ECON 214 - Probability.pdf

# Slide 24 multiplication rule there is a slight

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Slide 24

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Multiplication rule There is a slight complication when events are dependent. For instance, P(drawing two Aces from a pack of cards, without replacement)... If the first card drawn is an Ace ( P = 4 / 52 ), that leaves 51 cards, of which 3 are Aces. The probability of drawing the second Ace is 3 / 51 , different from the probability of drawing the first Ace. They are not independent events ; the probability changes. Thus P(two Aces) = 4 / 52 3 / 51 = 1 / 221 Slide 25
Conditional probability 3 / 51 is the probability of drawing a second Ace given that an Ace was drawn as the first card. This is the conditional probability and is written P (Second Ace | Ace on first draw) That is, the probability of a second Ace given that an Ace was drawn first. In general, it is written as P(B|A) That is, the probability of event B occurring, given that A has occurred. Slide 26

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Conditional probability Consider P (A2| not-A1)... That is, ‘not-Ace’ is drawn first, leaving 51 cards of which 4 are Aces. Here, we assumed that the first card drawn is not an Ace. Hence P (A2| not-A1) = 4 / 51 So P (A2| not-A1) P (A2|A1) They are not independent events. Slide 27
Conditional probability The general rule for multiplication is: P ( A and B ) = P ( A ) P ( B | A ) So that P(B|A) = P(A and B)/P(A) For independent events P ( B | A ) = P ( B | not- A ) = P ( B ) And so P ( A and B ) = P ( A ) P ( B ) Slide 28

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Example 4 The Dean of UGBS presented the following information on undergraduate students during a School Board meeting. (1) If a student is selected at random, what is the probability that the student is a female accounting major? (2) Given that the student is a female, what is the probability that she is an accounting major? Slide 29 M AJO R M ale Fem ale T otal A ccounting 170 110 280 Finance 120 100 220 M arketing 160 70 230 M anagem ent 150 120 270 T otal 600 400 1000
Combining the rules Consider that we wish to find the probability of one head in two tosses of a coin. We can solve this by combining the addition and multiplication rules P (1 Head in two tosses)... ...= P ( [H and T] or [T and H] ) = P ( [H and T] ) + P ( [T and H] ) = [1/2 1/2] + [1/2 1/2] = 1/4 + 1/4 = 1/2 Slide 30

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Tree diagram We can also solve the problem using a tree diagram Slide 31 H H H T T T {H,H} {H,T} {T,H} {T,T} P = 1 / 4 P = 1 / 4 P = 1 / 4 P = 1 / 4 ½ ½ ½ ½ ½ ½ P = ½
Gets complicated with larger number of outcomes What about calculating P (3 Heads in 5 tosses)? P (30 Heads in 50 tosses)? How many routes through the tree diagram? Drawing takes too much time, so we need a formula... In other words, if the number of possible outcomes in an experiment is large, it is difficult or cumbersome to list the total number of outcomes in either the event set or sample space. There are techniques for determining the number of outcomes. Slide 32

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