Probability

Study Action Plan (Sorted by Difficulty)
Mutually Exclusive and Exhaustive Events
Accuracy: 0%
Total Attempted0
Total Correct0
Total Incorrect0
Difficulty Level:
High
Probability of Three Events
Accuracy: 0%
Total Attempted0
Total Correct0
Total Incorrect0
Difficulty Level:
High
Combinatorics-Based Probability Problems
Accuracy: 33%
Total Attempted3
Total Correct1
Total Incorrect2
Difficulty Level:
High
Axiomatic Probability
Accuracy: 56%
Total Attempted9
Total Correct5
Total Incorrect4
Difficulty Level:
Medium
Probability of Events
Accuracy: 67%
Total Attempted9
Total Correct6
Total Incorrect3
Difficulty Level:
Medium
Valid Assignment of Probabilities
Accuracy: 67%
Total Attempted6
Total Correct4
Total Incorrect2
Difficulty Level:
Medium
Algebra of Events
Accuracy: 78%
Total Attempted9
Total Correct7
Total Incorrect2
Difficulty Level:
Low
Sample Space and Events
Accuracy: 78%
Total Attempted9
Total Correct7
Total Incorrect2
Difficulty Level:
Low
Equally Likely Outcomes and Classical Probability
Accuracy: 100%
Total Attempted3
Total Correct3
Total Incorrect0
Difficulty Level:
Low
Probability of Union and Complement of Events
Accuracy: 100%
Total Attempted3
Total Correct3
Total Incorrect0
Difficulty Level:
Low

Learning Objectives

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Learning Objectives

  • Understand the concept of Sample Space and Events as the universal set for an experiment and define events as subsets of this space, including impossible, sure, simple, and compound events.
  • Apply Algebra of Events using set operations like union, intersection, and complement, and comprehend mutually exclusive and exhaustive events.
  • Grasp the Axiomatic Probability approach, defining probability as a function satisfying non-negativity, normalization, and additivity for mutually exclusive events.
  • Calculate the Probability of Events using the axiomatic approach, including probabilities of 'A or B', 'not A', and equally likely outcomes.
  • Analyze Mutually Exclusive and Exhaustive Events to understand events that cannot occur simultaneously and sets of events covering all possible outcomes.
  • Utilize formulas for Probability of Union and Complement of Events, such as P(A∪B)=P(A)+P(B)−P(A∩B)P(A \cup B) = P(A) + P(B) - P(A \cap B) and P(A′)=1−P(A)P(A') = 1 - P(A), with Venn-diagram interpretation.
  • Determine probabilities in Equally Likely Outcomes and Classical Probability scenarios using P(E)=n(E)n(S)P(E) = \frac{n(E)}{n(S)} for dice, coins, cards, and other selections.
  • Validate Assignment of Probabilities ensuring 0≤P(ωi)≤10 \leq P(\omega_i) \leq 1 and total probability equals 1.
  • Solve Combinatorics-Based Probability Problems using arrangements and combinations for scenarios like committees, lotteries, and card hands.
  • Apply the formula for Probability of Three Events such as P(A∪B∪C)P(A \cup B \cup C) and related event-union problems.

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Revision Notes & Summary

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Chapter Notes

Sample Space and Events

  • Sample Space: The universal set for an experiment.
  • Events: Subsets of the sample space.
    • Impossible Event: The empty set Φ\Phi.
    • Sure Event: The whole sample space SS.
    • Simple Event: Contains only one sample point.
    • Compound Event: Contains more than one sample point.

Algebra of Events

  • Complementary Event: For event AA, the complement is A′=S−AA' = S - A.
  • Union of Events: A∪BA \cup B includes all outcomes in AA or BB or both.
  • Intersection of Events: A∩BA \cap B includes outcomes common to both AA and BB.
  • Difference of Events: A−B=A∩B′A - B = A \cap B' includes outcomes in AA but not in BB.

Axiomatic Probability

  • Axioms:
    1. P(E)≥0P(E) \geq 0 for any event EE.
    2. P(S)=1P(S) = 1.
    3. For mutually exclusive events EE and FF, P(E∪F)=P(E)+P(F)P(E \cup F) = P(E) + P(F).

Probability of Events

  • Probability Calculation:
    • For any event AA, P(A)=∑P(wi)P(A) = \sum P(w_i) for all wi∈Aw_i \in A.
    • Equally Likely Outcomes: P(E)=n(E)n(S)P(E) = \frac{n(E)}{n(S)}.

