PROBABILITY SEBASTIAN VATTAMATTAM 1. Basic Definitions • Random Experiment • Sample Space • Sample Points • Equally Likely Outcomes • Event Probability of an Event If S is the Sample Space and E is an Event, P(E) = n(E) n(S) Example 1.1. Find the probability of getting the sum 10 in a single throw of two dice. Solution: S = {(1, 1), (1, 2), (1, 3), ...(6, 5), (6, 6)} Number of Sample Points, n(S) = 6 × 6 = 36 If A is the required event, A = {(4, 6), (5, 5), (6, 4)} Therefore P(E) = 3 36 = 1 12 Example 1.2. Find the probability that a leap year, selected at random, will contain 53 Sundays. Solution: A leap year has 366 days. 366 = 7 × 52 + 2 12 SEBASTIAN VATTAMATTAM So, any leap year has at least 52 Sundays. We have to find the probability that the additional 2 days contains a Sunday. The possibilities for those two days are (1) Sunday, Monday (2) Monday, Tuesday (3) Tuesday, Wednesday (4) Wednesday, Thursday (5) Thursday, Friday (6) Friday, Saturday (7) Saturday, Sunday These 7 outcomes are equally likely. We will get 53 Sundays if the first or the last occurs. Therefore the required probability is 27 Example 1.3. Three coins are tossed. Find the probability of getting atleast two heads. Solution: S = {HHH,HHT,HTH, THH,HTT, THT, TTH, TTT} Number of Sample Points, n(S) = 23 = 8 If A is the required event, A = {HHH,HHT,HTH, THH} Therefore P(E) = 48 = 12 Example 1.4. In a lottery of 50 tickets, numbered 1 to 50, two are drawn simultaneously. Find the probability that (1) both are prime numbers (2) none them is a prime number. Solution: Number of sample points is 50C2 = 50.49 1.2 = 1225PROBABILITY 3 Prime numbers from 1 to 50 are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47 (1) Number of favorable outcomes is 15C2 = 105 Required probability is 105 1225 = 21 245 (2) Number of favorable outcomes is 35C2 = 595 Required probability is 595 1225 = 119 245 Example 1.5. An urn contains 8 red, 3 white, and 9 blue balls. If three balls are drawn at random, find the probability that (1) all three are red (2) all three are blue (3) one of each colour Solution: Total number of balls is 8 + 3 + 9 = 20 Number of sample points is20C3 (1) Number of favorable outcomes is 8C3 Required probability is 8C3 20C3 = 14 2854 SEBASTIAN VATTAMATTAM (2) Number of favorable outcomes is 9C3 Required probability is 9C3 20C3 = 7 95 (3) Number of favorable outcomes is 8.3.9 Required probability is 8.3.9 20C3 = 18 95 Example 1.6. Four-digit numbers are formed using the digits 1, 2, 3, 4, 6. A number is chosen at random. Find the probability that it is an odd number. Solution: Total number of four-digit numbers is 5 × 4 × 3 × 2 = 120 Total number of four-digit odd numbers is 4 × 3 × 2 × 2 = 48 Required probability is48 120 = 0.4