When a die is rolled once, how many outcomes are in the sample space
A 6 outcomes
B 4 outcomes
C 5 outcomes
D 12 outcomes
A standard die has possible outcomes {1,2,3,4,5,6}. The sample space size is 6 because exactly six distinct results can occur in one roll.
In probability, what does the symbol “∅” represent
A Whole sample space
B Common outcomes
C Empty event
D Union event
The empty set ∅ represents an event with no outcomes. Such an event cannot occur, so its probability is always 0 in any random experiment.
In set language, the sample space is usually denoted by
A A
B S
C B
D ∩
The sample space is commonly written as S (or Ω). It contains every possible outcome, and all events are subsets of this sample space.
In a Venn diagram for events, the rectangle usually shows
A Event A only
B Event B only
C Sample space
D Intersection only
In probability Venn diagrams, the outer rectangle represents the entire sample space. Circles or regions inside it represent events and their relationships.
If an event is a subset of the sample space, then its probability must lie in
A 0 to 1
B −1 to 1
C 1 to 2
D Any real number
Probabilities are always between 0 and 1 inclusive. Zero means impossible, one means certain, and any event probability must stay within these limits.
If A and B overlap, then A∩B is
A Always empty
B Whole sample space
C Common outcomes
D Complement region
The intersection A∩B contains outcomes that belong to both A and B. Overlap in a Venn diagram visually represents these common outcomes.
Which statement matches “mutually exclusive” events
A Can happen together
B Same probability
C Always independent
D Never overlap
Mutually exclusive events cannot occur simultaneously. They share no outcomes, so their intersection is empty and P(A∩B)=0.
Which statement defines “exhaustive” events
A Cover all outcomes
B No common outcomes
C Equal probability
D Single outcome
Exhaustive events together include every outcome of the sample space. Therefore, at least one of them must happen whenever the experiment is performed.
If A and A′ are complements, then A∩A′ equals
A S
B A
C ∅
D A′
A and its complement A′ cannot occur together. So their intersection has no outcomes, making it the empty set, and its probability is 0.
If A and A′ are complements, then A∪A′ equals
A ∅
B A
C A′
D S
Every outcome is either in A or not in A. So A union A′ covers the whole sample space, meaning it is always a sure event.
A coin is fair. Probability of head is
A 0
B 1/2
C 1/3
D 1
A fair coin has two equally likely outcomes, Head and Tail. One favourable outcome out of two gives probability 1/2.
A fair die is rolled. Probability of “odd number” is
A 1/6
B 2/3
C 1/2
D 5/6
Odd outcomes are {1,3,5}, which are 3 outcomes out of 6. Therefore P(odd)=3/6=1/2.
A fair die is rolled. Probability of “prime number” is
A 1/2
B 1/6
C 1/3
D 2/3
Prime outcomes on a die are {2,3,5}, total 3 outcomes. So probability is 3/6=1/2.
Two coins are tossed. Probability of getting two tails is
A 1/2
B 3/4
C 1/4
D 1/3
Outcomes are {HH, HT, TH, TT}. Only TT is favourable, so probability is 1 out of 4, i.e., 1/4.
Three coins are tossed. Probability of exactly three heads is
A 3/8
B 1/8
C 1/4
D 1/2
Total outcomes are 2³=8. Only HHH gives three heads, so favourable outcomes are 1, giving probability 1/8.
Three coins are tossed. Probability of at least one tail is
A 1/8
B 1/2
C 3/8
D 7/8
Use complement: “at least one tail” = 1 − P(all heads). P(all heads)=1/8, so answer is 1−1/8=7/8.
Two dice are thrown. Probability of getting doubles is
A 1/3
B 1/12
C 1/6
D 1/18
Doubles are (1,1) to (6,6): 6 outcomes. Total outcomes are 36, so probability is 6/36=1/6.
Two dice are thrown. Probability that sum is 12 is
A 1/36
B 1/6
C 1/12
D 1/18
Only (6,6) gives sum 12. There are 36 equally likely ordered outcomes, so probability is 1/36.
