Rules of Probability

The additive rule P[A  B] = P[A] + P[B] – P[A  B] and if P[A  B] =  P[A  B] = P[A] + P[B]

The additive rule for more than two events then and if A i  A j =  for all i ≠ j.

The Rule for complements for any event E

Conditional Probability, Independence and The Multiplicative Rue

Then the conditional probability of A given B is defined to be:

The multiplicative rule of probability and if A and B are independent. This is the definition of independent

The multiplicative rule for more than two events

and continuing we obtain Proof

Example What is the probability that a poker hand is a royal flush i.e. 1.10, J, Q, K,A 2.10, J, Q, K,A 3.10, J, Q, K,A 4.10, J, Q, K,A

Solution Let A 1 = the event that the first card is a “royal flush” card. 1.10, J, Q, K,A 2.10, J, Q, K,A 3.10, J, Q, K,A 4.10, J, Q, K,A Let A i = the event that the i th card is a “royal flush” card. i = 2, 3, 4, 5.

Another solution is by counting The same result

Independence for more than 2 events

Definition: The set of k events A 1, A 2, …, A k are called mutually independent if: P[A i 1 ∩ A i 2 ∩… ∩ A i m ] = P[A i 1 ] P[A i 2 ] …P[A i m ] For every subset {i 1, i 2, …, i m } of {1, 2, …, k } i.e. for k = 3 A 1, A 2, …, A k are mutually independent if: P[A 1 ∩ A 2 ] = P[A 1 ] P[A 2 ], P[A 1 ∩ A 3 ] = P[A 1 ] P[A 3 ], P[A 2 ∩ A 3 ] = P[A 2 ] P[A 3 ], P[A 1 ∩ A 2 ∩ A 3 ] = P[A 1 ] P[A 2 ] P[A 3 ]

0.12 P[A 1 ] =.4, P[A 2 ] =.5, P[A 3 ] =.6 P[A 1 ∩A 2 ] = (0.4)(0.5) = 0.20 P[A 1 ∩ A 3 ] = (0.4)(0.6) = 0.24 P[A 2 ∩ A 3 ] = (0.5)(0.6) = 0.30 P[A 1 ∩ A 2 ∩ A 3 ] = (0.4)(0.5)(0.6) = 0.12 A1A1 A2A2 A3A

Definition: The set of k events A 1, A 2, …, A k are called pairwise independent if: P[A i ∩ A j ] = P[A i ] P[A j ] for all i and j. i.e. for k = 3 A 1, A 2, …, A k are pairwise independent if: P[A 1 ∩ A 2 ] = P[A 1 ] P[A 2 ], P[A 1 ∩ A 3 ] = P[A 1 ] P[A 3 ], P[A 2 ∩ A 3 ] = P[A 2 ] P[A 3 ], It is not necessarily true that P[A 1 ∩ A 2 ∩ A 3 ] = P[A 1 ] P[A 2 ] P[A 3 ]

0.14 P[A 1 ] =.4, P[A 2 ] =.5, P[A 3 ] =.6 P[A 1 ∩A 2 ] = (0.4)(0.5) = 0.20 P[A 1 ∩ A 3 ] = (0.4)(0.6) = 0.24 P[A 2 ∩ A 3 ] = (0.5)(0.6) = 0.30 P[A 1 ∩ A 2 ∩ A 3 ] = 0.14 ≠ (0.4)(0.5)(0.6) = 0.12 A1A1 A2A2 A3A

Bayes Rule Due to the reverend T. Bayes Picture found on website: Portraits of Statisticians maths/histstat/people/welco me.htm#h

Proof:

Example: We have two urns. Urn 1 contains 14 red balls and 12 black balls. Urn 2 contains 6 red balls and 20 black balls. An Urn is selected at random and a ball is selected from that urn. If the ball turns out to be red what is the probability that it came from the first urn? Urn 1 Urn 2

