Practice Questions
Concept Explanation
At least one means one or more successes. The complement method is usually fastest: subtract P(none) from 1.
Key Formula
- P(at least one) = 1 − P(none)
- For independent trials: P(at least one) = 1 − P(fail)^n
- Works for coins, dice rolls, and repeated draws with replacement
Example
Three coins: P(at least one head) = 1 − (1/2)^3 = 7/8.
Common Trap
At least one is not the same as exactly one.
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Q1 EasyThree unbiased coins are tossed. What is the probability of getting at least one head?ExplanationTotal outcomes = 2^3 = 8. P(no head) = P(all tails) = 1/8. Using the complement: P(at least one head) = 1 − P(none) = 1 − 1/8 = 7/8. Hence the answer is 7/8.
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Q2 EasyFour fair coins are tossed. What is the probability of getting at least one head?ExplanationP(all tails) = (1/2)^4 = 1/16. P(at least one head) = 1 − 1/16 = 15/16. Hence the answer is 15/16.
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Q3 EasyTwo fair coins are tossed. What is the probability of getting at least one tail?ExplanationP(no tail) = P(HH) = 1/4. P(at least one tail) = 1 − 1/4 = 3/4. Hence the answer is 3/4.
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Q4 EasyTwo fair coins are tossed. What is the probability of getting at least one head?ExplanationP(no head) = P(TT) = 1/4. P(at least one head) = 1 − 1/4 = 3/4. Hence the answer is 3/4.
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Q5 MediumTwo dice are thrown simultaneously. What is the probability of getting at least one six?ExplanationTotal outcomes = 36. P(no six on either die) = (5/6) × (5/6) = 25/36. P(at least one six) = 1 − 25/36 = 11/36. Hence the answer is 11/36.
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Q6 MediumA fair die is rolled twice. What is the probability that at least one roll shows an even number?ExplanationP(odd on one roll) = 1/2. P(both odd) = (1/2)^2 = 1/4. P(at least one even) = 1 − 1/4 = 3/4. Hence the answer is 3/4.
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Q7 MediumThe probability of rain on any given day is 0.2 and days are independent. What is P(at least one rainy day in two days)?ExplanationP(no rain on a day) = 0.8. P(no rain both days) = 0.8^2 = 0.64. P(at least one rainy day) = 1 − 0.64 = 0.36. Hence the answer is 0.36.
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Q8 MediumA and B independently try to solve a problem. The probability that A solves it is 1/2 and that B solves it is 1/3. What is the probability that the problem is solved (at least one of them solves it)?ExplanationP(neither solves) = (1 − 1/2) × (1 − 1/3) = (1/2) × (2/3) = 1/3. P(at least one solves) = 1 − 1/3 = 2/3. Hence the answer is 2/3.
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Q9 MediumA bag is drawn from with replacement where P(red) = 0.4 each time. Two draws are made. What is P(at least one red)?ExplanationP(not red) = 0.6. P(no red in two draws) = 0.6^2 = 0.36. P(at least one red) = 1 − 0.36 = 0.64. Hence the answer is 0.64.
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Q10 MediumThree unbiased coins are tossed. What is the probability of getting at least two heads?ExplanationFavourable outcomes: HHT, HTH, THH, HHH — 4 out of 8. P = 4/8 = 1/2. Equivalently, P(exactly 2) + P(exactly 3) = 3/8 + 1/8 = 1/2. Hence the answer is 1/2.
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Q11 HardThe probability that a man hits a target is 1/4. If he fires 3 shots independently, what is the probability of hitting the target at least once?ExplanationP(miss) = 3/4. P(all 3 miss) = (3/4)^3 = 27/64. P(at least one hit) = 1 − 27/64 = 37/64. Hence the answer is 37/64.
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Q12 HardFive fair coins are tossed. What is the probability of getting at least one tail?ExplanationP(no tail) = P(all heads) = 1/32. P(at least one tail) = 1 − 1/32 = 31/32. Hence the answer is 31/32.
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Q13 HardThree fair dice are rolled. What is the probability of getting at least one six?ExplanationP(no six on one die) = 5/6. P(no six on all three) = (5/6)^3 = 125/216. P(at least one six) = 1 − 125/216 = 91/216. Hence the answer is 91/216.
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Q14 HardTwo cards are drawn at random from a pack of 52 cards. What is the probability that at least one of them is an ace?ExplanationTotal ways = C(52,2) = 1326. Ways with no ace = C(48,2) = 1128. P(no ace) = 1128/1326. P(at least one ace) = 1 − 1128/1326 = 198/1326 = 33/221. Hence the answer is 33/221.
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DS1 Easy
In a campus coin-toss drill, what is P(at least one head when two coins are tossed)?
Statement ITwo fair coins are tossed.
Statement IIP(both tails) = 1/4.
ExplanationStatement I alone: P(TT) = 1/4, so P(at least one head) = 1 − 1/4 = 3/4. Statement II alone gives the same complement result. Each alone is sufficient. Answer: D. -
DS2 Medium
What is P(at least one six when two dice are rolled)?
Statement ITwo fair dice are rolled.
Statement IIP(no six appears on either die) = 25/36.
ExplanationStatement I alone: P = 1 − (5/6)^2 = 11/36. Statement II alone: P = 1 − 25/36 = 11/36. Each alone is sufficient. Answer: D.
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DS3 Medium
A placement worksheet asks: A and B try a problem independently. What is P(at least one solves it)?
Statement IP(A solves) = 1/2 and P(B solves) = 1/3.
Statement IIA and B work independently.
ExplanationNeed both individual probabilities and independence: P(neither) = (1/2)(2/3) = 1/3, so P(at least one) = 2/3. Answer: C. -
DS4 Hard
For P(at least one hit in three independent shots). Can it be found?
Statement IEach shot has hit probability 1/4.
Statement IIThe three shots are independent.
ExplanationStatement I alone does not confirm independence for multiplying miss probabilities. Statement II alone lacks the hit probability. Together: P = 1 − (3/4)^3 = 37/64. Answer: C.