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**Conditional probability**

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Ordinary Probability You are dealt two cards from a deck. What is the probability the second card dealt is a Jack? We reason that if the two cards have been delt, the probabiity that the first is a jacek and the probabiity that the second is a jack are identical. Since there are 4 jacks in the deck, we compute P(second is jack) = 4/52 = 1/13.

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**Conditional Probability**

You are dealt two cards from a deck. What is the probability of the second card was a jack given the first card was not a jack is called card dealt is a Jack?

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**Conditional Probability**

You are dealt two cards from a deck. What is the probability that the second card was a jack given the first card was not a jack.

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**Conditional Probability**

You are dealt two cards from a deck. What is the probability of the second card was a jack given the first card was not a jack is called card dealt is a Jack?

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Reasoning Once the first card has been delt, there are 51 cards remaining, and, since the first was NOT a jack, there are 4 jacks in the set of 51 cards.

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**Conditional Probability**

The probability of drawing jack given the first card was not a jack is called conditional probability. A key words to look for is “given that.” We will use the notation: P(second a jack | first not a jack) = 4/51

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**General Conditional Probability**

The probability that the event A occurs, given that B occurs is denoted: This is read the probability of A given B.

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**Conditional Probability**

How would we draw the event A given B? A B A and B

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**Conditional Probability**

How would we draw the event A given B? Since we know B has occurred, we ignore everything else. A B A and B

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**Conditional Probability**

How would we draw the event A given B? Since we know B has occurred, we ignore everything else. A B A and B

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**Conditional Probability**

How would we draw the event A given B? Since we know B has occurred, we ignore everything else. With some thought this tells us: B A and B

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Additional notes In the case of a equi-probability space, we can reason that, since we know the outcome is in B, we can use the set B as our reduced sample space. The probability P(A|B) can then be computed as the number of points in A∩B as a fraction of the number of points in B.

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**Dividing top and bottom by the numbers of **

Continuing… This gives 𝑃 𝐴 𝐵 = 𝑛(𝐴∩𝐵) 𝑛(𝐵) Dividing top and bottom by the numbers of points in the original sample space S: 𝑃 𝐴|𝐵 = 𝑛(𝐴∩𝐵)/𝑛(𝑆) 𝑛(𝐵)/𝑛(𝑆) = 𝑃(𝐴∩𝐵) 𝑃(𝐵)

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**Example from well contamination**

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**Of private wells, 21% are contaminated. Therefore**

P(C|Private)=0.21 Of public wells, 40% are contaminated. Therefore P(C|Public)=0.40

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**Law of conditional probability**

𝑃 𝐴 𝐵 = 𝑃(𝐴∩𝐵) 𝑃(𝐵)

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Multiplication Rule 𝑃 𝐴 =𝑃 𝐵 𝑃(𝐴|𝐵)

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Example A local union has 8 members, 2 of whom are women. Two are chosen by a lottery to represent the union.

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Example A local union has 8 members, 2 of whom are women. Two are chosen by a lottery to represent the union. a) What is the probability they are both male?

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Example A local union has 8 members, 2 of whom are women. Two are chosen by a lottery to represent the union. a) What is the probability they are both male?

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Example A local union has 8 members, 2 of whom are women. Two are chosen by a lottery to represent the union. a) What is the probability they are both male?

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Example A local union has 8 members, 2 of whom are women. Two are chosen by a lottery to represent the union. b) What is the probability they are both female?

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Example A local union has 8 members, 2 of whom are women. Two are chosen by a lottery to represent the union. b) What is the probability they are both female?

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Example Find the probability of selecting an all male jury from a group of 30 jurors, 21 of whom are men.

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Example Find the probability of selecting an all male jury from a group of 30 jurors, 21 of whom are men. Solution: P(12 M) =P(M)*P(M|M)*P(M|MM) * …. = 21/30 * 20/29 * 19/28 * 18/27 * … 10/19 =

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Independent Events Two events A and B are independent if the occurrence of one does not affect the probability of the other.

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Independent Events Two events A and B are independent if the occurrence of one does not affect the probability of the other. Two events A and B are independent then P(A|B) = P(A).

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Independent Events Two events A and B are independent if the occurrence of one does not affect the probability of the other. Two events A and B are independent then P(A|B) = P(A). Two events which are not independent are dependent.

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**Multiplication Rule Multiplication Rule: For any pair of events:**

P(A and B) = P(A) * P(B|A)

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**Multiplication Rule Multiplication Rule: For any pair of events:**

P(A and B) = P(A) * P(B|A) For any pair of independent events: P(A and B) = P(A) * P(B)

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**Multiplication Rule For any pair of events: P(A and B) = P(A) * P(B|A)**

For any pair of independent events: P(A and B) = P(A) * P(B) If P(A and B) = P(A) * P(B), then A and B are independent.

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**Multiplication Rule Multiplication Rule:**

P(A and B) = P(A) * P(B) if A and B are independent. P(A and B) = P(B) * P(A|B) if A and B are dependent. Note: The multiplication rule extends to several events: P(A and B and C) =P(C)*P(B|C)*P(A|BC)

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**Example A study of 24 mice has classified the mice by two categories**

Fur Colour Black White Grey Eye Colour Red Eyes 3 5 2 Black Eyes 1 7 6

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**A study of 24 mice has classified the mice by two categories**

a) What is the probability that a randomly selected mouse has white fur? b) What is the probability it has black eyes given that it has black fur? c) Find pairs of mutually exclusive and independent events. Fur Colour Black White Grey Eye Colour Red Eyes 3 5 2 Black Eyes 1 7 6

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**A study of 24 mice has classified the mice by two categories**

a) What is the probability that a randomly selected mouse has white fur? 12/24=0.5 b) What is the probability it has black eyes given that it has black fur? c) Find pairs of mutually exclusive and independent events. Fur Colour Black White Grey Eye Colour Red Eyes 3 5 2 Black Eyes 1 7 6

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**A study of 24 mice has classified the mice by two categories**

a) What is the probability that a randomly selected mouse has white fur? 12/24=0.5 b) What is the probability it has black eyes given that it has black fur? 1/4=0.25 c) Find pairs of independent events. Fur Colour Black White Grey Eye Colour Red Eyes 3 5 2 Black Eyes 1 7 6

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**c) Find pairs of mutually exclusive and independent events.**

b) What is the probability it has black eyes given that it has black fur? 1/4=0.25 c) Find pairs of mutually exclusive and independent events. IND: White Fur and Red Eyes; Black Fur and Red Eyes Fur Colour Black White Grey Eye Colour Red Eyes 3 5 2 Black Eyes 1 7 6

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**Descriptive Phrases Descriptive Phrases require special care! At most**

At least No more than No less than

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**Review Conditional Probabilities Independent events**

Multiplication Rule Tree Diagrams

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Homework Answers 9) 6/24 + 6/24 = 12/24 or ½ 11) 12/24 + 12/24 = 24/24 or 1 23) P(2 and A) = (1/6 * 1/5) = 1/30 P(2 and B) = (1/6 * 1/5) = 1/30 P(2 and.

Homework Answers 9) 6/24 + 6/24 = 12/24 or ½ 11) 12/24 + 12/24 = 24/24 or 1 23) P(2 and A) = (1/6 * 1/5) = 1/30 P(2 and B) = (1/6 * 1/5) = 1/30 P(2 and.

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