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Discrete Probability

Discrete Probability

• Everything you have learned about counting constitutes the basis for computing the probability of events to happen.
• In the following, we will use the notion experiment for a procedure that yields one of a given set of possible outcomes.
• This set of possible outcomes is called the sample space of the experiment.
• An event is a subset of the sample space.

Discrete Probability

• Example I:
• An urn contains four blue balls and five red balls. What is the probability that a ball chosen from the urn is blue?
• Solution:
• There are nine possible outcomes, and the event “blue ball is chosen” comprises four of these outcomes. Therefore, the probability of this event is 4/9 or approximately 44.44%.

Discrete Probability

• Example II:
• What is the probability of winning the lottery 6/49, that is, picking the correct set of six numbers out of 49?
• Solution:
• There are C(49, 6) possible outcomes. Only one of these outcomes will actually make us win the lottery.
• p(E) = 1/C(49, 6) = 1/13,983,

Complimentary Events

• Example I: A sequence of 10 bits is randomly generated. What is the probability that at least one of these bits is zero?
• Solution: There are 2^10 = 1024 possible outcomes of generating such a sequence. The event –E, “none of the bits is zero” , includes only one of these outcomes, namely the sequence 1111111111.
• Therefore, p(-E) = 1/1024.
• Now p(E) can easily be computed as p(E) = 1 – p(-E) = 1 – 1/1024 = 1023/1024.

Complimentary Events

• Example II: What is the probability that at least two out of 36 people have the same birthday?
• Solution: The sample space S encompasses all possibilities for the birthdays of the 36 people, so |S| = 365^36.
• Let us consider the event –E (“no two people out of 36 have the same birthday”). –E includes P(365, 36) outcomes (365 possibilities for the first person’s birthday, 364 for the second, and so on).
• Then p(-E) = P(365, 36)/365 36 = 0.168, so p(E) = 0.832 or 83.2%

Discrete Probability

• Example: What is the probability of a positive integer selected at random from the set of positive integers not exceeding 100 to be divisible by 2 or 5?
• Solution:
• E 2 : “integer is divisible by 2” E 5 : “integer is divisible by 5”
• E 2 = {2, 4, 6, …, 100}
• |E 2 | = 50
• p(E 2 ) = 0.

Discrete Probability

• E 5 = {5, 10, 15, …, 100}
• |E 5 | = 20
• p(E 5 ) = 0.
• E 2 ∩ E 5 = {10, 20, 30, …, 100}
• |E 2 ∩ E 5 | = 10
• p(E 2 ∩ E 5 ) = 0.
• p(E 2 ∪ E 5 ) = p(E 2 ) + p(E 5 ) – p(E 2 ∩ E 5 )
• p(E 2 ∪ E 5 ) = 0.5 + 0.2 – 0.1 = 0.

Discrete Probability

• How can we obtain these probabilities p(s)?
• The probability p( s ) assigned to an outcome s equals the limit of the number of times s occurs divided by the number of times the experiment is performed.
• Once we know the probabilities p(s), we can compute the probability of an event E as follows:
• p(E) = ∑s∈E p(s)

Discrete Probability

• Example I: A die is biased so that the number 3 appears twice as often as each other number.
• What are the probabilities of all possible outcomes?
• Solution: There are 6 possible outcomes s 1 , …, s 6.
• p(s 1 ) = p(s 2 ) = p(s 4 ) = p(s 5 ) = p(s 6 )
• p(s 3 ) = 2p(s 1 )
• Since the probabilities must add up to 1, we have:
• 5p(s 1 ) + 2p(s 1 ) = 1
• 7p(s 1 ) = 1
• p(s 1 ) = p(s 2 ) = p(s 4 ) = p(s 5 ) = p(s 6 ) = 1/7, p(s 3 ) = 2/

Conditional Probability

• If we toss a coin three times, what is the probability that an odd number of tails appears (event E) , if the first toss is a tail (event F)?
• If the first toss is a tail, the possible sequences are TTT, TTH, THT, and THH.
• In two out of these four cases, there is an odd number of tails.
• Therefore, the probability of E, under the condition that F occurs, is 0.5.
• We call this conditional probability.

Conditional Probability

• If we want to compute the conditional probability of E given F, we use F as the sample space.
• For any outcome of E to occur under the condition that F also occurs, this outcome must also be in E ∩ F.
• Definition: Let E and F be events with p(F) > 0. The conditional probability of E given F, denoted by p(E | F), is defined as
• p(E | F) = p(E ∩ F)/p(F)

Independence

• Let us return to the example of tossing a coin three times.
• Does the probability of event E (odd number of tails) depend on the occurrence of event F (first toss is a tail)?
• In other words, is it the case that p(E | F) ≠ p(E)?
• We actually find that p(E | F) = 0.5 and p(E) = 0.5, so we say that E and F are independent events.

Independence

• Because we have p(E | F) = p(E ∩ F)/p(F), p(E | F) = p(E) if and only if p(E ∩ F) = p(E)p(F).
• Definition: The events E and F are said to be independent if and only if p(E ∩ F) = p(E)p(F).
• Obviously, this definition is symmetrical for E and F. If we have p(E ∩ F) = p(E)p(F), then it is also true that p(F | E) = p(F).