Clear Notion
Login

Concepts

Read through concept notes with worked explanations and diagrams, chapter by chapter.

Concept 08 / 08Statistics and Probability

Probability with Counting and Geometric Probability

Using permutations and combinations to count equally likely outcomes, standard problems with cards, balls, arrangements and committees, selecting with and without replacement, and geometric probability as a ratio of lengths or areas for continuous uniform choices.

Figure 1
Most JEE probability questions are counting questions in disguise: P(A)=n(A)n(S)P(A) = \frac{n(A)}{n(S)}, with both counts done by permutations and combinations. The only real decision is how to model the sample space so that its outcomes are equally likely, and then to count the favourable outcomes in the same model. Consistency of the model. If the total is counted with combinations (unordered selections), count the favourable outcomes with combinations too; if with ordered arrangements, use arrangements for both. Either choice gives the same probability, but mixing them does not. For drawing 33 balls at once from 55 red and 44 black, P(2 red,1 black)=(52)(41)(93)=4084=1021P(2 \text{ red}, 1 \text{ black}) = \frac{\binom{5}{2}\binom{4}{1}}{\binom{9}{3}} = \frac{40}{84} = \frac{10}{21}. Drawing one at a time without replacement gives the same answer: 3×59⋅48⋅47=10213 \times \frac{5}{9} \cdot \frac{4}{8} \cdot \frac{4}{7} = \frac{10}{21}, the factor 33 counting the positions of the black ball. Standard settings. Cards: 5252 cards, 44 suits of 1313; 44 aces, 1212 face cards. P(a hand of 5 has all hearts)=(135)(525)P(\text{a hand of } 5 \text{ has all hearts}) = \frac{\binom{13}{5}}{\binom{52}{5}}. Arrangements: P(two particular people sit together in a row of n)=2(n−1)!n!=2nP(\text{two particular people sit together in a row of } n) = \frac{2(n - 1)!}{n!} = \frac{2}{n}. Birthdays: P(all n people have different birthdays)=365⋅364⋯(365−n+1)365nP(\text{all } n \text{ people have different birthdays}) = \frac{365 \cdot 364 \cdots (365 - n + 1)}{365^{n}}, which drops below 12\frac{1}{2} already at n=23n = 23. Numbers: choosing integers at random and asking about divisibility or last digits; use the cyclicity of last digits (for example, 3n3^{n} ends in 3,9,7,13, 9, 7, 1 cyclically). Distributions: placing rr distinct balls into nn boxes gives nrn^{r} equally likely outcomes; placing identical balls does not give equally likely stars-and-bars configurations, a subtle trap. Symmetry arguments often replace counting. In a random arrangement, each particular object is equally likely to be in each position, and each of two particular objects is equally likely to come first. For example, when cards are dealt one by one, the probability that the ace of spades comes before the king of spades is 12\frac{1}{2}, with no counting at all. Similarly, the probability that the kkth card drawn (without replacement) is red is 12\frac{1}{2} for every kk. Geometric probability. When a point is chosen 'at random' from a continuous region (uniformly), probabilities are ratios of lengths, areas or volumes: P(A)=measure of favourable regionmeasure of whole regionP(A) = \frac{\text{measure of favourable region}}{\text{measure of whole region}}. The classic meeting problem: two people arrive independently and uniformly between 00 and 6060 minutes, and each waits 1515 minutes. Plotting their arrival times (x,y)(x, y) in the square, they meet iff ∣x−y∣≤15|x - y| \le 15, the band around the diagonal in the figure. The two corner triangles where they miss have total area 45245^{2}, so P(meet)=1−452602=1−916=716P(\text{meet}) = 1 - \frac{45^{2}}{60^{2}} = 1 - \frac{9}{16} = \frac{7}{16}. Other geometric examples: a point chosen in a square lies inside the inscribed circle with probability π4\frac{\pi}{4}; a stick broken at two random points forms a triangle with probability 14\frac{1}{4}; choosing x∈[0,1]x \in [0, 1] at random, P(x2<14)=12P(x^{2} < \frac{1}{4}) = \frac{1}{2}. Key Takeaways: - Model the sample space with equally likely outcomes; count favourable outcomes in the same model. - Combinations for simultaneous draws; ordered products for sequential draws; both agree. - Symmetry: each position or order is equally likely in a random arrangement. - Two particular people adjacent in a random row of nn: 2n\frac{2}{n}. - Geometric probability == favourable measure // total measure. - Meeting problem with waiting time ww in an interval TT: 1−(1−wT)21 - (1 - \frac{w}{T})^{2}. - Pitfall: treating arrangements of identical objects (stars and bars) as equally likely outcomes. Brain-Teaser 1: Five cards are drawn from a well-shuffled deck. Find the probability of exactly one pair (two cards of one rank, and three cards of three other distinct ranks). Solution: Choose the rank of the pair (1313) and its two suits ((42)=6\binom{4}{2} = 6); choose three other ranks ((123)=220\binom{12}{3} = 220) and a suit for each (43=644^{3} = 64). Favourable: 13×6×220×64=109824013 \times 6 \times 220 \times 64 = 1098240. Total: (525)=2598960\binom{52}{5} = 2598960. Probability ≈0.4226\approx 0.4226. Choosing the three single ranks as a combination (not in order) avoids overcounting. Brain-Teaser 2: Three numbers are chosen at random without replacement from 1,2,…,201, 2, \ldots, 20. Find the probability that their sum is even. Solution: There are 1010 odd and 1010 even numbers. The sum is even with 00 or 22 odd numbers: (103)+(102)(101)=120+450=570\binom{10}{3} + \binom{10}{2}\binom{10}{1} = 120 + 450 = 570. Total (203)=1140\binom{20}{3} = 1140. Probability =12= \frac{1}{2}. The exact half is no coincidence: pairing each selection with its 'complement-parity' partner is possible here because odd and even numbers are equally many. Brain-Teaser 3: A point is chosen at random inside a circle of radius RR. What is the probability that it is closer to the centre than to the circumference? Solution: A point at distance rr from the centre is at distance R−rR - r from the circle, so the condition is r<R−rr < R - r, i.e. r<R2r < \frac{R}{2}. The favourable region is the concentric disc of radius R2\frac{R}{2}, so the probability is πR2/4πR2=14\frac{\pi R^{2}/4}{\pi R^{2}} = \frac{1}{4}. Translating the condition into a region and comparing areas is the entire method; the intuitive but wrong answer 12\frac{1}{2} comes from comparing radii instead of areas.

Practice this concept

Free · no account required · instant solutions

DPP: Probability with Counting and Geometric Probability

15 questionsSee the solution the moment you answer

Practice