A brief note on Probability by Counting . Read it before a test. Try each question at the end before you open its answer.
Definition and rules Classical: P ( E ) = n ( E ) n ( S ) P(E) = \dfrac{n(E)}{n(S)} P ( E ) = n ( S ) n ( E ) for equally likely outcomes. Axioms: P ( E ) ≥ 0 P(E) \ge 0 P ( E ) ≥ 0 , P ( S ) = 1 P(S) = 1 P ( S ) = 1 , and P ( A ∪ B ) = P ( A ) + P ( B ) P(A \cup B) = P(A) + P(B) P ( A ∪ B ) = P ( A ) + P ( B ) for mutually exclusive events. Consequences: P ( ∅ ) = 0 P(\varnothing) = 0 P ( ∅ ) = 0 , 0 ≤ P ( E ) ≤ 1 0 \le P(E) \le 1 0 ≤ P ( E ) ≤ 1 , P ( E ′ ) = 1 − P ( E ) P(E') = 1 - P(E) P ( E ′ ) = 1 − P ( E ) . At least one: 1 − P ( none ) 1 - P(\text{none}) 1 − P ( none ) . Odds: odds of a : b a : b a : b in favour give P = a a + b P = \dfrac{a}{a + b} P = a + b a . The addition theorem Two events: P ( A ∪ B ) = P ( A ) + P ( B ) − P ( A ∩ B ) P(A \cup B) = P(A) + P(B) - P(A \cap B) P ( A ∪ B ) = P ( A ) + P ( B ) − P ( A ∩ B ) . Exactly one: P ( A ) + P ( B ) − 2 P ( A ∩ B ) P(A) + P(B) - 2P(A \cap B) P ( A ) + P ( B ) − 2 P ( A ∩ B ) . Neither: 1 − P ( A ∪ B ) 1 - P(A \cup B) 1 − P ( A ∪ B ) . Limits of the overlap: max { 0 , P ( A ) + P ( B ) − 1 } ≤ P ( A ∩ B ) ≤ min { P ( A ) , P ( B ) } \max\{0, P(A) + P(B) - 1\} \le P(A \cap B) \le \min\{P(A), P(B)\} max { 0 , P ( A ) + P ( B ) − 1 } ≤ P ( A ∩ B ) ≤ min { P ( A ) , P ( B )} . Three events: singles, minus pairs, plus triple. Same style: count the favourable and all outcomes both with order or both without order. Selections: n C r ^nC_r n C r . Arrangements: n P r ^nP_r n P r and n ! n! n ! . Group draw: K C k × N − K C r − k N C r \dfrac{^KC_k \times {}^{N-K}C_{r-k}}{^NC_r} N C r K C k × N − K C r − k . Together: glue into a block and multiply by the arrangements inside. Round table: ( n − 1 ) ! (n - 1)! ( n − 1 )! seatings, and two people are side by side with probability 2 n − 1 \dfrac{2}{n - 1} n − 1 2 . A worked example A card is drawn from a pack. Find the probability that it is a diamond or a queen.
The counts. 13 13 13 diamonds and 4 4 4 queens, and 1 1 1 card is both. The theorem. 13 52 + 4 52 − 1 52 = 16 52 \dfrac{13}{52} + \dfrac{4}{52} - \dfrac{1}{52} = \dfrac{16}{52} 52 13 + 52 4 − 52 1 = 52 16 . The answer. 4 13 \dfrac{4}{13} 13 4 . Practice with new numbers
A program to try The program checks the addition theorem and the complement rule for three throws of a die, with exact fractions.
Run it. Which pair of printed lines agree, and why?
1. Change o[1] == 6 to o[0] == 5. Do the first two lines still agree?
Mistakes to avoid Unequal outcomes. Counting works only for equally likely outcomes. Adding for at least one. Use the complement. Forgetting the overlap in the addition theorem. Mixing ordered and unordered counts in one fraction. Check yourself A number is chosen from 1 1 1 to 30 30 30 . Find the probability that it is a multiple of 4 4 4 . Stuck? Show a hint Show the answerThe probability of rain is 0.4 0.4 0.4 . Find the odds in favour of rain. Stuck? Show a hint Show the answerP ( A ) = 0.5 P(A) = 0.5 P ( A ) = 0.5 , P ( B ) = 0.3 P(B) = 0.3 P ( B ) = 0.3 and P ( A ∪ B ) = 0.65 P(A \cup B) = 0.65 P ( A ∪ B ) = 0.65 . Find P ( A ∩ B ) P(A \cap B) P ( A ∩ B ) . Stuck? Show a hint Show the answerA committee of 2 2 2 is chosen from 4 4 4 men and 3 3 3 women. Find the probability that it has one man and one woman. Stuck? Show a hint Show the answerFive people sit in a row. Find the probability that two particular people are side by side. Stuck? Show a hint Show the answerSources NCERT, Mathematics for Class XI (earlier edition), Probability. OpenStax, Introductory Statistics , Chapter 3, sections on terminology and on the two basic rules of probability.