19 red 11 Blue 20 Yellow given the number of red, blue, and yellow balls in...

Question

Question

19 red

11 Blue

20 Yellow

given the number of red, blue, and yellow balls in a box. In how many ways can one yellow and two red balls be taken from the box, with replacement?

please with the formula

Answer 1

Answer #1

Total balls = 19 + 11 + 20 = 50

P(yellow) = 20/50 = 0.4

P(red) = 19/50 = 0.38

one yellow and two red balls = 20C1 * 19C2

= (20! / (1! * 19!)) * (19! / (2! * 17!))

= 20 * 171

= 3420

Answer 2

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