Mutually Exclusive and Exhaustive Events

  • Mutually Exclusive: A∩B=ΦA \cap B = \Phi.
  • Exhaustive Events: E1∪E2∪…∪En=SE_1 \cup E_2 \cup \ldots \cup E_n = S.

Probability of Union and Complement of Events

  • Union Formula: P(A∪B)=P(A)+P(B)−P(A∩B)P(A \cup B) = P(A) + P(B) - P(A \cap B).
  • Complement Formula: P(A′)=1−P(A)P(A') = 1 - P(A).

Equally Likely Outcomes and Classical Probability

  • Classical Probability: P(E)=n(E)n(S)P(E) = \frac{n(E)}{n(S)} when all outcomes are equally likely.

Valid Assignment of Probabilities

  • Conditions:
    • 0≤P(wi)≤10 \leq P(w_i) \leq 1 for each wi∈Sw_i \in S.
    • ∑P(wi)=1\sum P(w_i) = 1.

Combinatorics-Based Probability Problems

  • Use of Combinations and Permutations: For problems involving arrangements and selections.

Probability of Three Events

  • Formula: P(A∪B∪C)=P(A)+P(B)+P(C)−P(A∩B)−P(B∩C)−P(C∩A)+P(A∩B∩C)P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A \cap B) - P(B \cap C) - P(C \cap A) + P(A \cap B \cap C).
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Practice Test – MCQs, True/False

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Multiple Choice Questions

A.

1930\frac{19}{30}

B.

710\frac{7}{10}

C.

1115\frac{11}{15}

D.

35\frac{3}{5}
Correct Answer: C

Solution:

Using the formula P(A∪B)=P(A)+P(B)−P(A∩B)P(A \cup B) = P(A) + P(B) - P(A \cap B), we have P(A∪B)=3060+3260−2460=3860=1930P(A \cup B) = \frac{30}{60} + \frac{32}{60} - \frac{24}{60} = \frac{38}{60} = \frac{19}{30}.

Chapter Concept:

Probability of Union and Complement of Events

A.

536\frac{5}{36}

B.

736\frac{7}{36}

C.

1136\frac{11}{36}

D.

1336\frac{13}{36}
Correct Answer: C

Solution:

The sample space has 36 outcomes. The favorable outcomes for the sum greater than 8 are {(3,6),(4,5),(4,6),(5,4),(5,5),(5,6),(6,3),(6,4),(6,5),(6,6)}\{(3,6), (4,5), (4,6), (5,4), (5,5), (5,6), (6,3), (6,4), (6,5), (6,6)\}, totaling 10 outcomes. Therefore, P(E)=1036=518P(E) = \frac{10}{36} = \frac{5}{18}.

Chapter Concept:

Equally Likely Outcomes and Classical Probability

A.

0.7

B.

0.8

C.

0.9

D.

0.6
Correct Answer: B

Solution:

The probability of the union of two events AA and BB is given by P(A∪B)=P(A)+P(B)−P(A∩B)P(A \cup B) = P(A) + P(B) - P(A \cap B). Therefore, P(A∪B)=0.5+0.3−0.1=0.7P(A \cup B) = 0.5 + 0.3 - 0.1 = 0.7.

Chapter Concept:

Probability of Union and Complement of Events

A.

122\frac{1}{22}

B.

111\frac{1}{11}

C.

144\frac{1}{44}

D.

133\frac{1}{33}
Correct Answer: A

Solution:

The number of ways to choose 3 red balls from 5 is (53)=10\binom{5}{3} = 10. The total number of ways to choose any 3 balls from 12 is (123)=220\binom{12}{3} = 220. Thus, the probability is 10220=122\frac{10}{220} = \frac{1}{22}.

Chapter Concept:

Combinatorics-Based Probability Problems

A.

0.9

B.

0.1

C.

0.2

D.

0.7
Correct Answer: A

Solution:

Since AA and BB are mutually exclusive, P(A∪B)=P(A)+P(B)=0.4+0.5=0.9P(A \cup B) = P(A) + P(B) = 0.4 + 0.5 = 0.9.

Chapter Concept:

Axiomatic Probability

A.

0.1

B.

0.2

C.

0.3

D.

0.4
Correct Answer: B

Solution:

Using the formula P(A∪B)=P(A)+P(B)−P(A∩B)P(A \cup B) = P(A) + P(B) - P(A \cap B), we have 0.7=0.4+0.5−P(A∩B)0.7 = 0.4 + 0.5 - P(A \cap B), thus P(A∩B)=0.2P(A \cap B) = 0.2.

Chapter Concept:

Algebra of Events

A.

A: 'an odd number appears' and B: 'an even number appears'

B.

