A letter is chosen from the word “PROBABILITY”. What is the probability of selecting a vowel
A 3/11
B 5/11
C 4/11
D 6/11
“PROBABILITY” has 11 letters. Vowels are O, A, I, I (4 vowels). With equally likely selection of a letter position, probability of vowel is 4/11.
A card is drawn from a deck. What is the probability that it is neither a heart nor a king
A 9/13
B 10/13
C 3/13
D 1/13
Hearts are 13, kings are 4, overlap (king of hearts) is 1. So P(heart or king)=16/52=4/13. “Neither” is 1−4/13=9/13.
Two dice are thrown. What is the probability that the sum is a prime number
A 1/3
B 5/12
C 7/12
D 1/2
Prime sums possible: 2,3,5,7,11. Number of outcomes: 1+2+4+6+2=15 out of 36. So probability is 15/36=5/12.
Two dice are thrown. Probability that at least one shows 5 is
A 1/6
B 5/36
C 25/36
D 11/36
Use complement: no 5 on either die has probability (5/6)²=25/36. So at least one 5 is 1−25/36=11/36.
Two coins are tossed repeatedly. Which is a Bernoulli trial example
A Single coin toss
B Drawing a card
C Rolling two dice
D Choosing a book
A Bernoulli trial has two outcomes (success/failure) with fixed probability. A coin toss has two outcomes and constant probability if the coin is fair.
A bag has 4 red, 3 blue, 2 green. One ball is drawn. P(not green) is
A 2/9
B 5/9
C 7/9
D 4/9
Total balls are 9. Green balls are 2. Not green means red or blue, which are 7 balls. So probability is 7/9.
From the same bag (4R,3B,2G), two balls are drawn without replacement. P(both green) is
A 1/36
B 1/12
C 2/9
D 1/9
First green: 2/9. Second green given first green: 1/8. Multiply: (2/9)(1/8)=2/72=1/36.
A bag has 4 red, 3 blue, 2 green balls. Two balls are drawn without replacement. What is the probability of getting at least one green
A 7/12
B 1/3
C 2/3
D 5/12
Use complement: “at least one green” = 1 − P(no green). No green means both from 7 non-green: (7/9)×(6/8)=7/12. So 1−7/12=5/12.
An urn has 5 white and 5 black balls. Two draws without replacement. Probability of different colors
A 1/2
B 4/9
C 5/9
D 2/9
Different colors means WB or BW. P(WB)=(5/10)(5/9)=25/90, P(BW)=(5/10)(5/9)=25/90. Total=50/90=5/9.
In the same urn (5W,5B), two draws with replacement. Probability of different colors
A 1/2
B 5/9
C 4/9
D 1/4
With replacement, draws are independent. P(W)=P(B)=1/2. Different colors = 2×(1/2)(1/2)=1/2.
A fair die is rolled twice. Probability that the second number is greater than the first
A 1/2
B 7/12
C 1/3
D 5/12
There are 36 ordered pairs. Ties are 6 pairs. Remaining 30 are unequal, split equally between second>first and second
A fair die is rolled twice. Probability that both numbers are the same
A 1/6
B 1/12
C 5/12
D 1/36
Matching pairs are (1,1) to (6,6): 6 outcomes out of 36. So probability is 6/36=1/6.
A die is rolled. Events A: “≤3”, B: “odd”. What is P(A|B)
A 1/3
B 1/2
C 2/3
D 3/4
Given odd outcomes {1,3,5}. Event A within odds is {1,3}. So conditional probability is 2 favourable out of 3 possible, giving 2/3.
A die is rolled. Events A: “prime”, B: “odd”. What is P(A|B)
A 1/3
B 2/3
C 1/2
D 1
Given odd outcomes {1,3,5}. Prime among these are {3,5}. So probability that number is prime given it is odd equals 2/3.
A card is drawn. Given it is red, probability it is a diamond
A 1/2
B 1/4
C 3/4
D 1/13
Given red, possible suits are hearts or diamonds (26 cards). Diamonds are 13 of these 26. So P(diamond | red)=13/26=1/2.
A card is drawn. Given it is a face card, probability it is a queen
A 1/4
B 1/12
C 1/13
D 1/3
Face cards are J, Q, K across 4 suits: 12 total. Queens are 4. So probability is 4/12=1/3.
A box has 3 defective and 7 good bulbs. Two chosen without replacement. Probability both are good
A 7/10
B 21/30
C 7/15
D 7/12
P(first good)=7/10. P(second good | first good)=6/9=2/3. Multiply: (7/10)(2/3)=14/30=7/15.
✅ Correct Answer: A. 7/15
Explanation: Without replacement: (7/10)(6/9)=42/90=7/15.
