When two coins are tossed, which event is a compound event
A Getting HH
B Getting TT
C Getting HT only
D Getting one head
“Getting one head” includes two outcomes, HT and TH, so it contains more than one sample point. An event with multiple outcomes is a compound event.
For any two events A and B, which set identity is always true
A (A∪B)′=A′∪B′
B (A∩B)′=A∩B
C (A∩B)′=A′∪B′
D (A∪B)′=A∪B
This is De Morgan’s law: the complement of an intersection equals the union of complements. It is frequently used to simplify probability expressions before applying rules.
For events A and B, which identity is always true
A (A∪B)′=A′∪B′
B (A∪B)′=A′∩B′
C (A∩B)′=A′∩B′
D (A′)′=A′
Another De Morgan’s law states: complement of union equals intersection of complements. This is useful when calculating “neither A nor B” probabilities.
If A and B are mutually exclusive, then which is correct
A P(A∪B)=P(A)P(B)
B P(A∩B)=P(A)+P(B)
C P(A′)=P(B′)
D P(A|B)=0
If B occurs, A cannot occur when events are mutually exclusive, so P(A∩B)=0. Hence P(A|B)=P(A∩B)/P(B)=0 when P(B)>0.
If P(A)=0.25 and P(A∪B)=0.60 and P(A∩B)=0.10, then P(B) is
A 0.45
B 0.35
C 0.25
D 0.75
Using union formula: P(B)=P(A∪B)−P(A)+P(A∩B)=0.60−0.25+0.10=0.45.
If P(A)=0.3, P(B)=0.5 and P(A∩B)=0.15, then A and B are
A Disjoint
B Independent
C Exhaustive
D Complements
Check independence: P(A)P(B)=0.3×0.5=0.15, which matches P(A∩B). Therefore, A and B are independent events.
A die is rolled. Events A: “even”, B: “multiple of 3”. P(A∩B) is
A 1/3
B 1/2
C 2/3
D 1/6
Even outcomes are {2,4,6}. Multiples of 3 are {3,6}. Intersection is {6} only. So probability is 1 outcome out of 6, i.e., 1/6.
A die is rolled. Events A: “even”, B: “multiple of 3”. P(A∪B) is
A 1/2
B 1/3
C 2/3
D 5/6
A={2,4,6} and B={3,6}. Union is {2,3,4,6} which has 4 outcomes. So probability is 4/6=2/3.
Two coins are tossed. Events A: “first is head”, B: “second is head”. P(A∩B) is
A 1/2
B 1/4
C 3/4
D 1/8
Sample space is {HH, HT, TH, TT}. Both first and second head means HH only. That is 1 favourable outcome out of 4, so probability is 1/4.
Two coins are tossed. P(first head | second head) is
A 1/4
B 3/4
C 1
D 1/2
Given second head, possible outcomes are {HH, TH}. In these, first head occurs only in HH. So probability becomes 1/2.
Two dice are thrown. Probability of getting sum 9 is
A 1/9
B 1/12
C 1/6
D 5/36
There are 4 ordered outcomes giving sum 9. With 36 total outcomes, probability is 4/36=1/9.
Two dice are thrown. Probability that sum is at least 10 is
A 1/12
B 1/9
C 1/6
D 1/4
Sums 10,11,12 have cases 3,2,1 respectively, total 6 outcomes. So probability is 6/36=1/6.
Two cards drawn without replacement from a deck. P(second ace | first ace) is
A 4/52
B 3/51
C 1/13
D 4/51
After drawing an ace first, remaining cards are 51 with 3 aces left. So conditional probability for second ace is 3/51.
In a deck, event A: “red”, B: “king”. P(A|B) is
A 1/4
B 3/4
C 1/13
D 1/2
Given the card is a king, possible kings are 4. Red kings are 2 (hearts, diamonds). So P(red | king)=2/4=1/2.
Use P(A∩B)=P(B)P(A|B). So 0.4×0.25=0.10. This is a direct use of the multiplication rule.
A bag has 5 red and 3 blue. Two draws without replacement. P(one red, one blue) is
A 5/14
B 3/8
C 15/28
D 1/2
Add both orders: 15/56 + 15/56 = 30/56 = 15/28.