Solution: Note: the desired conditional probability is in the reverse direction of the given conditional probabilities. This is the case when Bayes rule should be used Let A = the event that we select urn 1 = the event that we select urn 2 Let B = the event that we select a red ball

Bayes rule states

Example: Testing for a disease Suppose that 0.1% of the population have a certain genetic disease. A test is available the detect the disease. If a person has the disease, the test concludes that he has the disease 96% of the time. It the person doesn’t have the disease the test states that he has the disease 2% of the time. Two properties of a medical test Sensitivity = P[ test is positive | disease] = 0.96 Specificity = P[ test is negative | disease] = 1 – 0.02 = 0.98 A person takes the test and the test is positive, what is the probability that he (or she) has the disease?

Solution: Note: Again the desired conditional probability is in the reverse direction of the given conditional probabilities. Let A = the event that the person has the disease = the event that the person doesn’t have the disease Let B = the event that the test is positive.

Bayes rule states Thus if the test turns out to be positive the chance of having the disease is still small (4.58%). Compare this to (.1%), the chance of having the disease without the positive test result.

Let A 1, A 2, …, A k denote a set of events such that An generlization of Bayes Rule for all i and j. Then

If A 1, A 2, …, A k denote a set of events such that for all i and j. Then A 1, A 2, …, A k is called a partition of S. A1A1 AkAk A2A2 … S

Proof for all i and j. A1A1 AkAk A2A2 B Then

and

Example: We have three urns. Urn 1 contains 14 red balls and 12 black balls. Urn 2 contains 6 red balls and 20 black balls. Urn 3 contains 3 red balls and 23 black balls. An Urn is selected at random and a ball is selected from that urn. If the ball turns out to be red what is the probability that it came from the first urn? second urn? third Urn? Urn 1 Urn 2 Urn 3

Solution: Note: the desired conditional probability is in the reverse direction of the given conditional probabilities. This is the case when Bayes rule should be used Let A i = the event that we select urn i Let B = the event that we select a red ball

Bayes rule states

Example: Suppose that an electronic device is manufactured by a company. During a period of a week –15% of this product is manufactured on Monday, –23% on Tuesday, –26% on Wednesday, –24% on Thursday and –12% on Friday.

Also during a period of a week –5% of the product is manufactured on Monday is defective –3 % of the product is manufactured on Tuesday is defective, –1 % of the product is manufactured on Wednesday is defective, –2 % of the product is manufactured on Thursday is defective and –6 % of the product is manufactured on Friday is defective. If the electronic device manufactured by this plant turns out to be defective, what is the probability that is as manufactured on Monday, Tuesday, Wednesday, Thursday or Friday?

Solution: Let A 1 = the event that the product is manufactured on Monday A 2 = the event that the product is manufactured on Tuesday A 3 = the event that the product is manufactured on Wednesday A 4 = the event that the product is manufactured on Thursday A 5 = the event that the product is manufactured on Friday Let B = the event that the product is defective

Now P[A 1 ] = 0.15, P[A 2 ] = 0.23, P[A 3 ] = 0.26, P[A 4 ] = 0.24 and P[A 5 ] = 0.12 Also P[B|A 1 ] = 0.05, P[B|A 2 ] = 0.03, P[B|A 3 ] = 0.01, P[B|A 4 ] = 0.02 and P[B|A 5 ] = 0.06 We want to find P[A 1 |B], P[A 2 |B], P[A 3 |B], P[A 4 |B] and P[A 5 |B]. We will apply Bayes Rule

iP[Ai]P[Ai]P[B|Ai]P[B|Ai]P[Ai]P[B|Ai]P[Ai]P[B|Ai]P[Ai|B]P[Ai|B] Total

The sure thing principle and Simpson’s paradox

The sure thing principle Suppose Example – to illustrate Let A = the event that horse A wins the race. B = the event that horse B wins the race. C = the event that the track is dry = the event that the track is muddy

Proof:

Simpson’s Paradox Does Example to illustrate D = death due to lung cancer S = smoker C = lives in city, = lives in country

similarly Solution

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