OVERVIEW: It is often important andnecessary to provide a mathematical description or model forrandomness. This section provides insights into how this can bedone.
Sample space: Theset of all possible outcomes of a random phenomenon.
Example:
Consider the sum obtained when two dice are rolled.
The following table displays the sums that can be obtained from thisevent.
RED DIE
GREEN
DIE
1 2 3 4 5 6 1
2 3 4 5 6 7 2
3 4 5 6 7 8 3
4 5 6 7 8 9 4
5 6 7 8 9 10
5
6 7 8 9 10 11 6
7 8 9 10 11 12 The sample space contains 11 outcomes: S ={2,3,4,5,6,7,8,9,10,11,12}
Probability rules:
o If A is an event, then the probability P(A) is anumber between 0 and 1, inclusive.
o Prob(A does not happen) is 1 - P(A).
o Two events are
-In a roll of two dice, if A = "roll asum of 7" and B = "roll a sum of 2", then P(A or B) = 6/36 + 1/36 =7/36, since A and B are disjoint.
-If C = "roll a sum of 7" and D = "roll an oddsum" then C and D are not disjoint since 7 is an odd sum. In thiscase, P(C or D) = P(C) + P(D) - P(C and D) = 6/36 + 18/36 - 6/36 =18/36 = 1/2.
Two events are
-If A is the event "getting a head onthe first coin toss" and B is the event "getting a head on a secondtoss of the coin." then A and B are independent, and prob(A and B) =(1/2)(1/2) = 1/4.
-If C is the event "getting a red card by pickinga card from a randomly shuffled deck," and D is the event "getting ared card by picking a second card from the deck" (when the first cardhas not been replaced), then C and D are not independent. Prob(C) =26/52 = 1/2. If the first card is red, then prob(C and D) =(26/52)(25//51), which is not equal to (26/52)(26/52). If the firstcard was replaced in the deck, and the deck reshuffled before eventD, then C and D would be independent.
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