Three events A,B,C are mutually independent with P(A)=0.5, P(B)=0.4, P(C)=0.3. Probability at least one occurs
A 0.58
B 0.21
C 0.79
D 0.50
Use complement: none occur = (1−0.5)(1−0.4)(1−0.3)=0.5×0.6×0.7=0.21. So at least one = 1−0.21=0.79.
A box has 5 bulbs, 2 are defective. Two bulbs chosen without replacement. Probability exactly one defective
A 2/5
B 3/5
C 1/2
D 4/5
Exactly one defective: choose 1 defective and 1 good. Probability = [C(2,1)C(3,1)]/C(5,2) = (2×3)/10 = 6/10=3/5.
Two cards are drawn without replacement. Probability that both are of the same color
A 26/51
B 1/2
C 25/51
D 1/4
Same color = both red or both black. Probability = (26/52)(25/51)+(26/52)(25/51)=2×(1/2)(25/51)=25/51.
Two cards are drawn with replacement. Probability both are of the same suit
A 1/13
B 1/4
C 1/16
D 1/52
First card can be any suit. Second card must match that suit: probability 13/52=1/4. With replacement, suit distribution stays constant.
Two dice are thrown. Probability that their product is even
A 3/4
B 1/2
C 1/4
D 2/3
Product is odd only if both dice are odd. P(both odd)=(3/6)^2=1/4. So product even = 1−1/4=3/4.
Two dice are thrown. Probability that their product is divisible by 3
A 1/3
B 4/9
C 5/9
D 2/3
Product not divisible by 3 means neither die shows 3 or 6. Probability each not multiple of 3 is 4/6=2/3. So none multiple of 3 is (2/3)^2=4/9. Complement gives 1−4/9=5/9.
A die is rolled thrice. Probability of getting exactly one 6
A 25/216
B 75/216
C 125/216
D 36/216
Exactly one 6: choose which roll is 6 (3 ways), then other two are not 6: (5/6)^2. Probability = 3×(1/6)×(25/36)=75/216.
Two events A and B satisfy P(A)=0.4, P(B)=0.3, and P(A∪B)=0.6. Find P(A∩B)
A 0.2
B 0.3
C 0.0
D 0.1
Use union rule: P(A∪B)=P(A)+P(B)−P(A∩B). So 0.6=0.4+0.3−x, giving x=0.1.
A and B are independent with P(A)=0.6 and P(A∩B)=0.3. Find P(B)
A 0.4
B 0.3
C 0.5
D 0.2
Independence gives P(A∩B)=P(A)P(B). So 0.3=0.6×P(B). Thus P(B)=0.3/0.6=0.5.
If P(A|B)=0.5 and P(A|B′)=0.2, P(B)=0.6, then P(A) is
A 0.32
B 0.38
C 0.50
D 0.20
Use total probability on partition {B,B′}: P(A)=P(A|B)P(B)+P(A|B′)P(B′)=0.5×0.6+0.2×0.4=0.3+0.08=0.38.
A bag has 2 red, 2 blue, 1 green. Three balls drawn without replacement. Probability of getting all three colors
A 3/5
B 4/5
C 1/5
D 2/5
Need 1 green and 1 red and 1 blue. Ways: C(2,1)C(2,1)C(1,1)=4. Total ways C(5,3)=10. Probability=4/10=2/5.
A die is rolled twice. Probability that maximum of two numbers is 4
A 7/36
B 1/4
C 1/6
D 5/36
Max is 4 means both ≤4 and at least one is 4. P(both ≤4)=16/36. P(both ≤3)=9/36. Difference =7/36.
A die is rolled twice. Probability that minimum of two numbers is 3
A card is drawn. Probability it is a queen given it is a red face card
A 1/2
B 1/3
C 1/4
D 2/3
Red face cards are J,Q,K of hearts and diamonds: 6 cards. Red queens are 2. So conditional probability =2/6=1/3.
Two cards drawn without replacement. Given both are red, probability both are hearts
A 12/25
B 1/2
C 1/4
D 6/25
Given both red, sample space is choosing 2 from 26 red cards: C(26,2). Hearts are 13, so both hearts is C(13,2). Ratio = C(13,2)/C(26,2)=78/325=6/25.
Two dice are thrown. Probability that sum is 10 given product is even
A 1/12
B 1/9
C 1/6
D 1/18
Given product even, allowed outcomes are 27. Sum 10 with even product occurs in (4,6) and (6,4) only, so probability is 2/27.
