The sample space is the complete set of all possible outcomes of a random experiment. Every event is formed from outcomes taken from this full outcome set.
Which term means a subset of sample space
A Trial
B Randomness
C Event
D Observation
An event is any collection (subset) of outcomes from the sample space. It may contain one outcome (simple event) or many outcomes (compound event).
An event that can never occur is called
A Sure event
B Impossible event
C Certain event
D Equal event
An impossible event has no outcomes in it, so its probability is 0. Example: getting 7 on a standard die is impossible.
An event that always occurs is called
A Random event
B Empty event
C Rare event
D Sure event
A sure (certain) event equals the whole sample space, so it must happen. Its probability is 1, like getting a number from 1 to 6 on a die.
Probability of a sure event is
A 0
B 1/2
C 1
D Depends
A sure event includes all outcomes, so it always happens. Hence its probability is 1, which is the maximum possible probability value.
Probability of an impossible event is
A 0
B 1
C 1/3
D Depends
An impossible event has no favourable outcomes. Since there is nothing that can make it happen, its probability is 0.
If outcomes are equally likely, then probability is
A Total/Favourable
B Favourable/Total
C Total−Favourable
D Favourable×Total
For equally likely outcomes, probability is the ratio of favourable outcomes to total outcomes. This basic definition applies to fair coins, fair dice, and well-shuffled cards.
For a fair coin, sample space is
A {H}
B {T}
C {H, T}
D {0,1,2}
Tossing a coin has exactly two possible outcomes: Head or Tail. So the sample space is {H, T}.
A single outcome event is called
A Compound event
B Certain event
C Complement
D Simple event
A simple event contains exactly one outcome from the sample space. Example: getting “3” on a die is a simple event.
An event with more than one outcome is
A Sure event
B Compound event
C Empty event
D Single event
A compound event contains two or more outcomes. Example: getting an even number on a die includes {2,4,6}, so it is compound.
Complement of event A is written as
A A′
B A×B
C A∩B
D A∪B
A′ represents “not A”, meaning all outcomes in the sample space that are not in A. Complement helps in “at least one” type questions.
If P(A)=0.35, then P(A′) is
A 0.35
B 1.35
C 0.65
D 0.00
Complement rule says P(A′)=1−P(A). So P(A′)=1−0.35=0.65. This works for any event A.
Union of events A and B means
A A and B only
B A or B
C Not A
D Not B
A∪B includes outcomes that are in A, in B, or in both. It represents “A occurs or B occurs (or both)”.
Intersection of events A and B means
A A or B
B Only A
C Only B
D A and B
A∩B contains outcomes common to both A and B. It represents both events happening together.
If A and B are mutually exclusive, then
A A∪B=∅
B A=B
C A∩B=∅
D P(A)=P(B)
Mutually exclusive events cannot occur together, so they have no common outcome. Therefore their intersection is the empty set.
For mutually exclusive A and B, P(A∩B) is
A 1
B 0
C P(A)+P(B)
D P(A)P(B)
Since mutually exclusive events never happen together, the probability of their intersection is 0.
Addition theorem for two events is
A P(A∪B)=P(A)+P(B)−P(A∩B)
B P(A∪B)=P(A)P(B)
C P(A∩B)=P(A)+P(B)
D P(A′)=P(A)+P(B)
The subtraction removes double-counting of outcomes common to both events. If A and B overlap, P(A)+P(B) counts the overlap twice, so we subtract P(A∩B).
If A and B are mutually exclusive, then P(A∪B) equals
A P(A)P(B)
B P(A)−P(B)
C P(A)+P(B)
D 1−P(A)
When A and B are mutually exclusive, P(A∩B)=0. So the addition rule becomes P(A∪B)=P(A)+P(B).
Exhaustive events mean
A No overlap
B Cover sample space
C Equal probability
D Single outcome
Exhaustive events together include every outcome of the sample space. So at least one of them must occur in any trial.
