Probability
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.
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.
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.
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.
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.
| 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 |
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.
A bag contains 4 red, 5 blue and 4 green balls.
One ball is drawn at random.
What is the probability that it is red?
A box contains 5 defective and 6 good bulbs.
One bulb is picked at random.
What is the probability that it is NOT defective?
A box contains 3 defective and 3 good bulbs.
One bulb is picked at random.
What is the probability that it is NOT defective?
This page shows 3 worked out in full. The question bank has 15, every answer on the page, and the whole set as a free PDF you can print or work offline: probability questions with answers PDF. That page is the questions; this one is the method behind them.
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.
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.
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.
It is legal but slow, and each extra case is a chance to miss one. The complement is one calculation and a subtraction.
The specific wrong numbers this topic produces, and what each one tells you about the step you took. If you have just got a question wrong and want to know which habit did it, start here.
Because without replacement the pool shrinks. Drawing two reds from 5 red and 4 blue is 5/9 × 4/8, not 5/9 × 5/9. Whenever the question says one after another without replacement, every draw after the first has a smaller denominator, and often a smaller numerator too.
Subtract from 1. The probability of at least one is 1 − P(none), which is a single calculation. Adding up exactly one, exactly two and exactly three works, takes four times as long, and is wrong the moment you miss a case.
Yes — both halves of the ratio must count the same way. Mixing nCr on top with nPr underneath leaves a stray factor of r! in the answer. Choose one convention for the whole question; for anything phrased as choosing a group it is combinations.
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.
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.
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.
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