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Hints for Activity 10 and Individual 5PowerPoint Presentation

Hints for Activity 10 and Individual 5

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### Hints for Activity 10and Individual 5

LSP 121 hints!

Q #2 – at least once

- Formula
1 – (Not A in one)n

A = exposed to virus, the event

P(A) = 1.5% = .015

P(not A) = 1 – P(A) = 1 - .015 =

n = 10 for part a) and 20 for part b)

Q #3

- Expected value
- start with what you pay as (-1) * cost in $
- then add the probability or probabilities of gaining back each amount of $

Q #4

- Expected value
- In this example, start with the cost as 100% x $1, express as negative: -$1
- To this add the other winnings x probabilities
- -$1 x 1 + $20 x .012 + etc.
- Expected value = …

Q #5

- Possible outcomes
- Set this up with 10 ‘boxes’:

Area Codes Exchanges Extensions

In each box, write the number of possible #’s that can be found there.

Multiply these numbers to find possible outcomes for

particular event (set of phone numbers or area codes).

individual assignment #5

- #1: possible outcomes (ski shop packages) is simply the product of the different sets
- #2: (# of ways to get a 5)/(total outcomes)
- #5: Example of calculating ‘odds’, usually expressed as a ratio, such as 2:1 or 1:2, read ‘2 to 1’ or ‘1 to 2’. Note: if something has a probability of 50% (for 2 outcomes), then this event has 1:1 odds

indiv #5, more

- #6: this is an example of multiplication rule #1; the probability of multiple independent occurences is the product of the individual probabilities
- #7: (total ways of getting 3, 4 or 5) / (total possible outcomes for 2 dice); be careful when calculating ODDs for this same situation.

indiv #5, more

- #8:
- a) empirical probability; do the math
- b) apply the math and predict for next year
- c) use ‘at least once’ formula: 1 – (not A)n

- #9: Use addition of probabilities to calculate ‘expect value’. Note: since this is ‘for the company’, you start with $1000 x 100% + (-$20,000 … ) + (-$50,000 …) + …, since each claim is a ‘loss’ to the company.

indiv #5 last part

- #10: This problem is slightly different in that two events are ‘non-mutually exclusive’; the occurrence of one event does not exclude nor is it excluded by another event. Simply work out the formula given:
- P(A or B) = P(A)+P(B) – P(A and B),
- = P(king) + P(heart) – P(king and heart)
- = (# of kings/52) + (# of hearts/52) – (# of king of hearts/52) = …

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