A fair coin is tossed 6 times. Probability of getting no two consecutive heads
A 13/64
B 34/64
C 21/64
D 1/2
Count sequences of length 6 with no consecutive heads. This count follows Fibonacci: for n=6, count is 21. Total outcomes are 2^6=64, so probability 21/64.
A fair die is rolled. Event A: “even”. Event B: “≥4”. Check if independent
A Not independent
B Independent
C Mutually exclusive
D Complementary
P(A)=3/6=1/2, P(B)=3/6=1/2. A∩B is {4,6} so P=2/6=1/3. Since 1/3 ≠ (1/2)(1/2)=1/4, not independent.
A fair die is rolled. Events A: “odd”, B: “prime”. Which is true
A Independent
B Mutually exclusive
C Exhaustive pair
D Equal events
Since P(odd∩prime)=1/3 but P(odd)P(prime)=1/4, the events are not independent. They also are not mutually exclusive because 3 and 5 satisfy both.
Two children in a family, each equally likely boy/girl. Given at least one is a boy, probability both are boys
A 1/2
B 2/3
C 1/4
D 1/3
Sample space: {BB,BG,GB,GG}. Given at least one boy removes GG, leaving 3 outcomes. Only BB is favourable, so probability 1/3.
Same family problem. Given the older child is a boy, probability both are boys
A 1/3
B 2/3
C 1/2
D 1/4
If older is a boy, possible outcomes are BB or BG, equally likely. Only BB gives both boys, so probability is 1/2.
A die is rolled and a coin is tossed. Probability of getting a head and an even number
A 1/3
B 1/4
C 1/2
D 1/6
Independent events: P(head)=1/2 and P(even)=1/2. Multiply gives 1/4.
A random 2-digit number from 10 to 99 is selected. Probability it is divisible by 9
A 1/10
B 1/8
C 1/11
D 1/9
There are 90 two-digit numbers. Multiples of 9 between 10 and 99 are 18 to 99: 10 numbers. Probability is 10/90=1/9.
A fair die is rolled. Probability that number is greater than 2 given it is not 6
A 4/5
B 2/5
C 3/5
D 1/5
Restrict sample space to five outcomes excluding 6. Three of them are greater than 2, so conditional probability is 3/5.
In Bayes theorem, if prior probabilities are equal, the most likely hypothesis after evidence is the one with highest
A Likelihood
B Complement
C Sample space
D Intersection size
With equal priors, posterior is proportional to likelihood P(A|Bi). The hypothesis that makes the evidence most probable gets the largest posterior after normalization.
If P(A)=0.4, P(B)=0.5, P(A|B)=0.6, then P(B|A) is
A 0.60
B 0.75
C 0.48
D 0.30
First find P(A∩B)=P(A|B)P(B)=0.6×0.5=0.3. Then P(B|A)=P(A∩B)/P(A)=0.3/0.4=0.75.
A bag has 3 red and 3 blue. Balls drawn one by one without replacement until a red appears. Probability that first red appears on 3rd draw
A 1/5
B 2/5
C 3/10
D 1/10
Without replacement: P(BB then R)=(3/6)×(2/5)×(3/4)=18/120=3/20.
Two events A and B have P(A)=0.7, P(B)=0.6. What is the minimum possible P(A∩B)
A 0.1
B 0.0
C 0.3
D 0.4
Minimum intersection is max(0, P(A)+P(B)−1)=0.7+0.6−1=0.3. This lower bound comes from union not exceeding 1.
If A and B are independent with P(A)=0.2 and P(B)=0.3, then P(exactly one occurs) is
A 0.32
B 0.38
C 0.06
D 0.44
Exactly one = P(A)+P(B)−2P(A)P(B)=0.5−0.12=0.38. This counts A only plus B only in independent case.
A and B are events with P(A)=0.5 and P(A∪B)=0.8 and P(A|B)=0.5. Find P(B)
A 0.6
B 0.4
C 0.3
D 0.5
P(A|B)=0.5 means P(A∩B)=0.5P(B). Use union: 0.8=0.5+P(B)−0.5P(B)=0.5+0.5P(B). So 0.3=0.5P(B), P(B)=0.6.
A box contains 3 coins: two fair, one double-headed. A coin chosen randomly and tossed once. Given head occurs, probability chosen coin was double-headed
A 2/3
B 1/2
C 1/3
D 3/4
Bayes updating gives P(DH|H)= (1/3) / (2/3) = 1/2. Head makes double-headed more likely, but still only half because two fair coins also can produce heads.