Same box (3D,7G). Two chosen without replacement. Probability at least one defective
A 8/15
B 7/15
C 1/5
D 3/10
Use complement: at least one defective = 1 − P(both good). From earlier, P(both good)=7/15. So result is 1−7/15=8/15.
If P(A)=0.7, P(B)=0.5, and P(A∩B)=0.4, then P(A|B) is
A 0.4
B 0.8
C 0.2
D 0.6
P(A|B)=P(A∩B)/P(B)=0.4/0.5=0.8. It means given B happened, A is quite likely in this situation.
If P(A|B)=0.6 and P(B)=0.5, then P(A∩B) equals
A 0.10
B 0.60
C 0.30
D 0.50
Use multiplication rule: P(A∩B)=P(A|B)P(B)=0.6×0.5=0.30. This is the standard way to convert conditional to joint probability.
A and B are independent with P(A)=0.4, P(B)=0.3. Then P(A|B) is
A 0.4
B 0.3
C 0.12
D 0.7
Independence means B does not affect A. So conditional probability stays the same: P(A|B)=P(A)=0.4.
If A and B are independent, then P(A∩B′) equals
A P(A)+P(B′)
B P(A)−P(B)
C 1−P(A∩B)
D P(A)P(B′)
Independence extends to complements, so A and B′ are independent. Therefore P(A∩B′)=P(A)P(B′), useful for “A occurs but B does not” cases.
A test for a disease has sensitivity 0.9 meaning
A P(disease|positive)
B P(negative|disease)
C P(positive|disease)
D P(positive|no disease)
Sensitivity is the probability a test is positive given the person truly has the disease. It measures how well the test detects disease cases.
A test has specificity 0.8 meaning
A P(negative|no disease)
B P(positive|disease)
C P(disease|negative)
D P(positive|no disease)
Specificity is the probability the test is negative given the person does not have the disease. It relates to controlling false positives.
If disease prevalence is 0.1, this is
A Posterior probability
B Prior probability
C Likelihood only
D Evidence only
Prevalence is the probability of disease before seeing test results. In Bayes problems, such base rates are prior probabilities used for later updating.
If events B1,B2,B3 partition S, then in Bayes formula denominator is
A ΠP(A|Bi)P(Bi)
B P(Bi|A)
C ΣP(A|Bi)P(Bi)
D P(A|Bi) only
The denominator is P(A) computed via total probability over the partition. It normalizes the posteriors so probabilities across hypotheses sum to 1.
In Bayes theorem, which quantity is updated after evidence
A Posterior probability
B Prior probability
C Sample space
D Exhaustive set
After observing evidence A, the probability of hypothesis Bi becomes P(Bi|A), called the posterior. It is the “updated belief” after using data.
A box has 2 gold and 3 silver coins. Two coins drawn without replacement. Probability both are gold
A 1/5
B 2/5
C 3/10
D 1/10
P(first gold)=2/5. Then gold given gold: 1/4. Multiply: (2/5)(1/4)=2/20=1/10.
Same box (2G,3S). Two drawn without replacement. Probability exactly one gold
A 3/5
B 3/10
C 1/2
D 2/5
Exactly one gold occurs as GS or SG. P(GS)=(2/5)(3/4)=6/20 and P(SG)=(3/5)(2/4)=6/20. Total=12/20=3/5.
A fair die is rolled. Probability that number is divisible by 2 or 3
A 1/3
B 1/2
C 2/3
D 5/6
Divisible by 2: {2,4,6}. Divisible by 3: {3,6}. Union is {2,3,4,6} with 4 outcomes. So probability 4/6=2/3.
A die is rolled. Probability that number is divisible by 2 and 3
A 1/3
B 1/6
C 1/2
D 2/3
Divisible by both 2 and 3 means divisible by 6. Only outcome is 6, so probability is 1/6.
A fair coin is tossed 4 times. Probability of at least one head
A 15/16
B 1/16
C 1/4
D 3/4
Complement method: no head means all tails, probability (1/2)^4=1/16. So at least one head is 1−1/16=15/16.
A fair coin is tossed 4 times. What is the probability of getting exactly two heads
A 5/16
B 1/4
C 7/16
D 3/8
Exactly two heads can occur in C(4,2)=6 ways. Total outcomes are 2⁴=16. So probability = 6/16 = 3/8.
A coin is biased with P(H)=0.6. Two tosses. Probability of exactly one head
A 0.36
B 0.24
C 0.48
D 0.16
Exactly one head occurs as HT or TH. Probability = 2×(0.6)(0.4)=0.48. This uses independence of tosses and multiplication rule.