A: 'a number less than 4 appears' and B: 'a number greater than 2 appears'

C.

A: 'a prime number appears' and B: 'a composite number appears'

D.

A: 'a number greater than 3 appears' and B: 'a number less than 5 appears'
Correct Answer: A

Solution:

Events A and B are mutually exclusive if they cannot occur simultaneously. For option A, an odd number and an even number cannot appear at the same time.

Chapter Concept:

Mutually Exclusive and Exhaustive Events

A.

0.5

B.

0.7

C.

0.8

D.

1.0
Correct Answer: B

Solution:

The probability of event AA is P(A)=P(w1)+P(w3)=0.5+0.2=0.7P(A) = P(w_1) + P(w_3) = 0.5 + 0.2 = 0.7.

Chapter Concept:

Valid Assignment of Probabilities

A.

Yes, they are mutually exclusive.

B.

No, they are not mutually exclusive.

C.

They are mutually exclusive and exhaustive.

D.

They are neither mutually exclusive nor exhaustive.
Correct Answer: B

Solution:

Event A = {1, 2, 3} and Event B = {2, 4, 6}. Since 2 is common to both events, A and B are not mutually exclusive.

Chapter Concept:

Mutually Exclusive and Exhaustive Events

A.

0.7

B.

0.9

C.

0.5

D.

1.1
Correct Answer: B

Solution:

The probability of 'A or B' is calculated using the formula P(A∪B)=P(A)+P(B)−P(A∩B)P(A \cup B) = P(A) + P(B) - P(A \cap B). Substituting the given values, P(A∪B)=0.5+0.4−0.2=0.7P(A \cup B) = 0.5 + 0.4 - 0.2 = 0.7.

Chapter Concept:

Probability of Events

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Probability

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True or False

Correct Answer: False

Solution:

The union of two mutually exclusive events AA and BB is not an empty set; it includes all outcomes that are in either AA or BB. The intersection A∩BA \cap B is the empty set.

Chapter Concept :

Algebra of Events

Correct Answer: True

Solution:

The probability of an event EE in a sample space SS with equally likely outcomes is given by P(E)=n(E)n(S)P(E) = \frac{n(E)}{n(S)}, where n(E)n(E) is the number of favorable outcomes and n(S)n(S) is the total number of possible outcomes.

Chapter Concept :

Equally Likely Outcomes and Classical Probability

Correct Answer: True

Solution:

The event 'second toss is not head' includes outcomes where the second coin shows tails. The subset D={HT,TT}D = \{HT, TT\} correctly represents this event as per the sample space S={HH,HT,TH,TT}S = \{HH, HT, TH, TT\}.

Chapter Concept :

Combinatorics-Based Probability Problems

Correct Answer: False

Solution:

Mutually exclusive events cannot occur simultaneously, meaning their intersection is an empty set, A∩B=∅A \cap B = \emptyset. Therefore, the statement is false.

Chapter Concept :

Equally Likely Outcomes and Classical Probability

Correct Answer: True

Solution:

The formula P(E)=n(E)n(S)P(E) = \frac{n(E)}{n(S)} is used to calculate the probability of an event when all outcomes in the sample space are equally likely.

Chapter Concept :

Equally Likely Outcomes and Classical Probability

Correct Answer: True

Solution:

The event 'number of tails is exactly 2' corresponds to the outcome where both tosses result in tails, which is represented by the subset A={TT}A = \{TT\} of the sample space S={HH,HT,TH,TT}S = \{HH, HT, TH, TT\}.

Chapter Concept :

Sample Space and Events

Correct Answer: False

Solution:

In the axiomatic approach to probability, the probability of the empty set Φ\Phi is 0, as it represents an impossible event.

Chapter Concept :

Axiomatic Probability

Correct Answer: True

Solution:

The axiomatic approach to probability includes the axioms: non-negativity (P(E)≥0P(E) \geq 0 for any event EE), normalization (P(S)=1P(S) = 1 for the sample space SS), and additivity for mutually exclusive events (P(E∪F)=P(E)+P(F)P(E \cup F) = P(E) + P(F) if EE and FF are mutually exclusive).

Chapter Concept :

Axiomatic Probability

Correct Answer: True

Solution:

Mutually exclusive events are defined as events that cannot occur at the same time. This means that if one event occurs, the other cannot.

Chapter Concept :

Mutually Exclusive and Exhaustive Events

Correct Answer: True

Solution:

According to the axiomatic approach, one of the axioms states that the probability of the entire sample space SS is 1, i.e., P(S)=1P(S) = 1. This is a fundamental property of probability.

Chapter Concept :

Axiomatic Probability

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