Another question about combinations/permutations?

Ugh I still don't get this. Please answer and explain them/show your work!





3 Cards are seleccted at random from a group of 7. 2 of the cards have been marked with winning numbers.





a) what is the probability that exactly 1 of the 3 cards has a winning number?


b) what is the probability that at least 1 of the 3 cards has a winning number?


c) what is the probability that none of the 3 cards has a winning number?


d) what relationship is there between the answers to parts b and c?

Another question about combinations/permutations?
Let's do the easiest problem first:


C) What is the probability that none of the 3 cards has a winning number?


P(0 Wins) = (5/7) * (4/6) * (3/5) = 2/7 = .285714 or 28.6%





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B) What is the probability that at least 1 of the 3 cards has a winning number?


If the weatherman told you the probability of rain


was 40%, what is the possibility of NO RAIN?


Easy - 60%.


Same logic applies here:


P(Wins≥1) = 1 - P(0 Wins) = 1 - .285714 = .714286 or 71.4%





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D) What relationship is there between the answers to parts b and c?


These two probabilities are 'complements',


as they add up to 'one'.


(look up the complement rule in your book).





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Now the tough problem:


A) What is the probability that exactly 1 of the 3 cards has a winning number?





For this problem, use the Hypergeometric Probability Formula:





P(x) = [(sCx)*(N-sCn-x)] / NCn





where:


N = is the size of the population (7)


s = is the number of successes in the population (2)


x = is the number of successes in the sample (1)


n = is the size of the sample or the number of trials (3)


C = is the symbol for a Combination





P(x) = [(sCx) * (N-sCn-x)] / NCn


P(1) = [(2C1) * (7-2C3-1)] / 7C3


P(1) = [(2C1) * (5C2)] / 7C3


P(1) = [(2) * (10)] / 35


P(1) = [20] / 35 = 4/7 = .571429 or 57.14%





Yeah, the math gets really messy,


but you can use an online Hypergeometric Probability Calculator:


http://stattrek.com/Tables/Hypergeometri...





Or you can do it on your TI-83/84 Calculator:


http://mathbits.com/MathBits/TISection/S...





Good luck in your studies,


~ Mitch ~





P.S. - My apologies if my explanation is confusing...

flower girl