Probability for NDA

Probability questions in NDA are short, and nearly all of them are won or lost at the counting stage. If you list the sample space correctly, the answer is a fraction you can simplify in your head. The syllabus goes further than school probability: along with events and the addition rule, it names conditional probability, Bayes' theorem, random variables and the binomial distribution.

The chapter leans on permutations and combinations. Whenever outcomes are too many to list, you count them with nCr.

Typical question types are:

• dice (sums, doubles, products), coins and cards;
• "at least one" questions, solved through the complement;
• the addition rule with overlapping events;
• independent events and drawing with or without replacement;
• conditional probability, usually with a die or two dice;
• Bayes' theorem with two bags or two machines;
• binomial probability, or finding n and p from a mean and variance.

The language of events

• Sample space S: the set of all outcomes. Two dice give 36 outcomes; three coins give 8; a pack has 52 cards (4 suits of 13, with 12 face cards).
• Classical probability: P(E) = (favourable outcomes)/(total outcomes), when all outcomes are equally likely. So 0 ≤ P(E) ≤ 1.
• Complement: P(not E) = 1 − P(E).
• Mutually exclusive events cannot happen together: P(A ∩ B) = 0.
• Exhaustive events together cover the whole sample space.
• Odds in favour of E are a : b when P(E) = a/(a + b). Odds against are b : a.

The core rules

• Addition rule: P(A ∪ B) = P(A) + P(B) − P(A ∩ B). For mutually exclusive events, drop the last term.
• Independent events: P(A ∩ B) = P(A) × P(B). If A and B are independent, so are their complements, and P(neither) = [1 − P(A)] × [1 − P(B)].
• At least one: P(at least one) = 1 − P(none). This is almost always faster than adding cases.
• Conditional probability: P(A | B) = P(A ∩ B)/P(B), the chance of A once you know B has happened. In practice, shrink the sample space to B and count inside it.

Bayes' theorem

Suppose an outcome can arise from one of several causes E₁, E₂, …, which are mutually exclusive and exhaustive. If A is observed, the probability that it came from cause Eᵢ is

P(Eᵢ | A) = P(Eᵢ) P(A | Eᵢ) / [P(E₁) P(A | E₁) + P(E₂) P(A | E₂) + …]

In words: this cause's share of all the ways A can happen. A small table with one row per cause, and columns for P(cause), P(A given cause) and their product, keeps the work tidy.

Random variables and the binomial distribution

A random variable assigns a number to each outcome, such as the number of heads in three tosses.

A binomial situation has a fixed number n of independent trials, each with the same probability p of success (q = 1 − p). Then

• P(exactly r successes) = nCr pʳ qⁿ⁻ʳ;
• mean = np, variance = npq, and variance < mean always (since q < 1).

Tossing a fair coin n times is binomial with p = 1/2, so P(exactly r heads) = nCr/2ⁿ.

Worked NDA-style MCQs

Q1. Two dice are thrown. The probability that the sum is 8 is:
(a) 1/6 (b) 7/36 (c) 1/9 (d) 5/36

The favourable outcomes are (2, 6), (3, 5), (4, 4), (5, 3) and (6, 2): 5 of 36. Answer: (d).

Q2. One card is drawn from a pack of 52. The probability that it is a king or a heart is:
(a) 17/52 (b) 1/4 (c) 4/13 (d) 1/13

4 kings + 13 hearts − 1 king of hearts = 16. So 16/52 = 4/13. Option (a) forgets to subtract the overlap. Answer: (c).

Q3. A die is thrown twice and the sum is 6. The probability that one of the throws shows a 2 is:
(a) 2/5 (b) 1/3 (c) 1/5 (d) 1/6

Given sum 6, the sample space shrinks to (1, 5), (2, 4), (3, 3), (4, 2), (5, 1): 5 outcomes. Two of them contain a 2. Answer: (a).