Events that are both disjoint and cover S are
A Complement
B Intersection
C Simple event
D Partition
A partition is a collection of events that are mutually exclusive (disjoint) and exhaustive (cover the sample space). Partitions are used in total probability and Bayes’ theorem.
Conditional probability P(A|B) equals
A P(A∪B)/P(B)
B P(A)/P(B)
C P(A∩B)/P(B)
D P(B)/P(A)
Conditional probability measures probability of A when B has occurred. We restrict to outcomes in B, so we divide the joint probability P(A∩B) by P(B), assuming P(B)>0.
If P(B)=0, then P(A|B) is
A Always 0
B Not defined
C Always 1
D P(A)
P(A|B)=P(A∩B)/P(B). If P(B)=0, division is not possible, so conditional probability is undefined in standard probability.
Multiplication theorem for two events is
A P(A∩B)=P(A)P(B|A)
B P(A∩B)=P(A)+P(B)
C P(A∩B)=P(A)−P(B)
D P(A|B)=P(A)P(B)
The chance that both occur equals chance of A times chance of B after A occurs. This is very useful for sequential events like cards drawn without replacement.
Another form of multiplication theorem is
A P(A∩B)=P(A)+P(B)
B P(A∪B)=P(A)P(B)
C P(A∩B)=P(B)P(A|B)
D P(A′)=P(A|B)
Joint probability can be written in two ways depending on which event is treated as the first condition. Both forms are correct when the condition event has nonzero probability.
For independent events A and B, P(A|B) equals
A P(B)
B P(A)
C P(A∩B)
D 1−P(A)
Independence means B happening does not change the probability of A. So P(A|B)=P(A) and similarly P(B|A)=P(B).
Independence condition is
A P(A∪B)=P(A)+P(B)
B A∩B=∅
C P(A)=P(B)
D P(A∩B)=P(A)P(B)
If two events are independent, the probability that both occur equals the product of their separate probabilities. This is common in repeated coin tosses or independent trials.
Mutually exclusive events are generally
A Always independent
B Always equal
C Not independent
D Always exhaustive
If A and B are mutually exclusive with positive probabilities, then P(A∩B)=0 but P(A)P(B)>0, so they cannot be independent. Only possible if one event has probability 0.
On a fair die, P(even) is
A 1/2
B 1/3
C 2/3
D 1/6
Even outcomes are {2,4,6}, which are 3 outcomes out of 6 total. So P(even)=3/6=1/2.
On a fair die, P(number >4) is
A 1/2
B 1/3
C 2/3
D 1/6
Numbers greater than 4 are {5,6}, which is 2 outcomes out of 6. So probability is 2/6=1/3.
Two coins tossed, probability of exactly one head
A 1/4
B 3/4
C 1/2
D 1/3
Sample space is {HH, HT, TH, TT}. Exactly one head occurs in HT and TH, so favourable outcomes are 2 out of 4, giving probability 2/4=1/2.
Two coins tossed, probability of at least one head
A 1/4
B 1/2
C 2/3
D 3/4
Use complement method: “at least one head” = 1 − P(no head). No head means TT only, probability 1/4. So answer is 1−1/4=3/4.
Two dice thrown, probability of sum 7
A 1/6
B 1/12
C 1/9
D 1/3
There are 36 equally likely ordered outcomes. Sum 7 happens in 6 cases: (1,6),(2,5),(3,4),(4,3),(5,2),(6,1). So probability is 6/36=1/6.
Drawing one card, probability of a heart
A 1/13
B 1/2
C 1/4
D 3/4
A standard deck has 52 cards with 13 hearts. Probability of drawing a heart is 13/52=1/4.
From a deck, probability of a face card
A 1/13
B 3/13
C 1/4
D 4/13
Face cards are J, Q, K in each suit: 3×4=12 cards. Probability is 12/52=3/13.
