# Probability Questions — Free Timed Practice | ComputePrep > Practise probability questions free and timed: favourable over total, the complement shortcut, and when to add versus multiply. Worked solutions on every drill. Source: https://computeprep.com/drill-types/probability --- Probability # Probability questions — count the favourable, count the total Probability questions reduce to one ratio: favourable outcomes over total outcomes. The judgement is in the counting — whether order matters, whether an item is replaced, and whether the events are independent. When a question says at least one, count the complement instead and subtract from one. Probability is one or two marks in a bank prelims paper and rather more in mains, and it is the topic where candidates most often compute a correct ratio from a wrong count. The formula is never the difficulty; deciding what the sample space actually is, is. **Try it against the clock.** Probability runs at 45s easy, 55s medium and 70s hard. The clock is stamped and judged on our server, so the limit you see is the deadline that is actually enforced — and it starts when you tap Start, not while the question is loading. [Start a timed Probability drill →](https://computeprep.com/?drill=probability&src=seo-probability) ## How to solve probability questions by counting correctly - Write the total outcomes before anything else Choosing 2 from 10 balls has C(10,2) = 45 total outcomes. Fixing the denominator first stops the common error of counting favourable cases with order and total cases without, which quietly produces a ratio that is off by a factor. - Decide once whether order matters, and stay consistent Use combinations for both numerator and denominator, or permutations for both. Mixing them is the single largest source of wrong answers here, and the resulting value usually looks plausible. - Use the complement whenever you see “at least” The probability of at least one head in three tosses is 1 − P(no heads) = 1 − 1/8 = 7/8. Counting the complement is almost always fewer cases than counting the event, and the subtraction is free. - Add for OR, multiply for AND Mutually exclusive events add; independent events multiply. Where events overlap, subtract the intersection — P(A or B) = P(A) + P(B) − P(A and B). Deciding which connector the question uses is most of the work. ## Which rule the wording calls for | The question says | Use | | --- | --- | | At least one | 1 − P(none) | | Both / and (independent) | Multiply the probabilities | | Either / or (exclusive) | Add the probabilities | | Either / or (overlapping) | P(A) + P(B) − P(A and B) | | Drawn together / at once | Combinations — order does not matter | | Drawn one after another, not replaced | Denominator shrinks on the second draw | ## 3 real probability questions, with worked solutions Generated by the same engine that mints the ComputePrep daily. Each answer is computed from the numbers printed in the question, and the walkthrough below each one is the engine's own working — not a solution written afterwards. Reload this page's live drill and you get different numbers. ### Question 1 A bag contains 7 red, 5 blue and 4 green balls. One ball is drawn at random. What is the probability that it is red? - 1/2 - 1/16 - 9/16 - 7/16 - 7/9 - Total balls = 16; favourable (red) = 7. - P(red) = favourable ÷ total = 7/16 = 7/16. - Answer: 7/16. ### Question 2 A box contains 5 defective and 3 good bulbs. One bulb is picked at random. What is the probability that it is NOT defective? - 1/2 - 3/8 - 5/8 - 1/8 - 3/5 - P(not defective) = 1 − P(defective) = 1 − 5/8. - That is the 3 good bulbs out of 8: 3/8. - Answer: 3/8. ### Question 3 A bag contains 5 red, 2 blue and 3 green balls. One ball is drawn at random. What is the probability that it is red? - 1/2 - 1/3 - 1/4 - 1/5 - 1/10 - Total balls = 10; favourable (red) = 5. - P(red) = favourable ÷ total = 5/10 = 1/2. - Answer: 1/2. ## Common mistakes this drill is built from The wrong options are not random numbers. Each one is the result of a specific careless error for this topic, so picking one tells you which habit is costing you marks. ### Mixing permutations and combinations across the ratio If the numerator counts ordered outcomes the denominator must too. This produces answers that are wrong by a constant factor and look entirely reasonable, which is why it survives a quick check. ### Forgetting that the denominator shrinks without replacement Drawing two balls without replacement from ten gives 10 then 9, not 10 then 10. Read for the word replaced, and if it is absent assume it is not. ### Counting “at least one” directly It is legal but slow, and each extra case is a chance to miss one. The complement is one calculation and a subtraction. ## FAQ ### How do I solve “at least one” probability questions? Compute the probability that none of them happens and subtract from one. For at least one head in three coin tosses: P(no heads) = (1/2)³ = 1/8, so the answer is 7/8. Counting the event directly means enumerating one head, two heads and three heads separately — more work and more chances to miss a case. ### When do I multiply probabilities and when do I add them? Multiply when both events must happen and they are independent; add when either will do and they cannot happen together. If they can both happen, add and then subtract the overlap. The connector in the question — and versus or — is what tells you. ### Should I use combinations or permutations? Use combinations when the selection is simultaneous or order is irrelevant, which covers most exam questions about drawing balls or forming committees. What matters more than the choice is consistency: whichever you use for the favourable count must also be used for the total. ## Practise the next topic - [All 23 drill types](https://computeprep.com/drill-types) — the full rotation with every published time limit. - [Percentage questions for bank exam papers](https://computeprep.com/drill-types/percentages) — percentage questions for bank exam, timed. - [Average questions](https://computeprep.com/drill-types/averages) — average questions, timed. - [Vedic maths shortcuts](https://computeprep.com/vedic-maths) — the seven sutras that make a 15-second answer possible. - [How to improve calculation speed](https://computeprep.com/calculation-speed) — the six-week arc these drills fit into. ComputePrep is free, needs no account, and is not affiliated with IBPS, SBI, SSC, RRB or any exam authority.