Ben twice chooses a random integer between 1 and 50 inclusive, twice (and he may choose the same integer both times). What is the probability that at least one of the numbers Ben chooses is a multiple of \(8\)?
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Ben twice chooses a random integer between 1 and 50 inclusive, twice (and he may choose the same integer both times). What is the probability that at least one of the numbers Ben chooses is a multiple of 8?
There are 6 multiples of 8 between 1 and 50: namely 8, 16, 24, 32, 40, and 48
There are 50 numbers between 1 and 50.
The probability of choosing a multiple of eight with the first number is 6 / 50.
The probability of choosing a multiple of eight the second time is also 6 / 50.
So the total probability is 12 / 50, which, denoted as a percentage, is 24% likely.
This is only a probability; extrapolating it up to 100% does not create a certainty.
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