 \approx 11.11\%$.

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(ii) Case 2: Without Replacement

If the first ball is not replaced, the trials are dependent. The composition of the bag for the second draw depends on the outcome of the first.

• Probability of first red ball: $P(R_1) = \frac{3}{9} = \frac{1}{3}$
• After drawing a red ball, the remaining red balls are $2$ and the total balls are $8$: $P(R_2 \mid R_1) = \frac{2}{8} = \frac{1}{4}$
• Joint probability of drawing two red balls in succession: $P(R_1 \cap R_2) = P(R_1) \cdot P(R_2 \mid R_1) = \frac{1}{3} \times \frac{1}{4} = \frac{1}{12} \approx 0.0833$

Inference: Under the dependent trial condition (without replacement), the probability of drawing two red balls in succession decreases to $1/12 \approx 8.33\%$ due to the depletion of the target class in the first draw.

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Question 5: Drawing Red Balls Without Replacement

Given Data

A bag contains:

• Red balls ($R$): $3$
• White balls ($W$): $4$
• Total balls ($N$): $7$

We seek the joint probability of drawing two red balls in succession without replacement.

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Step-by-Step Derivation

Let $R_1$ and $R_2$ represent drawing a red ball on the first and second draws respectively.

• Probability of drawing red on the first draw: $P(R_1) = \frac{3}{7}$
• Since the draw is made without replacement, the conditional probability of drawing red on the second draw given that the first was red is: $P(R_2 \mid R_1) = \frac{3-1}{7-1} = \frac{2}{6} = \frac{1}{3}$
• Applying the multiplication rule for dependent events: $P(R_1 \cap R_2) = P(R_1) \cdot P(R_2 \mid R_1) = \frac{3}{7} \times \frac{1}{3} = \frac{1}{7} \approx 0.1429$

Inference: The probability that both drawn balls are red without replacement is $1/7 \approx 14.29\%$.

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Question 6: Bayes' Theorem and Bulbs from Selected Boxes

Given Data

Let $B_1$ denote selecting Box 1, and $B_2$ denote selecting Box 2. Since the box selection is random: $P(B_1) = P(B_2) = 0.5$

• Box 1: $1000$ bulbs total, $10\%$ defective ($100$ defective, $900$ non-defective).
• Box 2: $2000$ bulbs total, $5\%$ defective ($100$ defective, $1900$ non-defective).

Let $D_{both}$ denote the event that both drawn bulbs are defective (drawn without replacement).

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Step-by-Step Derivation

(i) Find the probability that both balls are defective ($P(D_{both})$)

Using the Law of Total Probability: $P(D_{both}) = P(D_{both} \mid B_1)P(B_1) + P(D_{both} \mid B_2)P(B_2)$

First, calculate the conditional probabilities of drawing two defective bulbs from each box:

• From Box 1: $P(D_{both} \mid B_1) = \frac{100}{1000} \times \frac{99}{999} = \frac{1}{10} \times \frac{11}{111} = \frac{11}{1110} \approx 0.009910$
• From Box 2: $P(D_{both} \mid B_2) = \frac{100}{2000} \times \frac{99}{1999} = \frac{1}{20} \times \frac{99}{1999} = \frac{99}{39980} \approx 0.002476$

Now, substitute these into the total probability formula: $P(D_{both}) = 0.5 \times \left(\frac{11}{1110}\right) + 0.5 \times \left(\frac{99}{39980}\right)$ $P(D_{both}) = \frac{11}{2220} + \frac{99}{79960} = \frac{11 \times 3998 + 99 \times 111}{4437780 \times 2}$ Let's calculate decimals to preserve absolute clarity: $P(D_{both}) \approx 0.00495495 + 0.00123812 = 0.00619307$ Or in exact fractional representation: $P(D_{both}) = \frac{54967}{8875560} \approx 0.0061931$

Inference: The total probability that both drawn bulbs are defective is approximately $0.619\%$.