From a deck, probability of drawing a red card is
A 1/4
B 3/4
C 1/2
D 1/13
There are 26 red cards (hearts and diamonds) out of 52 total. Probability of red is 26/52=1/2.
From a deck, probability of drawing a king is
A 1/26
B 1/13
C 1/4
D 3/13
A standard deck has 4 kings. So probability is 4/52=1/13.
Two cards drawn without replacement. P(both red) is
A 25/102
B 13/51
C 1/4
D 1/2
First red: 26/52=1/2. Then red given red: 25/51. Multiply: (1/2)×(25/51)=25/102.
Two cards drawn with replacement. P(both red) is
A 1/2
B 3/4
C 1/4
D 25/102
With replacement, draws are independent. P(red)=1/2 each time. So P(both red)=(1/2)×(1/2)=1/4.
If P(A)=0.42, then P(not A) equals
A 0.42
B 0.58
C 1.42
D 0.00
Complement rule: P(A′)=1−P(A). So 1−0.42=0.58. This is useful in “at least one” type probability problems.
If P(A)=0.30 and P(B)=0.40 independent, then P(A∩B) is
A 0.70
B 0.10
C 0.34
D 0.12
For independent events, multiply probabilities: P(A∩B)=P(A)P(B)=0.30×0.40=0.12.
If P(A∩B)=0.08 and P(B)=0.20, then P(A|B) is
A 0.40
B 0.60
C 0.16
D 0.10
Conditional probability formula is P(A|B)=P(A∩B)/P(B). So 0.08/0.20=0.40, assuming P(B)>0.
If P(A|B)=P(A), then A and B are
A Mutually exclusive
B Complementary
C Independent
D Exhaustive only
If knowing B does not change the probability of A, then A is independent of B. This matches the basic test P(A|B)=P(A).
If P(A)=0.5, P(B)=0.6 and independent, then P(A∪B) is
A 1.1
B 0.8
C 0.3
D 0.6
Use P(A∪B)=P(A)+P(B)−P(A∩B). Here P(A∩B)=0.5×0.6=0.3. So union is 0.5+0.6−0.3=0.8.
If A and B are mutually exclusive with P(A)=0.2, P(B)=0.5 then P(A∪B) is
A 0.7
B 0.3
C 0.1
D 1.0
For mutually exclusive events, P(A∩B)=0. So P(A∪B)=P(A)+P(B)=0.2+0.5=0.7.
A bag has 3 red and 2 blue balls. P(red) in one draw is
A 2/5
B 1/5
C 3/5
D 4/5
Total balls are 5. Favourable red balls are 3. With equally likely draws, probability is 3/5.
Same bag: two draws without replacement. P(both red) is
A 9/25
B 2/5
C 1/5
D 3/10
First red: 3/5. Then red given red: 2/4=1/2. Multiply: (3/5)×(1/2)=3/10.
Same bag: two draws with replacement. P(both red) is
A 3/10
B 6/25
C 9/25
D 1/2
With replacement, draws are independent. P(red)=3/5 each draw. So P(both red)=(3/5)²=9/25.
A card is drawn from a deck. Event “not a spade” has probability
A 3/4
B 1/4
C 1/2
D 13/52
There are 13 spades out of 52, so P(spade)=1/4. Complement gives P(not spade)=1−1/4=3/4.
A die is rolled. Event “number ≤ 4” has probability
A 1/3
B 1/2
C 2/3
D 1/6
Outcomes ≤4 are {1,2,3,4} which are 4 outcomes. So probability is 4/6=2/3.
For events A and B, the maximum possible value of P(A∩B) is
A P(A)+P(B)
B 1−P(A)
C 0 always
D min(P(A),P(B))
Intersection cannot exceed either event’s probability because A∩B is contained in both A and B. So the maximum possible value is the smaller of P(A) and P(B).
