Introduction to Probability By Dr. Carol A. Marinas

Sample Space Definition: a SET of all possible outcomes. Example #1: Flipping a coin Sample Space or S = {heads, tails} Example #2: Rolling a die Sample Space or S = { 1, 2, 3, 4, 5, 6}

Probability The probability of an event A is: P(A) = Number of elements of A = n(A) Number of elements of S n(S) Example: Rolling of a die P(getting a 6) = n(sixes) = 1 n(Sample Space) 6

“OR” When events A and B are mutually exclusive, P(A or B) = P(A U B) = P(A) + P(B) [Mutually exclusive means A ∩ B = Ø.] Example: P(rolling a 4 or a 6) Since rolling a 4 or rolling a 6 are mutually exclusive events, P(4 or 6) = P(4) + P(6) = 1/6 + 1/6 = 1/3

“OR” When events A and B are not mutually exclusive, then P(A U B) = P(A) + P(B) – P(A ∩ B) P( roll is even or roll is multiple of 3)  Roll is even: 2, 4, 6  Roll is multiple of 3: 3, 6 P(roll is even) + P(roll is mult 3) – P (even & mult of 3)= = 3/6 + 2/ 6 – 1/6 = 4/6 or 2/3

Other Probabilities _ _ P(A) + P(A) = 1 so P(A) = 1 – P(A)  P(roll is 5) + P(roll is NOT a 5) = 1 P(Ø) = 0 (impossible event)  P(roll is greater than 12) = 0 P(S) = 1 (certain event)  P(roll is less than 7) = 1 0 < P(A) < 1

Rolling 2 die See the Sample Space on Page 382 The die are white and orange so there are 36 outcomes in the sample space. Example: P(rolling a sum of 7) Sums of 7: (1,6), (6,1), (2,5), (5, 2), (3, 4), (4, 3) P(rolling a sum of 7) = 6/36 or 1/6 Where are the sums of 7 found in the chart?

More Dice Problems P(sum is 8) = P(sum is 2 or sum is 11) = P(sum is 4 or sum is greater than 10) = P(sum is multiple of 3 or sum is greater than 8) =

Other Dice Problems P(sum is 8) = 5/36 P(sum is 2 or sum is 11) = Sum is 2: (1, 1) Sum is 11: (5, 6), (6,5) Mutually exclusive so P(sum is 2 or sum is 11) = P(sum is 2) + P(sum is 11) = 1/36 + 2/36 = 3/36 = 1/12

Other Dice Problems P(sum is 4 or sum is greater than 10) = Sum is 4: (1,3), (3, 1), (2, 2) Sum > 10: (5, 6), (6, 5), ( 6, 6) Mutually exclusive so P(sum is 4 or sum is greater than 10) = P(sum is 4) + P(sum is greater than 10) = 3/36 + 3/36 = 6/36 = 1/6

More Dice Problems P(sum is multiple of 3 or sum > 8) = Sum is multiple of 3: (1, 2), (2, 1), (5, 1), (1, 5), (2, 4), (4, 2), (3, 3), (6, 3), (3, 6), (5, 4), (4, 5), (6, 6) Sum is greater than 8: (6, 3), (3, 6), (5, 4), (4, 5), (6, 4), (4, 6), (5, 5), ( 6, 5), (5, 6), (6,6) P(sum is multiple of 3 or sum > 8) = P(sum is mult 3) + P(sum > 8) – P(sum is mult 3 & > 8) = 12/ /36 – 5/36 = 17/36

Hope you enjoyed Probability