Question

Bag 1 contains six red balls, seven blue balls, and three

green balls. Bag 2 contains eight red balls, eight blue

balls, and two green balls. Bag 3 contains two red balls,

nine blue balls, and eight green balls. Bag 4 contains

four red balls, seven blue balls, and no green balls.

Bag 1 is chosen with a probability of 0.15, bag 2 with a

probability of 0.20, bag 3 with a probability of 0.35, and

bag 4 with a probability of 0.30, and then a ball is chosen

at random from the bag. Calculate the probabilities:

(a) A blue ball is chosen.

(b) Bag 4 was chosen if the ball is green.

(c) Bag 1 was chosen if the ball is blue.

Answer #1

a)P(blue ball)=P(bag 1 and blue ball)+P(bag 2 and blue ball)+P(bag 3 and blue ball)+P(bag 4 and blue ball)

=0.15*(7/16)+0.2*(8/18)+0.35*(9/19)+0.3*(7/11)=**0.5112**

b)

P(green ball)=P(bag 1 and green ball)+P(bag 2 and green ball)+P(bag 3 and green ball)+P(bag 4 and green ball)

=0.15*(3/16)+0.2*(2/18)+0.35*(8/19)+0.3*(0/11)=0.1977

hence P(Bag 4 was chosen if the ball is green)=P(bag 4 and green
ball)/P(green ball)=0.3*(0/11)/0.1977 =**0**

**c)**

P( Bag 1 was chosen if the ball is blue)=P(bag 4 and blue
ball)/P(blue ball)=0.15*(7/16)/0.5112=**0.1284**

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