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4. An urn contains 5 blue balls, 3 green balls, and 7 white balls. (a) A ball is selected from the urn. What are the odds the
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Answer #1

(a). The total number of balls = 5 + 3 + 7 = 15

P(G) = 15 = 5 3

(b). Two balls can be blue and green in two ways because the first ball drawn is not replaced, therefore the order of the balls matter.

Case 1: First ball is blue, second is green

P(B then G) =-× 15 1414

Case 2: First ball is green, second is blue

P(G then B) 15 1414

Total probability is given by

Pl B and G) = 14 14 7

(c). P( At least one if blue or at least one is green) = 1 - P(Both are white)

P(W,W) = 15 14 5

P(required)-1--=

(d). Case 1: Both are blue

PlB.B) _ 15 × 11-21

Case 2 : Both are green

15 14 35

2 13 P(required) 35105 21 35105

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