Two draws with replacement from same bag (5 red,3 blue). P(one red, one blue) is
A 15/28
B 15/32
C 5/8
D 3/8
With replacement, probabilities stay 5/8 and 3/8 each draw. One red and one blue: 2×(5/8)(3/8)=30/64=15/32.
If A and B are independent, then P(A′∩B) equals
A P(A′)+P(B)
B 1−P(A∩B)
C P(A)P(B)
D P(A′)P(B)
Independence extends to complements. If A and B are independent, then A′ and B are also independent, so intersection equals product P(A′)P(B).
If A and B are independent, then A′ and B′ are
A Mutually exclusive
B Exhaustive only
C Independent
D Equal events
For independent events, complements also remain independent. This follows from probability rules using P(A′∩B′)=1−P(A∪B) and product relations.
If P(A)=0.2 and P(B)=0.3 independent, then P(neither) is
A 0.14
B 0.56
C 0.70
D 0.50
“Neither” means A′∩B′. With independence, P(A′∩B′)=P(A′)P(B′)=(0.8)(0.7)=0.56.
If P(A)=0.2 and P(B)=0.3 independent, then P(at least one) is
A 0.56
B 0.06
C 0.50
D 0.44
“At least one” = 1 − P(neither). From previous, P(neither)=0.56, so answer is 1−0.56=0.44.
A partition {B1,B2,B3} satisfies which condition
A Overlap allowed
B Equal probabilities
C Disjoint and exhaustive
D Same outcomes only
A partition breaks the sample space into non-overlapping events whose union is the whole sample space. Each outcome belongs to exactly one Bi.
If {B1,B2} is a partition, then P(A) equals
A P(A|B1)P(B1)+P(A|B2)P(B2)
B P(A|B1)+P(A|B2)
C P(B1|A)+P(B2|A)
D P(A)P(B1)P(B2)
This is the law of total probability. It computes P(A) by adding probabilities of A occurring through each mutually exclusive case Bi.
Bayes theorem mainly converts
A P(A|B) from P(A∩B)
B P(A) from P(A′)
C P(B) from P(B′)
D P(B|A) from P(A|B)
Bayes reverses conditioning. It lets you compute the probability of a hypothesis B given evidence A using likelihood P(A|B), prior P(B), and total probability P(A).
A factory has two machines M1 and M2. If 60% items from M1 and 40% from M2, then these represent
A Likelihood values
B Posterior values
C Prior probabilities
D Conditional sample
The production shares 0.6 and 0.4 are probabilities before any defect information is used. In Bayes problems, such initial probabilities are called priors.
If events B1 and B2 form a partition, then B1∩B2 equals
A Empty set
B Sample space
C Event A
D Union set
In a partition, events are mutually exclusive. So they have no common outcomes, meaning their intersection is the empty set.
If events B1 and B2 form a partition, then B1∪B2 equals
A Empty set
B Event A only
C Event B only
D Sample space
Partition events must be exhaustive. So their union covers the full sample space, ensuring exactly one of them happens in each trial.
If P(A)=0.7 and P(B)=0.6, then maximum P(A∪B) is
A 0.7
B 1
C 0.6
D 0.3
The union probability cannot exceed 1. Maximum occurs when A and B together cover the sample space fully, making P(A∪B)=1.
If P(A)=0.7 and P(B)=0.6, then minimum P(A∪B) is
A 0.6
B 0.1
C 1.0
D 0.7
The union is at least the larger probability. Minimum happens when the smaller event lies completely inside the larger, so P(A∪B)=max(0.7,0.6)=0.7.
If P(A∩B)=0, then A and B are
A Independent always
B Complements always
C Mutually exclusive
D Exhaustive always
Zero intersection probability indicates they cannot occur together (for standard events). That matches mutual exclusivity, where A∩B is empty and P(A∩B)=0.
Which expression gives P(A−B)
A P(A)−P(A∩B)
B P(A)+P(B)
C P(A∩B)−P(B)
D 1−P(A)
A−B means outcomes in A but not in B. That equals A∩B′, so probability is P(A)−P(A∩B).
A and B are independent with P(A)=0.4, P(B)=0.5. Then P(A|B′) is
A 0.5
B 0.4
C 0.6
D 0.2
If A is independent of B, then A is also independent of B′. So conditioning on B′ does not change A: P(A|B′)=P(A)=0.4.