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(ii) Assuming both are defective, find the probability they came from Box 1 ($P(B_1 \mid D_{both})$)

Using Bayes' Theorem: $P(B_1 \mid D_{both}) = \frac{P(D_{both} \mid B_1)P(B_1)}{P(D_{both})}$ $P(B_1 \mid D_{both}) = \frac{\frac{11}{2220}}{\frac{54967}{8875560}} = \frac{11}{2220} \times \frac{8875560}{54967} = \frac{11 \times 3998}{54967} = \frac{43978}{54967} \approx 0.80008$

Inference: Given that both drawn bulbs are defective, there is an $80.01\%$ posterior probability that they originated from Box 1. This higher probability is due to Box 1 having a significantly higher density of defective items ($10\%$) than Box 2 ($5\%$).

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Question 15: Course Enrollment Contingency Table Analysis

Given Data

We construct a formal contingency table from the given data ($N = 200$):

Gender	Mathematics ($M$)	Science ($S$)	Arts ($A$)	Total
Male ($M_1$)	$30$	$50$	$20$	$100$
Female ($F$)	$20$	$30$	$50$	$100$
Total	$50$	$80$	$70$	$200$

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Step-by-Step Probability Calculations

A. Probability that a student is enrolled in Science ($P(S)$)

$P(S) = \frac{n(S)}{N} = \frac{80}{200} = 0.40$

B. Probability that a student is Female ($P(F)$)

$P(F) = \frac{n(F)}{N} = \frac{100}{200} = 0.50$

C. Probability that a student is a Male enrolled in Mathematics ($P(M_1 \cap M)$)

$P(M_1 \cap M) = \frac{n(M_1 \cap M)}{N} = \frac{30}{200} = 0.15$

D. Probability that a student is a Female enrolled in Arts ($P(F \cap A)$)

$P(F \cap A) = \frac{n(F \cap A)}{N} = \frac{50}{200} = 0.25$

E. Probability that a student is Male given they are enrolled in Science ($P(M_1 \mid S)$)

$P(M_1 \mid S) = \frac{P(M_1 \cap S)}{P(S)} = \frac{n(M_1 \cap S)}{n(S)} = \frac{50}{80} = 0.625$

F. Probability that a student is enrolled in Mathematics given they are Female ($P(M \mid F)$)

$P(M \mid F) = \frac{P(M \cap F)}{P(F)} = \frac{n(M \cap F)}{n(F)} = \frac{20}{100} = 0.20$

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Question 16: Family Vehicle Ownership Contingency Table

Given Data

We construct the formal contingency table ($N = 300$):

Family Class	Cars ($Ca$)	Motorcycles ($Mc$)	Bicycles ($Bi$)	Total
With Children ($C$)	$90$	$30$	$20$	$140$
Without Children ($NC$)	$80$	$50$	$30$	$160$
Total	$170$	$80$	$50$	$300$

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Step-by-Step Probability Calculations

a. Probability that a selected family owns a motorcycle ($P(Mc)$)

$P(Mc) = \frac{n(Mc)}{N} = \frac{80}{300} = \frac{4}{15} \approx 0.2667$

b. Probability that a selected family has children ($P(C)$)

$P(C) = \frac{n(C)}{N} = \frac{140}{300} = \frac{7}{15} \approx 0.4667$

c. Probability that a selected family owns a car and has children ($P(Ca \cap C)$)

$P(Ca \cap C) = \frac{n(Ca \cap C)}{N} = \frac{90}{300} = 0.30$

d. Probability that a selected family owns a bicycle and has no children ($P(Bi \cap NC)$)

$P(Bi \cap NC) = \frac{n(Bi \cap NC)}{N} = \frac{30}{300} = 0.10$

e. Probability that a family has children given they own a car ($P(C \mid Ca)$)

$P(C \mid Ca) = \frac{P(Ca \cap C)}{P(Ca)} = \frac{n(Ca \cap C)}{n(Ca)} = \frac{90}{170} \approx 0.5294$

f. Probability that a family owns a motorcycle given they have no children ($P(Mc \mid NC)$)

$P(Mc \mid NC) = \frac{P(Mc \cap NC)}{P(NC)} = \frac{n(Mc \cap NC)}{n(NC)} = \frac{50}{160} = 0.3125$

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Question 22: Drawing King or Queen (Addition Rule)

Given Data

A standard playing deck contains $N = 52$ cards.

