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## PowerPoint Slideshow about ' MATH104 Ch. 11: Probability Theory part 3' - angelo

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- Assignment by intuition – based on intuition, experience, or judgment.
- Assignment by relative frequency –
- P(A) = Relative Frequency =
- Assignment for equally likely outcomes

One Die

- Experimental Probability (Relative Frequency)
- If the class rolled one die 300 times and it came up a “4” 50 times, we’d say P(4)= 50/300
- The Law of Large numbers would say that our experimental results would approximate our theoretical answer.
- Theoretical Probability
- Sample Space (outcomes): 1, 2, 3, 4, 5, 6
- P(4) = 1/6
- P(even) = 3/6

Two Dice

- Experimental Probability
- “Team A” problem on the experiment: If we rolled a sum of “6, 7, 8, or 9” 122 times out of 218 attempts, P(6,7,8, or 9)= 122/218= 56%
- Questions: What sums are possible?
- Were all sums equally likely?
- Which sums were most likely and why?
- Use this to develop a theoretical probability
- List some ways you could get a sum of 6…

Outcomes

- For example, to get a sum of 6, you could get:
- 5, 1 4,2 3,3 …

Two Dice – Theoretical Probability

- Each die has 6 sides.
- How many outcomes are there for 2 sides? (Example: “1, 1”)
- Should we count “4,2” and “2,4” separately?

Sample Space for 2 Dice

1, 1 1, 2 1, 3 1, 4 1,5 1,6

2,1 2,2 2,3 2,4 2,5 2,6

3,1 3,2 3,3 3,4 3,5 3,6

4,1 4,2 4,3 4,4 4,5 4,6

5,1 5,2 5,3 5,4 5,5 5,6

6,1 6,2 6,3 6,4 6,5 6,6

If Team A= 6, 7, 8, 9, find P(Team A)

Two Dice- Team A/B

- P(Team A)= 20/36
- P(Team B) = 1 – 20/36 = 16/36
- Notice that P(Team A)+P(Team B) = 1

Some Probability Rules and Facts

- 0<= P(A) <= 1
- Think of some examples where
- P(A)=0 P(A) = 1
- The sum of all possible probabilities for an experiment is 1. Ex: P(Team A)+P(Team B) =1

One Coin

- Experimental
- If you tossed one coin 1000 times, and 505 times came up heads, you’d say P(H)= 505/1000
- The Law of Large Numbers would say that this fraction would approach the theoretical answer as n got larger.
- Theoretical
- Since there are only 2 equally likely outcomes, P(H)= 1/2

Two Coins

- Experimental Results
- P(0 heads) =
- P(1 head, 1 tail)=
- P(2 heads)=
- Note: These all sum to 1.
- Questions:
- Why is “1 head” more likely than “2 heads”?

2 Coins- Theoretical answer

P(0 heads) = 1/4

P(1 head, 1 tail)= 2/4 = 1/2

P(2 heads)= ¼

Note: sum of these outcomes is 1

Three Coins

- Are “1 head” , “2 heads”, and “3 heads” all equally likely?
- Which are most likely and why?

3 coins

- P(0 heads)=
- P(1 head)=
- P(2 heads)=
- P(3 heads)=

Theoretical Probabilities for 3 Coins

- P(0 heads)= 1/8
- P(1 head)= 3/8
- P(2 heads)= 3/8
- P(3 heads)= 1/8
- Notice: Sum is 1.

Cards

- 4 suits, 13 denominations; 4*13=52 cards
- picture = J, Q, K

When picking one card, find…

- P(heart)=
- P(king)=
- P(picture card)=
- P(king or queen)=
- P(king or heart)=

Theoretical Probabilities- Cards

- P(heart)= 13/52 = ¼ = 0.25
- P(king)= 4/52= 1/13
- P(picture card)= 12/52 = 3/13
- P(king or queen)= 4/52 + 4 /52 = 8/52
- P(king or heart)= 4/52 + 13/52 – 1/52 = 16/52

11.6 Not, Mutually Exclusive, Odds

P(E)= 1-P(E ‘ ) where E’ = not E=complement of E

1. If there is a 20% chance of snow tomorrow, what is the chance it will not snow tomorrow?

2. When choosing one card from a deck, find the probability of selecting:

a. A heart

b. A card that is not a heart

c. A king

d. A card that is not a king

P(A or B)

1. When selecting one card, find the probability of:

a. king or queen

b. king or a heart

c. king or a 5

d. 5 or a diamond

e. Picture card or a 7

f. Picture card or a red card

P(A or B)

- Mutually exclusive events—cannot occur together
- If A and B are mutually exclusive,

P(A or B) = P(A) + P(B)

- If A and B are not mutually exclusive,

P(A or B) = P(A) + P(B) – P(A and B)

Odds

Basic idea: If, when drawing one card from a deck, the probability of getting a heart is ¼, then

The odds in favor of drawing a heart are 1:3 and the odds against a heart are 3:1.

Another example: If, when drawing one card from a deck, the probability of getting a king is 1/13, then

The odds in favor of drawing a king are 1:12, and the odds against a king are 12:1.

Odds to Probability if odds in favor of E are a:b, then P(E)=

Given probabilities, find odds

1. Recall probabilities

a. P(heart)

b. P(not a heart)

c. P(king)

d. P(picture card)

e. P(red card)

1. Find the odds in favor of:

a. A heart

b. A card that is not a heart

c. A king

d. A picture card

e. A red card

…odds

2. If the odds in favor of winning the lottery are 1:1,000,000, find the probability of winning the lottery

3. If the odds in favor of getting a certain job are 3:4, find the probability of getting the job.

11.7: And, Conditional

Independent events- two events are independent events if the occurrence of either of them has no effect on the probability of the other

If A and B are independent events, then P(A and B) = P(A)*P(B)

2 kids

1. Assuming it’s equally likely that boys and girls are born, in a family with 2 kids, find the probability of getting:

a. 2 girls

b. 2 boys

2. In a family with 3 kids, find the probability of getting:

Assuming P(B)=P(G)

a. 3 girls

b. 3 boys

c. At least 1 boy

4. Two cards

4. If you pick two cards out of a deck of cards and replace them in between picks, find:

a. P( 2 red cards)

b. P(2 hearts)

c. P(2 kings)

Dependent events—

the occurrence of one of them has an effect on the occurrence of the other

If A and B are dependent, P(A and B) = P(A)*P(B, given A)

Without replacement:

1. If you pick two cards out a deck without replacement, find the probability of getting:

a. 2 red cards

b. 2 kings

2. pick 3 cards without replacement

find the probability of getting:

a. 3 red cards

b. 3 kings

c. A king, then a queen, then a jack (in that order)

Conditional Probability

Find:

P(driver died)=

P(driver died/given no seat belt)=

P(no seat belt)=

P(no seat belt/given driver died)=

P(driver died)= 2111/577,006 = .00366

- P(driver died/given no seat belt)= 1601/164,128 = .0097
- P(no seat belt)= 164,128/577,006= .028
- P(no seat belt/given driver died)= 1602/2111= .76

Birthday problem

What is the probability that two people in this class would have the same birth date?

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