An urn contains 6 red, 6 white, 6 blue, and 6 green balls. a) Balls are drawn (w

Question

An urn contains 6 red, 6 white, 6 blue, and 6 green balls.

a) Balls are drawn (without replacement) until the first green ball is drawn. What is the probability that a total of 4 balls are drawn?

b) Ten balls are drawn (without replacement). What is the probability that exactly 1 red, 2 white, 3 blue and 4 green balls are drawn?

c) Ten balls are drawn with replacement. What is the probability that exactly 1 red, 2 white, 3 blue and 4 green balls are drawn?

d) Five balls are drawn (without replacement). What is the probability that at least one of the four colors is not present among those that are drawn?

Explanation / Answer

We have given 6 red, 6 white, 6 blue and 6 green balls .

b) we have to select 10 balls without replacement ,

And we have to find probability for exactly 1 red, 2 white , 3 blue and 4 green balls .

For without replacment we have to use pmutation formula .

P( 1 red , 2 white , 3 blue and 4 green balls ) =   ( 6P1 * 6P2 * 6P3*6P4) / (24P10) = 0.0000

c) We have to raw 10 balls with replacement , and we have to find probability for exactly 1 red , 2 white , 3 blue and 4 green balls .

Here we have to use formula of combination , (6^1 * 6^2 * 6^3 * 6^4)/(6^10) = 0.0000

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