A fair die is rolled twice. Probability of getting at least one 6 is
A 1/6
B 1/36
C 5/36
D 11/36
Use complement: no 6 in both rolls has probability (5/6)²=25/36. So at least one 6 is 1−25/36=11/36.
A fair coin is tossed three times. Probability of exactly two heads is
A 1/8
B 1/4
C 3/8
D 1/2
Exactly two heads occurs in HHT, HTH, THH: 3 outcomes. Total outcomes are 8. So probability is 3/8.
If P(A)=0.5, P(B)=0.4, and A and B are mutually exclusive, then P(A|B) equals
A 0.5
B 0
C 0.4
D 1
Mutually exclusive means A∩B is impossible, so P(A∩B)=0. Hence P(A|B)=0/P(B)=0, provided P(B)>0.
If P(A|B)=0.7 and P(B)=0.2, then P(A∩B) is
A 0.50
B 0.90
C 0.07
D 0.14
Multiplication rule: P(A∩B)=P(A|B)P(B)=0.7×0.2=0.14. This is commonly used in conditional probability questions.
In total probability, weights are the probabilities of
A Target event only
B Partition events
C Complement events
D Impossible events
In P(A)=ΣP(A|Bi)P(Bi), each P(Bi) is a weight. These weights come from the partition events that split the sample space into cases.
In Bayes theorem, the “evidence” probability usually means
A P(B)
B P(A|B)
C P(A)
D P(B|A)
Evidence is the overall probability of observing A. It is computed using total probability across all hypotheses and is used in the denominator to normalize posteriors.
A and B are independent with P(A)=0.3. Then P(A′|B) is
A 0.3
B 0.5
C 0.2
D 0.7
Independence implies P(A|B)=P(A). Then P(A′|B)=1−P(A|B)=1−0.3=0.7. Complements keep the relation simple.
If a fair die is rolled, probability that number is not prime is
A 1/2
B 1/3
C 2/3
D 1/6
Prime numbers on a die are {2,3,5} (3 outcomes). Not prime are {1,4,6} (3 outcomes). So probability is 3/6=1/2.
If P(A)=0.6 and P(B)=0.5, then maximum P(A∩B) is
A 0.3
B 0.5
C 1.1
D 0.1
Intersection cannot exceed the smaller event probability. So maximum P(A∩B)=min(0.6,0.5)=0.5, achieved when B is contained in A.
If P(A)=0.6 and P(B)=0.5, then minimum P(A∩B) is
A 0.0
B 0.3
C 0.1
D 0.5
Use bound P(A∩B)≥P(A)+P(B)−1. Here 0.6+0.5−1=0.1, and it is positive, so minimum is 0.1.
An event with probability 0.2 is
A Sure event
B Impossible event
C Empty event
D Possible event
Any probability strictly between 0 and 1 means the event can happen but is not certain. It is neither impossible nor sure.
If P(A|B)=1, then which must be true
A A ⊆ B
B B ⊆ A
C A and B disjoint
D A′ ⊆ B
P(A|B)=1 means whenever B happens, A definitely happens too. That implies all outcomes of B lie inside A, so B is a subset of A.
If P(A|B)=0, then which must be true
A A∪B is empty
B A equals B
C A∩B is empty
D B is empty
If probability of A given B is 0 (with P(B)>0), then A never occurs when B occurs. So A and B share no outcomes, making A∩B empty.
A fair die is rolled. Probability of getting a number less than 3 is
A 1/3
B 1/2
C 2/3
D 1/6
Numbers less than 3 are {1,2}, giving 2 favourable outcomes out of 6. So probability is 2/6=1/3.
If P(A)=0.4, P(B)=0.3, and A and B are disjoint, then P(A|B) is
A 0.4
B 0
C 0.3
D 0.7
Disjoint means mutually exclusive, so P(A∩B)=0. Therefore P(A|B)=0/P(B)=0 when P(B)>0.
If P(A)=0.2 and P(B)=0.3, then the largest possible P(A∪B) is
A 0.3
B 0.2
C 0.5
D 1.0
Maximum union happens when events do not overlap, so P(A∪B)=P(A)+P(B)=0.2+0.3=0.5. Union cannot exceed this without exceeding total probability.