A biased coin has P(H)=0.7. Three tosses. Probability of all heads
A 0.343
B 0.49
C 0.21
D 0.70
Tosses are independent. All heads probability is 0.7×0.7×0.7=0.7^3=0.343.
From a deck, two cards drawn without replacement. Probability both are face cards
A 3/13
B 1/17
C 1/13
D 11/221
Face cards are 12. P(first face)=12/52=3/13. Then P(second face)=11/51. Multiply: (3/13)(11/51)=33/663=11/221.
From a deck, two cards drawn without replacement. Probability of getting at least one ace
A 33/221
B 1/13
C 4/13
D 15/221
Use complement: no ace in two draws. P(no ace first)=48/52=12/13, then 47/51. Product = (12/13)(47/51)=188/221. So at least one ace = 1−188/221=33/221.
If A and B are mutually exclusive with P(A)=0.4 and P(B)=0.3, then P(A∩B) is
A 0.12
B 0.7
C 0
D 1
Mutually exclusive events cannot happen together. So their intersection is impossible, giving P(A∩B)=0.
If P(A)=0.4, P(B)=0.3, and P(A∩B)=0.2, then P(A∪B) is
A 0.7
B 0.5
C 0.9
D 0.2
Apply union rule: P(A∪B)=P(A)+P(B)−P(A∩B)=0.4+0.3−0.2=0.5. This avoids double counting overlap.
If P(A|B)=0.5 and P(B|A)=0.4 and P(A)=0.2, then P(B) is
A 0.16
B 0.25
C 0.10
D 0.08
Use equality P(A∩B)=P(A)P(B|A)=0.2×0.4=0.08. Also P(A∩B)=P(B)P(A|B)=P(B)×0.5. So P(B)=0.08/0.5=0.16.
A die is rolled. Probability that number is at most 5 given it is odd
A 2/3
B 1/2
C 1/3
D 1
Given odd outcomes {1,3,5}, all are at most 5. So conditional probability equals 3/3=1.
A die is rolled. Probability of getting 6 given number is even
A 1/2
B 2/3
C 1/3
D 1/6
Given even outcomes {2,4,6}, only one of these is 6. So P(6 | even)=1/3.
In a group, 60% speak Hindi, 40% speak English, 20% speak both. Probability a person speaks at least one is
A 0.80
B 1.00
C 0.60
D 0.40
Use union: P(H∪E)=0.6+0.4−0.2=0.8. This includes people speaking at least one language and avoids double counting bilingual speakers.
Using the same data (Hindi 0.6, English 0.4, both 0.2), probability a person speaks only Hindi is
A 0.20
B 0.40
C 0.60
D 0.80
Only Hindi means Hindi but not English. So P(only Hindi)=P(H)−P(H∩E)=0.6−0.2=0.4.
Three events are mutually independent. Then P(A∩B∩C) equals
A P(A)+P(B)+P(C)
B P(A)P(B|C)
C P(A)P(B)P(C)
D P(A|B|C)
Mutual independence means all combinations are independent. So probability that all three occur together equals the product of their individual probabilities.
If P(A)=0.3, P(B)=0.4, P(C)=0.5 and they are mutually independent, then P(A∩B∩C) is
A 0.06
B 0.12
C 0.20
D 0.30
Multiply probabilities: 0.3×0.4×0.5=0.06. This is the simplest application of multiplication rule under independence for three events.
In a class, 30% like tea, 50% like coffee, 10% like both. Probability a student likes neither is
A 0.40
B 0.50
C 0.10
D 0.30
P(tea or coffee)=0.3+0.5−0.1=0.7. Neither is complement: 1−0.7=0.3.
✅ Correct Answer should be A. 0.30
Explanation: Using complement: neither = 1 − (0.3+0.5−0.1)=1−0.7=0.3.
If P(A)=0.6 and P(B)=0.5, and A contains B (B⊆A), then P(A∩B) equals
A P(B)
B P(A)
C P(A)+P(B)
D 0
If B is a subset of A, then whenever B happens, A also happens. Thus intersection A∩B is exactly B, so P(A∩B)=P(B).
If P(B|A)=1 and P(A)>0, then relation between events is
A B ⊆ A
B A and B disjoint
C A ⊆ B
D A and B independent
P(B|A)=1 means if A occurs then B must occur. This implies every outcome of A lies inside B, so A is a subset of B.
For any events A and B, which inequality is always true
A P(A∪B) ≥ P(A)+P(B)
B P(A∪B) ≤ P(A)+P(B)
C P(A∩B) ≥ P(A)
D P(A) > 1
Union counts outcomes in A or B. Adding P(A)+P(B) may double-count overlap, so union cannot exceed this sum. Equality holds when events are disjoint.