Probability questions, with answers

The questions, with answers

1. 1.What is the basic probability formula, and what counts as an outcome?

   Probability = favourable outcomes / total outcomes, when every outcome is equally likely. The work is in counting both correctly. A die has 6 outcomes; two dice have 36 ordered outcomes, not 21 — (1, 2) and (2, 1) are different — so the probability of a sum of 7 is 6 / 36 = 1/6. Rolling a number greater than 2 on one die is 4 / 6 = 2/3. Probabilities lie between 0 and 1, and the probabilities of all outcomes sum to 1. The word "or" between mutually exclusive events means add; the word "and" for independent events means multiply; everything on this page is a way of counting or of using those two rules.

2. 2.How does the complement rule handle "at least one" questions?

   P(at least one) = 1 - P(none), and it turns a many-case count into a single case. For two coins, at least one head is 1 - P(both tails) = 1 - 1/4 = 3/4. For three dice, at least one six is 1 - (5/6)³ = 1 - 125/216 = 91/216. Any question containing "at least one", "not all" or "at most n - 1" is asking for a complement; counting the cases directly works but is where the arithmetic slips occur. The reverse also appears: the probability that none of four independent 20% events happen is 0.8⁴ = 0.4096, so at least one is 0.5904.

3. 3.How do you add probabilities when the events can overlap?

   P(A or B) = P(A) + P(B) - P(A and B): subtract the overlap, or you count it twice. From a standard 52-card deck, P(king or heart) = 4/52 + 13/52 - 1/52 = 16/52 = 4/13, because the king of hearts is both. Only when the events are mutually exclusive — king or queen — is the overlap zero and the simple sum right: 8/52 = 2/13. The trap questions are the ones where the overlap is easy to miss: "a multiple of 2 or 3 from 1 to 30" is 15 + 10 - 5 = 20, so 2/3, the 5 being the multiples of 6.

4. 4.How do you handle drawing without replacement?

   Multiply the probabilities of each draw, updating the totals after every draw. A bag has 5 red and 7 blue balls; two are drawn without replacement. P(both red) = 5/12 × 4/11 = 20/132 = 5/33 — the second draw sees 4 reds among 11 balls. P(one of each) counts both orders: red then blue is 5/12 × 7/11, blue then red is 7/12 × 5/11, together 70/132 = 35/66. The combinations version gives the same answer: C(5,2) / C(12,2) = 10 / 66 = 5/33 for both red. With replacement the fractions do not change, so both red would be (5/12)² = 25/144. Say which case you are in before you multiply.

5. 5.What is the difference between independent and mutually exclusive events?

   They are almost opposites, and candidates confuse them constantly. Independent events do not affect each other: P(A and B) = P(A) × P(B), like two coin tosses, and both can happen together. Mutually exclusive events cannot both happen: P(A and B) = 0, like a card being a king and a queen, so P(A or B) is just the sum. A pair of events with non-zero probability cannot be both independent and mutually exclusive: if they cannot co-occur, knowing one happened tells you the other did not, which is dependence. Drawing without replacement makes successive draws dependent; with replacement they are independent.

6. 6.What are odds, and how do they relate to probability?

   Odds in favour of an event are the ratio of favourable to unfavourable outcomes; probability is favourable over total. So a probability of 1/4 is odds of 1 : 3 in favour, and odds of 3 : 1 against. To convert odds a : b in favour to probability, take a / (a + b). Questions that give odds and ask for a probability, or give odds for two independent events and ask for the probability that both occur, are testing exactly this conversion: odds 2 : 3 in favour of A and 3 : 5 in favour of B give P(both) = 2/5 × 3/8 = 6/40 = 3/20. Convert first, then apply the usual rules — never multiply the odds directly.

7. 7.How do you solve questions on two dice?

   Lay out the 36 equally likely ordered pairs, then count. Sums run from 2 to 12; the number of ways to make a sum s is s - 1 for s up to 7 and 13 - s beyond, so 7 is the most likely sum (6 ways, 1/6) and 2 and 12 the least (1 way each, 1/36). Doubles are 6 of 36, 1/6. "At least one die shows a 5" is 11 of 36, not 12 — (5, 5) is one outcome — or by the complement, 1 - (5/6)² = 11/36. "The product is even" is easier via the complement too: both odd is (1/2)² = 1/4, so 3/4. The only real skill is remembering that the pairs are ordered and counting from that grid.

8. 8.What is conditional probability, and how do you spot it in a question?

   P(A given B) = P(A and B) / P(B): the probability of A once you know B has happened, computed by shrinking the sample space to the outcomes where B holds. A family has two children and you learn at least one is a boy; the probability both are boys is not 1/2 but 1/3 — of the outcomes BB, BG, GB with at least one boy, only one is BB. The signal words are "given that", "if it is known that" and "of those who". Tests rarely go as far as Bayes' theorem, but they do like the tree version: choose a bag at random, then draw a ball, and ask for the probability of a colour — multiply along each branch and add the branches.

How the diagnostic asks it

One question from the Aptitude bank, exactly as a sitting would show it. The bank has 4 on probability and 60 across Aptitude.