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You have a bag of marbles that has 12 red, 5 green, 6 yellow, and 4 blue. If you draw one at random. what is P(red)? (What would the percent of choosing red be)

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konrad509

[tex]|\Omega|=12+5+6+4=27\\|A|=12\\\\P(A)=\dfrac{12}{27}=\dfrac{4}{9}=44.4\%[/tex]

Answer:  44.44% approximately

Work Shown:

P(red) = (number of red)/(number total)

P(red) = (12 red)/(12 red + 5 green + 6 yellow + 4 blue)

P(red) = (12 red)/(27 total)

P(red) = 12/27

P(red) = 4/9

P(red) = 0.4444 approximately

P(red) = 44.44% approximately

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