• Number of Kings ($K$) = $4$
• Number of Queens ($Q$) = $4$

We seek the probability of drawing either a King or a Queen in a single draw ($P(K \cup Q)$).

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Step-by-Step Derivation

• Probability of drawing a King: $P(K) = \frac{4}{52} = \frac{1}{13}$
• Probability of drawing a Queen: $P(Q) = \frac{4}{52} = \frac{1}{13}$
• Since a card cannot be both a King and a Queen simultaneously, these events are mutually exclusive: $P(K \cap Q) = 0$
• Applying the general Addition Rule: $P(K \cup Q) = P(K) + P(Q) - P(K \cap Q) = \frac{4}{52} + \frac{4}{52} - 0 = \frac{8}{52} = \frac{2}{13} \approx 0.1538$

Inference: The probability of drawing a King or a Queen is $2/13 \approx 15.38\%$.

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Question 23: Club Membership Probability (Addition Rule)

Given Data

Let $A$ denote membership in the Art Club, and $M$ denote membership in the Music Club.

• Probability of being in Art Club: $P(A) = 0.4$
• Probability of being in Music Club: $P(M) = 0.5$
• Probability of being in both clubs: $P(A \cap M) = 0.2$

We seek the probability of being a member of at least one club, i.e., $P(A \cup M)$.

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Step-by-Step Derivation

Applying the General Addition Rule of Probability: $P(A \cup M) = P(A) + P(M) - P(A \cap M)$ Substitute the given values into the equation: $P(A \cup M) = 0.4 + 0.5 - 0.2 = 0.7$

Inference: The probability that a randomly selected student is a member of at least one of the clubs is $0.70$ (or $70\%$). This accounts for the overlap of students who are members of both clubs, avoiding double-counting.

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Quick-Facts & Formula Sheet (Fundamental Probability)

Concept / Axiom	Mathematical Formulation	Description / Note
Probability Bounds	$0 \le P(A) \le 1$	Probability must lie in the range $[0, 1]$.
Complement Rule	$P(A^c) = 1 - P(A)$	Probability of event $A$ not occurring.
Addition Rule (Mutually Exclusive)	$P(A \cup B) = P(A) + P(B)$	If events cannot occur simultaneously ($P(A \cap B) = 0$).
General Addition Rule	$P(A \cup B) = P(A) + P(B) - P(A \cap B)$	For any two overlapping events $A$ and $B$.
Conditional Probability	$P(A \mid B) = \frac{P(A \cap B)}{P(B)}$	Probability of $A$ occurring given that $B$ has occurred.
Multiplication Rule (Dependent)	$P(A \cap B) = P(B) \cdot P(A \mid B)$	Joint probability of dependent events.
Multiplication Rule (Independent)	$P(A \cap B) = P(A) \cdot P(B)$	If occurrence of one doesn't affect the other ($P(A \mid B) = P(A)$).
Law of Total Probability	$P(A) = \sum_{i=1}^k P(A \mid B_i)P(B_i)$	Where $B_1, \dots, B_k$ partition the sample space $\Omega$.
Bayes' Theorem	$P(B_j \mid A) = \frac{P(A \mid B_j)P(B_j)}{\sum_{i=1}^k P(A \mid B_i)P(B_i)}$	Calculates posterior probability of partition $B_j$ given event $A$.