Lesson 10 of 11 · 8 min
Probability by counting
NCERT §14.2.4, Examples 8 to 12
Two volunteers are to be picked at random from the Class XI stall team of 5 boys and 3 girls to run the lucky draw. The girls want to know how likely it is that at least one of them is chosen. Listing all the pairs by hand would take a while.
The lesson in notes
In short
When outcomes are equally likely, P(E) = n(E)/n(S), and permutations and combinations do the counting. Use ⁿCᵣ when order does not matter (committees, hands of cards) and ⁿPᵣ when it does (finishing orders).
NCERT's Example 8: a committee of two from two men and two women can be formed in ⁴C₂ = 6 ways. P(no man) = 1/6, P(one man) = (2 × 2)/6 = 2/3, P(two men) = 1/6.
NCERT's Example 9: Veena visits four cities A, B, C, D in random order, so n(S) = 4! = 24. A comes before B in 12 of the orders, P = 1/2; A before B before C in 4 orders, P = 4/24 = 1/6.
NCERT's Example 10: a hand of 7 from 52 cards, n(S) = ⁵²C₇. All four kings: ⁴C₄ × ⁴⁸C₃ ways, P = 1/7735. Exactly 3 kings: ⁴C₃ × ⁴⁸C₄ ways, P = 9/1547. At least 3 kings: 9/1547 + 1/7735 = 46/7735.
NCERT's Example 12: five relay teams. Ordered first three places: ⁵P₃ = 5 × 4 × 3 = 60 points. P(A, B, C finish first, second, third in that order) = 1/60; P(A, B and C are the first three in any order) = 3!/60 = 1/10.
Mela volunteers (made up): 2 are picked at random from 5 boys and 3 girls, ⁸C₂ = 28 ways. P(both girls) = ³C₂/28 = 3/28, P(one of each) = (5 × 3)/28 = 15/28, P(at least one girl) = 1 − ⁵C₂/28 = 1 − 10/28 = 9/14.
Mela lucky draw (made up): 3 prize tickets among 20, and a visitor takes 2. P(no prize) = ¹⁷C₂/²⁰C₂ = 136/190, so P(at least one prize) = 54/190 = 27/95.
History: the subject grew out of games of chance. Jerome Cardan (1501–1576) wrote a book on them, published in 1663; in 1654 Pascal and Fermat independently solved a dice problem put to Pascal; Kolmogorov's 1933 book set out the axiomatic theory.