For events A and B, the minimum possible value of P(A∩B) is
A min(P(A),P(B))
B max(0, P(A)+P(B)−1)
C P(A)+P(B)
D 1
Using bounds from probability rules, intersection cannot be negative. Also P(A∪B)≤1 implies P(A∩B)≥P(A)+P(B)−1, so minimum is max of these.
If P(A)=0.7 and P(B)=0.6, then smallest possible P(A∪B) is
A 0.6
B 0.3
C 0.7
D 1.0
The union is at least as large as the larger event. Minimum union happens when one event is fully inside the other, giving P(A∪B)=max(P(A),P(B))=0.7.
If two events are complements, then they are
A Mutually exclusive
B Independent always
C Same event
D Equal probability
A and A′ never occur together, so they are mutually exclusive and their union is the full sample space. They also satisfy P(A)+P(A′)=1.
Which statement is true for any event A
A P(A)+P(A′)=0
B P(A)=P(A′) always
C P(A)+P(A′)=1
D P(A) can exceed 1
A and its complement cover the whole sample space without overlap. Therefore their probabilities add to 1. This rule holds for every event.
In a tree diagram, probabilities along a single path are
A Added
B Multiplied
C Subtracted
D Ignored
A path represents a sequence of events. The probability of that sequence is found by multiplying the branch probabilities along the path, matching the multiplication theorem.
In a tree diagram, probabilities of different final paths for the same event are
A Multiplied
B Divided
C Squared
D Added
If an event can happen through different disjoint paths, total probability is the sum of probabilities of those paths. This matches “case by case” addition.
Law of total probability requires events B1, B2, … to be
A Equal in size
B Independent always
C Partition of S
D All impossible
The theorem uses a partition: events must be mutually exclusive and exhaustive. Then P(A)=ΣP(A|Bi)P(Bi), covering all possible cases.
Bayes theorem needs which quantity in denominator
A P(A)
B P(B)
C P(A∩B)
D P(A′)
Bayes formula is P(B|A)=P(A|B)P(B)/P(A). The denominator P(A) normalizes probabilities so the posterior values are valid and sum properly across hypotheses.
In Bayes, “likelihood” usually refers to
A P(B|A)
B P(A|B)
C P(A)
D P(B)
Likelihood measures how probable the evidence A is when hypothesis B is true. Bayes uses this likelihood with prior P(B) to compute posterior P(B|A).
If {B1,B2} is a partition, then P(B1|A)+P(B2|A) equals
A 0
B P(A)
C 1
D P(B1)
Given A has occurred, the hypotheses B1 and B2 still form a partition of the restricted sample space. Their conditional probabilities must add to 1.
A diagnostic test has false positive meaning
A Positive, no disease
B Negative, has disease
C Positive, has disease
D Negative, no disease
A false positive occurs when the test indicates positive but the person does not actually have the condition. It is important in Bayes-type medical probability problems.
“Posterior probability” means
A Before any evidence
B Always 1/2
C Same as likelihood
D Updated after evidence
Posterior probability is the revised probability of a hypothesis after observing evidence. Bayes’ theorem converts prior information and likelihood into this updated belief.
Two independent trials with success probability p. Probability of two successes is
A 2p
B p(1−p)
C p²
D 1−p²
Independence means multiply probabilities for successive successes. So success then success has probability p×p=p², matching the multiplication rule for independent events.
For independent events, probability of at least one occurring is often found using
A Complement rule
B Intersection only
C Venn shading
D Division rule
“At least one” equals 1 minus “none”. For independent events, “none” probability is the product of complements, making calculations simple and less error-prone.
If A⊆B, then which is always true
A P(A) > P(B)
B P(A)=1 always
C P(A) ≤ P(B)
D P(B)=0 always
If A is contained in B, every outcome of A is also in B. So A cannot be more likely than B, giving P(A)≤P(B).
If A⊆B and P(B)=0.4, then maximum possible P(A) is
A 0.6
B 0.4
C 1.0
D 0.2
Since A is a subset of B, P(A) cannot exceed P(B). The maximum occurs when A and B are the same event, so maximum P(A)=0.4.