What is the probability that a person would match none of the numbers?

In a lottery game, participants select six numbers from 1 to 49. You win a prize if you can match 3, 4, 5, or 6 numbers.





What is the probability that a person would match none of the numbers?


Answers must be expressed as a decimal (correct to four decimal places like .xxxx) OR as a percent (correct to 2 decimal places like xx.xx%).What is the probability that a person would match none of the numbers?
(43/49)*(42/48)*(41/47)*(40/46)*(39/45)*鈥?= 0.4360





I'm assuming that by ';not match any of the numbers'; you mean that he matches zero numbers, not that he matches 0, 1, or 2 (that is, he loses).What is the probability that a person would match none of the numbers?
P(not matching any number) = 1 - P(matching a number)





= 1 - P(matching 6) - P(matching 5)- P(matching 4) - P(matching 3)





= 1 - 1/(49C6) - 1/(49C5) - 1/(49C4) - 1/(49C3)





= 0.999940407





=0.9999
As per condition of the question, when a person is not in winning position, he will have 0,1 or 2 matching number.


for zero
i'm just starting probability. yay!

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