A bag contains 5 red balls and 5 white balls. 4 balls are randomly selected from

ID: 3300855 • Letter: A

Question

A bag contains 5 red balls and 5 white balls.

4 balls are randomly selected from the bag with replacement and the color of each

selection recorded. Give the probabilities below.

Pr(the 4 balls contain 2 red balls and 2 white balls regardless of sequence) =

Pr(the 4 balls contain all red balls)

Pr(sequence is r,r,w,w)

Pr(sequence is w,r,w,w)

4 balls are randomly selected from the bag without replacement and the color of each selection recorded. Give the probabilities below.

Pr(the 4 balls contain 2 red balls and 2 white balls regardless of sequence) =

Pr(the 4 balls contain all red balls)

Pr(sequence is r,r,w,w)

Pr(sequence is w,r,w,w)

A) P(2 red and two white) = 5C1/10C1 * 5C1 / 10C1 * 5C1/10C1 * 5C1 / 10C1 = 5/10 * 5/10 * 5/10 * 5/10 = 1/2 * 1/2 * 1/2 * 1/2 = 1/16 = 0.063

p(4 balls contcontainall red) = 5/10 * 5/10 * 5/10 * 5/10 = 1/16 = 0.063

P(r,r,w,w) = 5/10 * 5/10 * 5/10 * 5/10 = 1/16 = 0.063

P(w,r,w,w,) = 5/10 * 5/10 * 5/10 * 5/10 = 1/16 = 0.063

B) Total sample space of drawing 4balls = 10C4 = 10! / (4! * 6!) = 210

P(4 balls contain 2red and 2 white) = (5C2 * 5C2) / 10C4 = 10 * 10 / 210 = 100/210 = 0.48

P(all are red) = 5C4 / 10C4 = 5/210 = 0.024

P(r,rw,w,) = 5C1/10C1 * 4C1/9C1 * 5C1/8C1 * 4C1/7C1

= 5/10 * 4/9 * 5/8 * 4/7 =0.08

P(w,r,w,w) = 5C1/10C1 * 5C1/9C1 * 4C1/8C1 * 3C1/7C1 = 5/10 * 5/9 * 4/8 * 3/7 = 0.06

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