Q4. Bag I has 4 red and 1 black ball; Bag II has 2 red and 3 black. A bag is chosen at random and a ball is drawn. It is red. The probability that it came from Bag I is:
(a) 4/5 (b) 1/2 (c) 2/3 (d) 3/5

P(I) × P(red | I) = 1/2 × 4/5 = 2/5, and P(II) × P(red | II) = 1/2 × 2/5 = 1/5. Bayes gives (2/5)/(2/5 + 1/5) = 2/3. Answer: (c).

Q5. A fair coin is tossed 6 times. The probability of at least 5 heads is:
(a) 3/32 (b) 1/64 (c) 6/64 (d) 7/64

P(5 heads) + P(6 heads) = (6C5 + 6C6)/2⁶ = (6 + 1)/64 = 7/64. Answer: (d).

Q6. A binomial distribution has mean 4 and variance 3. Its parameters n and p are:
(a) 16, 1/4 (b) 12, 1/3 (c) 8, 1/2 (d) 16, 3/4

npq/np = q = 3/4, so p = 1/4 and n = 4/(1/4) = 16. Answer: (a).

Common mistakes

• Counting (2, 6) and (6, 2) as one outcome with two dice. They are different outcomes of a 36-point sample space.
• Forgetting the overlap in "or" questions.
• Multiplying probabilities of dependent events without adjusting for the first draw (without replacement, the second draw is from 51 cards, not 52).
• Adding many cases for "at least one" instead of using 1 − P(none).
• Setting up Bayes upside down. The question gives P(red given Bag I) and asks for P(Bag I given red). The given probabilities go into the products; the one asked for is the ratio.

Practice set

1. Three coins are tossed. The probability of at least one head is: (a) 1/8 (b) 3/8 (c) 7/8 (d) 1/2
2. P(A) = 0.3, P(B) = 0.5 and P(A ∩ B) = 0.1. P(A ∪ B) is: (a) 0.8 (b) 0.6 (c) 0.9 (d) 0.7
3. Two cards are drawn without replacement from a pack of 52. The probability that both are aces is: (a) 1/221 (b) 1/169 (c) 1/13 (d) 4/663
4. The odds in favour of an event are 3 : 5. Its probability is: (a) 3/5 (b) 5/8 (c) 2/5 (d) 3/8
5. A and B are independent, with P(A) = 1/2 and P(B) = 1/3. The probability that neither happens is: (a) 1/6 (b) 1/3 (c) 2/3 (d) 5/6
6. For a binomial distribution with n = 10 and p = 0.4, the variance is: (a) 2.4 (b) 4 (c) 6 (d) 1.6
7. P(A ∩ B) = 0.2 and P(B) = 0.5. Then P(A | B) is: (a) 0.1 (b) 0.25 (c) 0.4 (d) 0.7
8. A leap year is chosen at random. The probability that it has 53 Sundays is: (a) 1/7 (b) 2/7 (c) 3/7 (d) 53/366

Answers:

1. (c). 1 − P(no head) = 1 − 1/8.
2. (d). 0.3 + 0.5 − 0.1.
3. (a). (4/52) × (3/51) = 12/2,652 = 1/221.
4. (d). 3/(3 + 5).
5. (b). (1/2) × (2/3) = 1/3.
6. (a). npq = 10 × 0.4 × 0.6.
7. (c). 0.2/0.5.
8. (b). A leap year has 52 full weeks plus 2 extra days. Of the 7 possible pairs (Sun–Mon, Mon–Tue, … Sat–Sun), 2 contain a Sunday.

What to do next

• Write the 36-outcome table for two dice once, and read off the number of ways to make each sum (2 to 12: 1, 2, 3, 4, 5, 6, 5, 4, 3, 2, 1).
• Solve five Bayes questions using the one-row-per-cause table until the set-up is automatic.
• Revise nCr from permutations and combinations, and see statistics for the mean and variance ideas behind the binomial results.