There are three bags of marbles. Bag X contains 2 blue, 1 red, and 4 green marbl

ID: 1828762 • Letter: T

Question

There are three bags of marbles. Bag X contains 2 blue, 1 red, and 4 green marbles; bag Y

contains 3 blue, 2 red, and 4 green marbles; and bag Z contains 1 blue, 4 red, and 3 green

marbles. There is a 15 percent chance of drawing a marble from bag X, a 30 percent chance

of drawing a marble from bag Y, and a 55 percent chance of drawing a marble from bag Z.

a) red,

b) blue, and

c) green? (Note: These are three separate cases)

probability that the red marble came from:

d) bag X,

e) bag Y, and

f) bag Z? (Note: These are also three separate cases)

(Hint: Use the Total Probability concept)

bag X= { 2B, 1R, 4G} and have 30% chance

bag Y= {3B, 2R, 4G} and have 15% chance

bag Z= {1B,4R, 3G} and have 55% chance

a) red

Total number of marbles = 2B+1R+4G+3B+2R+4G+1B+4B+3G =24 marbles

Probability of drawing a red marble from any bag = (1/24)(0.3)+(2/24)(0.15)+(4/24)(0.55)

= 0.0125+0.0125+0.916

= 0.94

b) Blue

Probability of drawing a Blue marble from any bag = (2/24)(0.3)+(3/24)(0.15)+(1/24)(0.55)

= 0.025+0.018+0.229

= 0.27

c) Green

Probability of drawing a Green marble from any bag = (4/24)(0.3)+(4/24)(0.15)+(3/24)(0.55)

= 0.05+0.6+0.068

= 0.71

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