Hints for Activity 10 and 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) = = 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 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) = …