Two cards drawn without replacement, both aces
A 1/169
B 2/221
C 1/52
D 1/221
P(first ace)=4/52. Then P(second ace | first ace)=3/51. Multiply: (4/52)×(3/51)=12/2652=1/221.
Two cards drawn with replacement, both aces
A 1/221
B 1/52
C 1/169
D 2/169
With replacement, draws are independent. P(ace)=4/52=1/13 each time. So P(both aces)=(1/13)×(1/13)=1/169.
If P(A)=0.6 and P(B)=0.5 independent, then P(A∩B)
A 0.10
B 0.30
C 0.55
D 0.11
For independent events, P(A∩B)=P(A)P(B). So 0.6×0.5=0.30.
If P(A)=0.4, P(B)=0.7 and P(A∩B)=0.28, then A and B are
A Independent
B Disjoint
C Exhaustive
D Complementary
Check independence: P(A)P(B)=0.4×0.7=0.28, which matches P(A∩B). So A and B are independent.
Multiplication for three events is
A P(A)+P(B)+P(C)
B P(A)P(B)P(C)+1
C P(A)P(B|A)P(C|A∩B)
D P(A|B|C)
For sequential events, multiply probability of first event, then second given first, then third given first two. This is the chain rule and works for dependent events.
A partition must be
A Overlapping and exhaustive
B Disjoint and exhaustive
C Disjoint and equal
D Random and disjoint
Events in a partition cannot overlap, and together they must cover the entire sample space. This ensures every outcome belongs to exactly one partition event.
Total probability theorem finds
A P(A|B) directly
B P(A∩B) only
C P(A′) always
D P(A) from cases
The law of total probability calculates P(A) by splitting the sample space into a partition of cases and adding conditional probabilities weighted by case probabilities.
Total probability formula uses
A Π P(A|Bi)P(Bi)
B Σ P(A|Bi)P(Bi)
C P(A)−P(B)
D P(A)+P(B)
If {B1, B2, …} is a partition, then P(A)=Σ P(A|Bi)P(Bi). Each term represents probability of A occurring through that particular case.
Bayes’ theorem mainly finds
A P(Bi|A)
B P(A∩Bi)
C P(A∪Bi)
D P(A′|Bi)
Bayes’ theorem reverses conditioning: it updates probability of a cause (Bi) after observing evidence (A). It uses prior probabilities and likelihoods to compute posterior probabilities.
Bayes formula for one hypothesis B is
A P(B|A)=P(A)P(B)
B P(B|A)=P(B)/P(A|B)
C P(B|A)=P(A|B)P(B)/P(A)
D P(B|A)=P(A|B)+P(B)
Bayes’ rule divides the joint probability by P(A). It converts likelihood P(A|B) and prior P(B) into the updated probability after observing A.
In Bayes, P(B) is called
A Posterior probability
B Prior probability
C Likelihood
D Evidence
Prior probability is the initial chance of hypothesis B before observing new information. After observing evidence A, it updates to posterior P(B|A).
In Bayes, P(B|A) is called
A Prior probability
B Random probability
C Equal probability
D Posterior probability
Posterior probability is the updated probability of a hypothesis after considering evidence. It combines the prior probability and the likelihood through Bayes’ theorem.
If events are independent, then P(A|B)=
A P(B)
B P(A∩B)
C P(A)
D 0
Independence means B provides no new information about A. So the probability of A remains unchanged even after B occurs.
“At least one” probability is often solved using
A Intersection rule
B Complement method
C Venn only
D Total probability
“At least one” is easier by finding “none” and subtracting from 1. This avoids complex counting and works well in repeated independent trials.
If P(A)=0.2, then P(not A) is
A 0.8
B 0.2
C 1.2
D 0.0
Not A is the complement of A. Using P(A′)=1−P(A), we get 1−0.2=0.8.
If A and B are exhaustive, then
A A∩B = S
B A = B
C A∪B = S
D A′ = B′
Exhaustive events together cover the entire sample space. So their union equals the sample space, meaning at least one of the events must occur.