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Suppose you have an urn (a bucket) with 10 green balls and 7 red balls. (a) suppose you draw 8 balls out of the bucket all at once. how many ways are there to draw 4 red balls and 4 green ones

Respuesta :

kritter1
Calculate the probability of drawing 4 red balls and then 4 green balls as shown; 

P(4R+4G)=([tex] \frac{7}{17} [/tex])([tex] \frac{6}{16} [/tex])([tex] \frac{5}{15} [/tex])([tex] \frac{4}{14} [/tex])([tex] \frac{10}{14} [/tex])([tex] \frac{9}{13} [/tex])([tex] \frac{8}{12} [/tex])([tex] \frac{7}{11} [/tex])

P(4R+4G)=0.00308 chance 

The denominators are decreasing as the balls get taken out. 

Hope I helped :) 

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