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Basic Probability Section 7.1. Definitions Sample Space: The set of all possible outcomes in a given experiment or situation Often denoted by S Event:

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Presentation on theme: "Basic Probability Section 7.1. Definitions Sample Space: The set of all possible outcomes in a given experiment or situation Often denoted by S Event:"— Presentation transcript:

1 Basic Probability Section 7.1

2 Definitions Sample Space: The set of all possible outcomes in a given experiment or situation Often denoted by S Event: A subset of outcomes from a sample space. Usually denoted by E Probability: The quotient: Usually denoted by p(E) or just p. Note: 0 ≤ p(E) ≤ 1

3 Examples 1 1) What is the probability of choosing a blue ball from a bag containing 4 blue and 5 black balls? |S| = 9, |E| = 4, so p(E) = 4/9 =.44444… 2) What is the probability of rolling a 7 with a pair of fair dice? |S| = 6 x 6 = 36, E = {(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)} p(E) = 6/36 = 1/6 =.16666666…

4 Examples 2 3) What is the probability of guessing a 4-digit PIN? |S| = 10 x 10 x 10 x 10 = 10000, |E| = 1 p(E) = 1/10000 =.0001 4) What is the probability of winning a Pick-Six Lottery chosen from 40 numbers? |S| = C(40,6) = 40!/(34!6!) = 3,838,380 p(E) = 1/|S| =.0000002605265…

5 Examples 3 5) What is the possibility of getting dealt a Royal Flush in 5- card poker? |S| = C(52,5) = 52!/(5!47!) = 2,598,960 |E| = 4 p(E) =.000001539…

6 Examples 4 6) What is the possibility of getting dealt a Full House? First consider getting 3 Queens and 2 Aces This is different than 3 Aces and 2 Queens (order matters!) How many ways can you get 3 Queens? C(4,3) = 4 How many ways can you get 2 Aces? C(4,2) = 6 p(3Q2A) = (4)(6)/2,598,360 How many ways can you choose two kinds, where order matters? P(13,2) = (13)(12) So p(E) = (13)(12)(4)(6)/2,598,960 =.0014…

7 Probabilities of Complements of Events A complement of an event E, denoted by or E', is the complement of the set E relative to S. Therefore, |E| + |E'| = |S| ⇒ p(|E|) = 1 – P(E'), and vice-versa 7) What is the probability that a sequence of 10 random bits contains a 0-bit? p(E) = 1 – p(E') = 1 – 2 -10 = 1023/1024

8 Probabilities of Unions Remember that |A ∪ B| = |A| + |B| - |A ∩ B| Therefore, if A and B are events in S, then p(A ∪ B) = |A|/|S| + |B|/|S| - |A ∩ B|/|S| = p(A) + p(B) – p(|A ∩ B|) 8) What is the probability of selecting a number from [1,100] that is divisible by 2 or 5? p(E) = 50/100 + 20/100 – 10/100 = 3/5 =.6


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